tag:blogger.com,1999:blog-37414104180967168272020-04-02T01:09:04.413-07:00Unizor - Creative Mind through Art of MathematicsUnizor is a site where students can learn the high school math (and in the future some other subjects) in a thorough and rigorous way. It allows parents to enroll their children in educational programs and to control the learning process.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.comBlogger403125tag:blogger.com,1999:blog-3741410418096716827.post-27235475065224806412020-03-29T08:10:00.001-07:002020-03-29T08:10:22.839-07:00Unizor - Physics4Teens - Electromagnetism - Votage Drop<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ueLvyFe-Y24" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Voltage Drop</u><br /><br />Imagine a waterfall going down the rocks. As it hits each rock, it loses some of its potential energy.<br />Analogously, consider several radiators in a series connected to a boiler that supplies hot water for them. As the water goes from one radiator to another, it cools down, losing its heat energy.<br />The same happens with electrons, as they go along the circuit and meet one resistance after another. Going through each of them, they lose their energy.<br /><br />Consider the following circuit<br /><img src="http://www.unizor.com/Pictures/ResistorSeries.jpg" style="height: 120px; width: 200px;" /><br />Assume, we know how to measure the difference in potential (<i>voltage</i>) between any two points on this circuit. Measuring the voltage between the terminals of the battery gives some value <i><b>U</b></i>.<br />Measuring the voltage between the positive terminal of the battery (the longer thin line with a plus sign) and the point in-between the resistors gives some other value <i><b>U<sub>1</sub></b></i>.<br />Measuring the voltage between the negative terminal of the battery (the shorter thick line with a minus sign) and the point in-between the resistors gives yet other value <i><b>U<sub>2</sub></b></i>.<br /><br />Notice that the amount of electricity going through this circuit per unit of time (<i>electric current</i> or <i>amperage</i>) is the same everywhere since it's a closed loop and equals <i><b>I</b></i>.<br /><br />Now let's apply the Ohm's Law to an entire circuit, keeping in mind that the voltage between the terminals of the battery is <i><b>U</b></i> and the total resistance of an entire circuit is <i><b>R=R<sub>1</sub>+R<sub>2</sub></b></i>:<br /><i><b>I = U <span style="font-size: medium;">/</span> R = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Applying the Ohm's Law to a segment of a circuit that includes only resistor <i><b>R<sub>1</sub></b></i> and knowing the voltage on its two ends <i><b>U<sub>1</sub></b></i>, we obtain<br /><i><b>I = U<sub>1</sub> <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br /><br />Equating two different expressions for the electric current <i><b>I</b></i>, we can find the voltage <i><b>U<sub>1</sub></b></i>:<br /><i><b>I = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U<sub>1</sub> <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br />from which the value of <i><b>U<sub>1</sub></b></i> is<br /><i><b>U<sub>1</sub> = U·R<sub>1</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Let's do the same calculation for the second resistor.<br /><i><b>I = U <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U<sub>2</sub> <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br />from which the value of <i><b>U<sub>2</sub></b></i> is<br /><i><b>U<sub>2</sub> = U·R<sub>2</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>)</b></i><br /><br />Notice that <nobr><i><b>U<sub>1</sub> + U<sub>2</sub> = U</b></i>.</nobr><br />Indeed,<br /><i><b>U·R<sub>1</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) +<br />+ U·R<sub>2</sub> <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) =<br />= U·(R<sub>1</sub>+R<sub>2</sub>) <span style="font-size: medium;">/</span> (R<sub>1</sub>+R<sub>2</sub>) = U</b></i><br /><br />Completely analogous calculations can be provided with three or more resistors connected in a <i>series</i>.<br />So, the difference in electric potential or <b>voltage drop</b> between the terminals of a battery <i><b>U</b></i> is split between the voltage drops on each resistor in a <i>series</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-10918279177176819342020-03-28T12:52:00.001-07:002020-03-28T12:52:18.732-07:00Unizor - Physics4Teens - Electromagnetism - Ohm's Law<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/B9xIOeHul_0" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Direct Current - Ohm's Law</u><br /><br />Simply speaking, experiment shows that the current in a conductor (<i>amperage</i>) is proportional to a difference in potentials (<i>voltage</i>) on its ends. This is the <b>Ohm's Law</b>.<br /><br />Obviously, some physical explanation of this law is appropriate, and this is what this lecture is about.<br /><br />Electric current is the flow of electrons inside a conductor. Just as a reminder, the traditionally defined direction of the current is opposite to the direction of the flow of electrons.<br /><br />Even without any electric field electrons inside a conductor are not exactly spin each around its own nucleus, staying on the orbit forever. Some of them, especially those on the outer orbit, can tear off their orbit, jump to another nucleus' orbit, then going somewhere else etc., thus creating certain chaotic environment.<br />When an electric field is present, the chaotic movement of electrons becomes directional to a degree, depending on the strength of the field, thus creating a flow of electrons - an <i>electric current</i>.<br /><br />This flow of electrons occurs when there is an excess of electrons on one end and deficiency (or less of an excess) of electrons on the other end, which creates an electric field inside a conductor that forces some light electrons to leave their atoms and move, while heavy nuclei with remaining electrons stay put.<br />Electrons are repelled from the end, where there is excess of them and attracted (or repelled less) from the other end.<br /><br />On their way from one end of a conductor to another these electrons must go through a maze of atoms, many of which have lost some electrons because of the electric field and, therefore, are positively charged, attracting negative electrons in an attempt to compensate lost electrons. Some succeed by capturing free electrons, some not, some lose more electrons in their chaotic but directional movement.<br /><br />The flow of electrons inside a conductor can be compared with a waterfall from the high level, where potential energy relative to gravitational field of the Earth is greater to a lower level with lesser potential energy.<br /><br />As water falling down the waterfall, hitting all the stones on its way, electrons hit atoms and lose some of their energy, some of them get absorbed by atoms and don't reach the other end of a conductor, thus diminishing the flow.<br /><br />The very important aspect in this movement of electrons is the properties of the material a conductor is made of.<br /><br />Different atoms of different materials have different characteristics of capturing and releasing electrons, when subjected to the forces of an electric field. However, certain common laws of conductivity can be logically explained and used in practical applications.<br /><br /><i>Ohm's Law</i><br /><br />Stronger electric field produced by a greater difference in electric potential (<i>voltage <b>U</b></i>) between the ends of a conductor should cause more intense movement of electrons, greater number of electrons going from one end of a conductor and reaching the other end per unit of time (<i>amperage <b>I</b></i>). This qualitative property is perfectly understood. Mathematically speaking, <i><b>I</b></i> is a <i>monotonically increasing</i> function of <i><b>U</b></i>.<br /><br />It was an experimental fact that this monotonic function to a very high precision is <i>linear</i>:<br /><i><b>I = σ·U</b></i><br />where <i><b>σ</b></i> is called a <b>conductivity</b> of a conductor.<br />In more common cases this law is written not in terms of <b>conductivity</b>, but in terms of <b>resistivity</b>:<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br />where <i><b>R</b></i> is called a <b>resistance</b> of a conductor.<br /><br />In most cases we will use the Ohm's Law in that last form, using a <b>resistance</b> as the characteristic of a conductor.<br /><br /><i>Resistance</i><br /><br />In practical life we use some conductors with a very low resistance, like copper wire, to connect some electrical appliance, like a regular incandescent lamp bulb, to a source of electric power. In this arrangement we usually consider wiring as having no resistance and concentrate on the properties of an appliance as the one with the electrical resistance <i><b>R</b></i>.<br /><br />Schematically, resistor is pictured as a rectangle or a zigzag line connected by straight lines of wires to a source of electric power<br /><img src="http://www.unizor.com/Pictures/Resistor.jpg" style="height: 120px; width: 200px;" /><br />Arrows on this picture show the <u>traditionally</u> defined direction of electricity from positive to negative terminal of the source of electric power - opposite to a direction of electrons' movement.<br /><br />Consider an extreme case when the resistor is non-conductive, like there is only vacuum in between its two ends. Symbolically, it's equivalent to <i><b>R=∞</b></i>. The Ohm's Law in this case results in <i><b>I=U/R=0</b></i>, which means that the circuit is broken and there is no current in it.<br /><br />In an opposite extreme case with <i><b>R=0</b></i> we have <i><b>I=U/R=∞</b></i>, which is the so called "short".<br />In practical life it happens when you detach two wires coming to a lamp and connect them. Without the bulb, which has some significant resistance, the electric current in a circuit would almost instantaneously grow very high and you might get electrocuted.<br /><br />Let's use an incandescent lamp as an example of a resistor. More precisely, the resistor is a tangent spiral (filament) inside the lamp.<br /><br />Consider what happens with the resistance of this tangent spiral if we will make it longer.<br />Obviously, the electrons will have to go through more obstacles on their way from one end of a spiral to another, which will slow their movement more than when a spiral was shorter. It's logical to expect that doubling the length of a resistor will double its resistance, and it is confirmed by experiments.<br />In general, for linearly-shaped resistors their resistance should be proportional to the length.<br /><br />The immediate consequence of this consideration is that <b>two resistors, sequentially connected one after another in a <i>series</i>, will have a combined resistance equal to a sum of resistance of each one</b>.<br /><img src="http://www.unizor.com/Pictures/ResistorSeries.jpg" style="height: 120px; width: 200px;" /><br /><br />Now, instead of making a spiral longer, let's make it thicker. Electrons will have more space to move, more freedom of direction and more electrons can travel across the spiral per unit of time. Doubling the thickness of a spiral is similar to doubling the width of a highway with more cars moving on it per unit of time. So, the resistance of a thicker spiral will decrease by the factor of increased cross-section area of a tangent spiral.<br />In general, for linearly-shaped resistors their resistance should be inversely proportional to the cross-section area.<br /><br />Increasing the thickness of a spiral is logically equivalent to using two spirals connected parallel to each other, as on this circuit diagram. <img src="http://www.unizor.com/Pictures/ResistorParallel.jpg" style="height: 120px; width: 200px;" /><br />The number of electrons per unit of time (electric current or <i>amperage</i>) coming from the common wire is split into two parallel flows, and all the electrons passing through a common part of a circuit per unit of time should be equal to a sum of numbers of electrons passing through each of the parallel segments.<br />The immediate consequence of this consideration is that <b>two resistors, connected parallel to each other, will result in the electric current in a common wire to be a sum of electric currents in each one of them</b>.<br /><i><b>I = I<sub>1</sub> + I<sub>2</sub></b></i><br /><br />Now let's apply the Ohm's Law to an entire circuit, and each of two parallel branches, keeping in mind that the voltage between the terminals of the battery is <i><b>U</b></i>, the same as the voltage between the ends of each resistor, and the resistance of each branch is known as <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i>.<br /><br />Assuming the total resistance of two parallel branches is <i><b>R</b></i>, the Ohm's Law gives<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br /><br />Applying the same principle to one branch with resistor <i><b>R<sub>1</sub></b></i>, we have<br /><i><b>I<sub>1</sub> = U <span style="font-size: medium;">/</span> R<sub>1</sub></b></i><br /><br />The same for another branch:<br /><i><b>I<sub>2</sub> = U <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br /><br />Now the original equation about a current in the common segment of a wire equaled to a sum of currents in two branches looks like<br /><i><b>U <span style="font-size: medium;">/</span> R = U <span style="font-size: medium;">/</span> R<sub>1</sub> + U <span style="font-size: medium;">/</span> R<sub>2</sub></b></i><br />or<br /><i><b>1<span style="font-size: medium;">/</span>R = 1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub></b></i><br /><br />From the last equation follows the expression for a combined resistance of two resistors installed parallel to each other<br /><i><b>R = 1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i><br />This formula looks more cumbersome and the one for <i><b>1<span style="font-size: medium;">/</span>R</b></i> above is more commonly occurs.<br /><br />Instead of resistance, it can be expressed in terms of <i>conductivity</i> of an entire assembly <i>σ=1/R</i> of two parallel resistors with conductivity <i>σ<sub>1</sub>=1/R<sub>1</sub></i> and <i>σ<sub>2</sub>=1/R<sub>2</sub></i>:<br /><i><b>σ = σ<sub>1</sub> + σ<sub>2</sub></b></i><br /><br />For two identical resistors of resistance <i><b>r</b></i> each the combined resistance <i><b>R</b></i> would be<br /><i><b>1/R = 1/r + 1/r = 2/r</b></i><br /><i><b>R = r/2</b></i><br /><br />More generally, we can derive the resistance of <i><b>n</b></i> identical resistors of resistance <i><b>r</b></i> each connected parallel to each other. In this case the electric current <i><b>I</b></i> in the common wire in split into <i><b>n</b></i> identical flows, each having an amperage of <i><b>I/n</b></i>. From the Ohm's Law the voltage at the ends of each resistor should be <i><b>U=I·r/n</b></i>, from which follows that <i><b>R=r/n</b></i> is the resistance of an entire assembly of <i><b>n</b></i> identical resistors with resistance <i><b>r</b></i> each.<br /><br /><i>Example</i><br /><br />Consider a circuit with resistors <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i> connected <i>parallel</i> to each other and resistor <i><b>R<sub>3</sub></b></i> installed after both of them in a <i>series</i>.<br />What would be the resistance <i><b>R</b></i> of a combined assembly of these three resistors?<br /><br />First, determine the resistance of two <i>parallel</i> resistors as a unit.<br /><i><b>R<sub>12</sub> = 1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i><br />This this unit of two parallel resistors with resistance <i><b>R<sub>12</sub></b></i> is in a <i>series</i> with resistor <i><b>R<sub>3</sub></b></i>, their combined resistance is the sum of their individual resistances<br /><i><b>R = R<sub>12</sub> + R<sub>3</sub> = </b></i>[<i><b>1 <span style="font-size: medium;">/</span> (1<span style="font-size: medium;">/</span>R<sub>1</sub> + 1<span style="font-size: medium;">/</span>R<sub>2</sub>)</b></i>]<i><b> + R<sub>3</sub></b></i><br /><br /><br /><i>Resistance Units<br />of Measurement</i><br /><br />The Ohm's Law allows to easily establish the units of measurement for the resistance of any component in an electric circuit - <b>ohms Ω</b>.<br />From the original form of the Ohm's Law<br /><i><b>I = U <span style="font-size: medium;">/</span> R</b></i><br />follows that<br /><i><b>R = U <span style="font-size: medium;">/</span> I</b></i><br /><br />This allows to establish a unit of measurement for resistance of any component of an electric circuit.<br />If the difference in potential (<i>voltage</i>) between one end of such a component and another is <i><b>1V </b>(volt)</i> and the electric current going through it is <i><b>1A </b>(ampere)</i>, this component by definition has a resistance of <i><b>1Ω </b>(ohm)</i>.<br /><br />Consequently, the resistance of a component in an electric circuit with the current going through it <i><b>I</b> (ampere)</i> with voltage on its end <i><b>U </b>(volt)</i> equals to <i><b>U <span style="font-size: medium;">/</span> I</b> (ohm)</i>.<br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-62565975928265431572020-03-11T09:41:00.001-07:002020-03-11T09:41:49.722-07:00Unizor - Physics4Teens - Electromagnetism - Electric Current - Speed of ...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/36iXMPKENRs" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Speed of Electrons</u><br /><br /><i>Electric current</i> is a movement of electrons. We know from experience that, when we turn on the switch, the lights in the room are lit practically immediately. Does it mean that electrons from one terminal of a switch go to the light fixture and back to another terminal of a switch that fast?<br />No.<br /><br />Let's calculate the real speed of electrons, first, theoretically and then in some practical case.<br /><br />Assume, the <i>amperage</i> of the electric current going through a wire, that is the number of <i>coulombs</i> of electric charge going through a wire per second, is <i><b>I</b></i>, and the wire has cross-section area <i><b>A</b></i>.<br />Assume further that we know all the physical characteristics of a material our wire is made of, which will be introduced as needed.<br /><br />Based on this information, our plan is to determine the number of electrons going through the wire per unit of time and, knowing the density of electrons per linear unit of length in the wire, determine the linear speed of these electrons.<br /><br />Obviously, to determine the linear density of electrons, we will need physical characteristics of a wire.<br /><br />The number of electrons going through a wire per unit of time is easily determined from the <i>amperage</i> <i><b>I</b></i>. Since <i><b>I</b></i> represents the number of <i>coulombs</i> of electric charge going through a wire per second, we just have to divide this by the charge of a single electron in <i>coulombs</i> <i><b>e=−1.60217646·10<sup>−19</sup></b>C</i>.<br />So, the number of electrons going through a wire per second is<br /><i><b>N<sub>e</sub> = I <span style="font-size: medium;">/</span> e</b></i>.<br /><br />Now we will determine the linear density of electrons in the wire.<br /><br />First of all, we have to know how many active electrons in an atom of material our wire is made of participate in the transfer of electric charge, because not all electrons of each atom are freely moving in the electric field, but only those on the outer orbit. Let's assume, this number is <i><b>n<sub>e</sub></b></i>.<br />Using this number, we convert the number of electrons participating in the transfer of electric charge <i><b>N<sub>e</sub></b></i> into the number of atoms <i><b>N<sub>atoms</sub></b></i> in that part of a wire occupied by all electrons transferring the given charge per second <i><b>I</b></i>.<br /><i><b>N<sub>atoms</sub> = N<sub>e</sub> <span style="font-size: medium;">/</span> n<sub>e</sub> = I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e)</b></i>.<br /><br />Next, from the number of atoms we will find their mass and, using the density of wire material, the volume.<br />Dividing the volume by a cross-section of a wire, we will get the length of a segment of wire occupied by those electrons transferring charge per second, which is the <b>speed of electrons</b> or <b>drift</b>.<br /><br />Knowing the number of atoms, to get to their mass, we will use the <i>Avogadro number</i><br /><i><b>N<sub>A</sub>=6.02214076·10<sup>23</sup></b></i><br />that represents the number of particles in one <i>mole</i> of a substance. One <i>mole</i> of material our wire is made of is the number of grams equal to its <i>atomic mass</i> <i><b>m<sub>a</sub></b></i>, known for any material used for a wire.<br />If <i><b>N<sub>A</sub></b></i> atoms have mass of <i><b>m<sub>a</sub></b></i> gram, <i><b>N<sub>atoms</sub></b></i> have total mass<br /><i><b>M<sub>atoms</sub> = m<sub>a</sub> · N<sub>atoms</sub> <span style="font-size: medium;">/</span> N<sub>A</sub> =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>)</b></i>.<br /><br />Knowing the total mass <i><b>M<sub>atoms</sub></b></i> of all atoms that contain all electrons traveling through a wire per second, we calculate the volume <i><b>V<sub>atoms</sub></b></i> by using the <i>density</i> <i><b>ρ</b></i> of a material the wire is made of.<br /><i><b>V<sub>atoms</sub> = M<sub>atoms</sub> <span style="font-size: medium;">/</span> ρ =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ)</b></i><br /><br />Dividing the volume <i><b>V<sub>atoms</sub></b></i> by the cross-section area of a wire, we will get the length of the wire <i><b>L</b></i> occupied by electrons traveling through it in one second<br /><i><b>L = V<sub>atoms</sub> <span style="font-size: medium;">/</span> A =<br />= m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ·A)</b></i><br />where<br /><i><b>m<sub>a</sub></b></i> - <i>atomic mass</i> of wire's material (assuming it's one atom molecules, like copper)<br /><i><b>I</b></i> - <i>electric current - amperage</i> in the wire<br /><i><b>n<sub>e</sub></b></i> - <i>number of active electrons in each atom</i> of wire's material that participate in the transfer of electric charge<br /><i><b>e</b></i> - <i>electric charge of one electron</i><br /><i><b>N<sub>A</sub></b></i> - <i>number of atoms in 1 mol of conducting material - Avogadro Number</i><br /><i><b>ρ</b></i> - <i>density</i> of wire's material<br /><i><b>A</b></i> - <i>cross-section area</i> of a wire<br /><br />The above formula represents the length of a wire occupied by all active electrons traveling through it during one second, <b>which is the speed of movement of electrons making up an electric current</b>, called <b>drift</b>.<br /><br />Let get to practical examples.<br /><br />Assume, the <i>voltage</i> or the difference of <i>electric potential <b>E</b></i> between two ends of a copper wire is maintained at <b>110V</b> (standard voltage for apartments in the USA).<br />This wire connects a lamp that consumes <b>120W</b> of electric power <i><b>P</b></i> (or <i>wattage</i>).<br />Let the cross-section area of a wire be <b>3 mm²</b>.<br /><br />First of all, let's calculate the amount of electricity moving through the wire per unit of time - <i>amperage <b>I</b></i>.<br />As we know, the <i>amperage <b>I</b></i>, multiplied by <i>voltage <b>E</b></i>, is the electric power <i><b>P</b></i> (<i>wattage</i>).<br />Therefore,<br /><i><b>E = 110</b>V</i><br /><i><b>P = 120</b>W</i><br /><i><b>I = P/E = 60W/110V ≅ 1.09</b>A</i><br /><br />The <i>atomic number</i> of copper is <i><b>Z=29</b></i>. It means, the atom of copper has 29 protons in the nucleus and 29 electrons orbiting a nucleus. These 29 electrons are in four orbits: 2+8+18+1.<br />The outer orbit has only one electron that participates in the movement of electric charge, so the number of active electrons in an atom of copper is<br /><i><b>n<sub>e</sub>=1</b></i>.<br /><br />The nucleus of an atom of copper has 29 protons and 34 or 36 neutrons, its <i>atomic mass</i> is<br /><i><b>m<sub>a</sub> = 63.546</b>g/mol</i><br /><br />The electric charge of one electron is<br /><i><b>e = −1.60217646·10<sup>−19</sup></b>C</i>.<br /><br />The Avogadro number is<br /><i><b>N<sub>A</sub>=6.02214076·10<sup>23</sup></b></i><br /><br />Density of copper is<br /><i><b>ρ = 8.96</b> g/cm³ <b>= 0.00896</b> g/mm³</i><br /><br />Cross-section area of a wire is<br /><i><b>A = 3</b> mm²</i><br /><br />Using the formula above with values listed, we obtain<br /><i><b>L = m<sub>a</sub>·I <span style="font-size: medium;">/</span> (n<sub>e</sub>·e·N<sub>A</sub>·ρ·A)</b></i><br />we get<br /><i><b>L ≅ 0.0267</b> mm/sec</i><br />This is the speed of electrons traveling along a copper wire in this case.<br />Pretty slow!Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-67528497808565416752020-02-24T20:42:00.001-08:002020-02-24T20:42:02.809-08:00Unizor - Physics4Teens - Electromagnetism - Electric Current<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/IygOGBg68mg" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Current</u><br /><br /><i>Conductors and Dielectrics</i><br /><br /><i>Electric current</i> is a flow of electric charge. Since the actual carrier of electric charge is excess or deficiency of electrons, we need certain material where electrons can travel. So, vacuum cannot be a <i>conductor</i> of electricity because there is no electrons in it, but many metals, like copper, can. But we know that the electrons are orbiting the nuclei of the atoms. So, why do they travel?<br /><br />The answer is: <b>the force of an external electric field</b> pushes or pulls electrons off their orbits and, as a result, they move inside the material where electric field is present towards or away from the source of the electric field, depending on whether the source is positively (deficiency of electrons) or negatively (excess of electrons) charged.<br /><br />Consider a copper wire. It contains atoms of copper with 29 electrons in each atom, orbiting on different orbits around corresponding nuclei with 29 protons and from 34 to 36 neutrons in each.<br />Electrons stay on their orbit until some outside electric field comes into play. When it does, if its intensity is sufficient to push or pull light electrons off their orbits, while heavy nuclei stay in place, these electrons move in one or another direction as a result of different forces acting on them, the major of which is the intensity of the outside electric field. General direction of electrons is defined by the vector of intensity of the electric field. That makes copper a good <i>conductor</i> of electricity.<br /><br />On the other hand, there are materials, like glass, where electrons are connected stronger to their nuclei, which makes more difficult to push them off their orbits. these materials do not conduct electricity, they are called <i>insulators</i> or <i>dielectrics</i>.<br /><br />Ideal <i>conductor</i>, connected to an electrically charged object, makes an extension of this object. Since electrons are freely moving between the original object and an attached conductor, both constitute a new object with an electric charge evenly distributed between its parts.<br />Ideal <i>dielectric</i>, attached to an electrically charged object, does not share its electrons with this object, so the object remains the only one charged.<br /><br />In practical cases there are no ideal conductors (except under certain conditions of superconductivity under temperatures close to absolute zero) and no ideal dielectrics (except absolute vacuum that has no electrons at all).<br /><br />Metals are usually good conductors because their nuclei are relatively not easily moved from their places, while electrons are easily pushed off their orbits.<br />We use this property of <i>conductors</i> to direct the electrons to perform some work, like lighting the bulbs or moving electrical cars.<br /><br />IMPORTANT NOTICE:<br /><i>Conductivity</i> is related to movement of electrons and is a measure of how easily electrons are pushed form their orbits by outside electric field.<br />This should not be mixed with <i>permittivity</i> defined for electric fields and is a measure of propagation of electric field inside some substance.<br /><br /><i>Electric Current</i><br /><br />If the source of the field is a positive charge located near one end of a copper wire, electrons inside the wire would go towards that end. If the negative charge is the source of the field, electrons will move towards the opposite end.<br /><br />If there is nothing on the opposite end of a copper wire, electrons, after being pushed towards one of the edges, will stop. If, however, there is an opposite charge on the other end of a wire, electrons will move from the negatively charged end to the positively charged one until both charges neutralize each other and whatever end was missing electrons (positively charged) will be compensated by electrons that are in excess on the negatively charged end.<br /><br />Imagine now that we manage to keep one end of the wire constantly charged positively, while another end constantly charged negatively. Then electrons from the negatively charged end will flow to the positively charged end as long as we can keep these constant opposite charges on both ends. We will have a constant flow of electricity, which is called <i>electric current</i> (or simply <i>current</i> in the context of electricity).<br /><br />This process of maintaining constant flow of electricity is analogous to maintaining constant flow of water down the water slide using a pump that constantly pumps the water from a pool to the top of a slide, from which it flows down because of the difference in heights and gravity.<br /><br />While the presence of the electric field is felt almost instantaneously (actually, with a speed close to a speed of light), the electrons that carry electrical charge are not moving from a negatively charged end of a copper wire to the positively charged end with this speed.<br /><br />A good analogy is the pipe filled with water and a pump connected to one of its ends. As soon as the pump starts working, the water it pumps starts its trip along the pipe and pushes the neighboring molecule of water. Those, in turn, push the next ones etc. So, the water will come from another end of a pipe almost instantaneously (actually, after a time interval needed for the sound waves in the water to cover the length of a pipe), but it's the "old" water already present in the pipe before the pump started working. "New" water that is physically pushed into a pipe by a pump will eventually reach the other end, but not that fast.<br /><br />Finally, let's talk about measurement of the <i>electric current</i>.<br />The natural way of measurement of the flow of water in the pipe, as exemplified above, would be amount of water flowing out of a pipe per unit of time.<br />In our case of electric charge we can do the same - measure the flow by amount of electricity (in <i>coulombs</i>) traveling from one source of electric field to another (with opposite charge) per unit of time.<br /><br />The unit of measurement of the <i>electric current</i> is <i>ampere</i>, where <i>1 ampere</i> is the flow of electricity, when <i>1 coulomb</i> of electricity is moving across the wire within <i>1 second</i>.<br /><b><i>1 A = 1 C / 1 sec</i></b>.<br /><br />Recall the definition of a unit <i>volt</i> as a difference in electric potential between points <i>A</i> and <i>B</i> such that moving one <i>coulomb</i> of electric charge between these points requires one <i>joule</i> of work. Therefore,<br /><i><b>1 J = 1 V · 1 C</b></i><br />From the definition of <i>ampere</i> above<br /><b><i>1 C = 1 A · 1 sec</i></b>.<br />Therefore,<br /><i><b>1 J = 1 V · 1 A · 1 sec</b></i><br /><i><b>1 V · 1 A = 1 J / 1 sec</b></i><br />As we know,<br /><i><b>1 J / 1 sec = 1 W (watt)</b></i><br />So, electric current of <i>1 ampere</i> between points with difference of potential <i>1 volt</i> performs work of <i>1 watt</i>, that is <i>1 joule per second</i>.<br /><br />There is a direct analogy between electricity and mechanics with <i>force</i> analogous to <i>voltage</i> and <i>speed</i> analogous to <i>amperage</i><br /><i><b>Force · Distance = Work</b></i><br /><i><b>Force · Distance / Time =<br />= Work / Time = Power</b></i><br /><i><b>Force · (Distance / Time) =<br />= Force · Speed = Power</b></i><br /><i><b>Voltage · Amperage = Power</b></i><br /><br />Let's consider a slightly more complicated example of the <i>electric current</i>.<br />Assume that at one end of a copper wire we have a source of electric field with negative charge and at another end of this wire we have another source of electric field also with negative charge. Both ends will repel electrons inside a wire. However, if the charges are not equal, the larger one will push stronger, and electrons will move away from it towards the other end of a wire.<br /><br />The situation with two unequal negative charges is analogous to a water pipe with two pumps of different power pumping water into it from both ends. The stronger pump will overcome the weaker and the water will move from a stronger pump to the weaker.<br /><br />So, the most important factor in determining the direction of electrons in the wire is the <i>intensity of electric field</i> produced by electric charges. For multiple sources of electric field their vectors of intensity are added. From a general viewpoint, if there is a difference in intensity of electric fields, electrons will travel in the direction defined by a stronger force. In practical situation, when two sources of electricity are applied to two ends of a wire, one positive and one negative, one end attracts electrons and another pushes them away, the flow of electrons will be always from negative to positive charge.<br /><br />Assume, the intensity of electric field at the end <i>A</i> of a wire is <i><b>E<sub>A</sub></b></i> and intensity at the other end <i>B</i> is <i><b>E<sub>B</sub></b></i>. If both charges at points <i>A</i> and <i>B</i> are positive or both negative, the vectors <i><b>E<sub>A</sub></b></i> and <i><b>E<sub>B</sub></b></i> inside a wire are oppositely directed. If the charges are of different sings (which is a typical situation in practical applications of electricity), these vectors are directed the same way.<br /><br />The force acting on each coulomb of electricity inside a wire is a vector sum of both intensities:<br /><i><b>E = E<sub>A</sub> + E<sub>B</sub></b></i><br />The work needed to move one coulomb of electricity is, therefore,<br /><i><b>W = E·L</b></i>,<br />where <i><b>L</b></i> is the length of a wire.<br />This value <i><b>W</b></i> represents the difference of <i>electric potentials</i> of the electric field between points A and B, that is the <i>voltage</i> <i><b>V<sub>AB</sub></b></i> between them.<br />The difference in <i>intensity of an electric field</i> corresponds to the non-zero <i>voltage</i> between these points.<br /><br />If we can maintain the difference in electric field's potential between the two ends of a wire (non-zero <i>voltage</i> between them), the intensity of an electric field will push electrons from one end of a wire to another. This is how <b>direct electric current</b> is maintained.<br /><br />As electrons move from one end to another, they leave "holes" - spots where they used to be, which are "moving" in the opposite direction. Since we conditionally associate "negative" charge with electrons and "positive" charge with the absence of electrons ("holes"), we can say that the direction of positive charges is opposite to that of negative.<br /><br />For historical reasons, because electrons were not discovered yet, the direction of positive charges (that is, "holes" that are left, when electrons leave their places) was defined as a direction of the electric current.<br /><br />The word <i>direct</i> means that the direction of the flow of electrons does not change with time and goes from the end with negative charge to the end with positive charge, which implies that the direction of the electric current (the direction of "holes" left by electrons) is opposite, from positive to negative end. For practical reasons we will not consider the case of the same sign of charges on both ends.<br /><br />In most practical cases there is a device that separates the electrons from the neutral atoms within some object, thereby producing negative and positive charges on its terminals. If there is some conductor of electricity between these terminals, electrons will move from one terminal to another along this conductor, which constitutes a <b>direct electric current</b> in it.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-14102784939056414392020-02-20T14:11:00.001-08:002020-02-20T14:11:09.732-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Capacitors<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/obgGC5itTsk" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Capacitors</u><br /><br /><i>Electric Field of<br />a Uniformly Charged Disk</i><br /><br />Please refer to Problems 4 of "Electric Field", as we will use its results.<br />The problem there derives a formula of intensity of the electric field produced by an infinitely thin disk of radius <i><b>R</b></i> charged with surface density of electricity <i><b>σ</b></i> at point <i><b>P</b></i> positioned at height <i><b>h</b></i> above the center of this disk in case the space around this disk is filled with material with <i>dielectric constant</i> (also known as <i>relative permittivity</i>) <i><b>ε<sub>r</sub></b></i>.<br /><br />The direction of the vector of electric field intensity is along the perpendicular from point <i><b>P</b></i> to a charged disk and its magnitude equals to<br /><i><b>E(h) =<br />= </b></i>[<i><b>σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b>·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />From the above formula for any media filling the space around a charged disk we see that the greater <i>dielectric constant</i> <i><b>ε<sub>r</sub></b></i> for the media that fills the space around the charged disk - the smaller intensity of the field around it.<br /><br />Assume now that our goal is to generate electric charge (excess or deficiency of electrons) and store it somehow. Since electrons are not easily produced from nothing, we should take some electrically neutral object <i>X</i>, separate part of its electrons from the atoms and place them into a different object <i>X<sub>−</sub></i>, which becomes negatively charged, leaving old object positively charged, which we can label now <i>X<sub>+</sub></i>.<br /><br />For practical reasons objects <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i> should be near each other and we should prevent any kind of exchange of electric charge, like a spark, between them, which would negate our efforts to separate negative and positive charges. Considering the proximity of <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i>, it would be beneficial for avoiding any exchange of the charge between them to have a uniform distribution of charge in each object.<br /><br />The best configuration of objects <i>X<sub>−</sub></i> and <i>X<sub>+</sub></i> that allows approximately uniform distribution of charge in the presence of opposite charge nearby is when both objects are thin flat plates positioned parallel to each other. Other configurations will cause the concentration of charges in places close to the opposite charge and higher intensity of the electric field between them, which might result in a discharging spark. Another configuration might be of two concentric spheres, but it's not very practical.<br /><br />Consider now two relatively large infinitely thin disks of radius <i><b>R</b></i> with opposite charges <i><b>+Q</b></i> and <i><b>−Q</b></i> positioned parallel to each other, perpendicular to a line connecting their centers and at distance <i><b>d</b></i> from each other. Let's measure an intensity of the electric field at any point between these disks on a center line at distance <i><b>h</b></i> from positively charged disk <nobr>(<i><b>0 ≤ h ≤ d</b></i>).</nobr><br /><br />The intensity of a combined field of two disks is a vector sum of intensities of all components of this field. The direction of the intensity of both fields is along the center line between them. The probe charge is <i><b>+1C</b></i>, so the positively charged disk will repel it, while negatively charged one will attract it. So, we can calculate the magnitude of each field and add them together to get the magnitude of the combined field.<br /><br />The density of the electric charge for these disks equals to a total charge <i><b>Q</b></i> divided by a surface area <i><b>A</b></i>. There are two opposite surfaces of each disk, but in case we have two close to each other parallel disks with opposite charges the electric charge of each disk (excess or deficiency of electrons) concentrates on a surface that is close to another disk with an opposite charge. So for each disk the absolute value of charge density is constant that equals to<br /><i><b>σ = Q/A = Q/(πR²)</b></i><br /><br />The magnitude of the field intensity from a positively charged disk at distance <i><b>h</b></i> from its surface is<br /><i><b>E(h) =<br />= </b></i>[<i><b>σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b>·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />The magnitude of the field intensity from a negatively charged disk at distance <i><b>d−h</b></i> from its surface is <i><b>E(d−h)</b></i>.<br /><br />Just as an observation, let's notice that, in case the disks are very large and the distance between them very small, both <i><b>E(h)</b></i> and <i><b>E(d−h)</b></i> are approximately equal to<br /><i><b>E = σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br /><br />Since both vectors of intensity are directed from positive to negative disk (positively charged disk repels the probe charge of <i><b>+1C</b></i> positioned in-between the disks, and the negatively charged disk attracts it), the magnitude of the intensity of the combined electric field produced by both disks equals to<br /><i><b><span style="text-decoration-line: overline;">E</span>(h) = E(h)+E(d−h)</b></i><br />Expressed in all its details, the formula is quite large and difficult to analyze.<br /><br />At this point we will do what physicists usually do with a cumbersome formula - assume that in practice certain really small values can be assumed as infinitely small and certain large values to be infinitely large.<br />Indeed, if <i><b>h→0</b></i> or <i><b>h→d</b></i> the assumption above is correct. Representing graphically intensity <i><b>E</b></i> as a function of <i><b>h</b></i> for <i><b>R=10</b></i> and <i><b>d=0.1</b></i> on a segment <nobr><i><b>0 ≤ h ≤ d</b></i></nobr> shows hardly visible rise in the middle of a segment with <i><b>h=d/2</b></i>.<br /><br />So, for practical reasons we will assume that the distance between the disks <i><b>d</b></i> is very small, while the radius of disks <i><b>R</b></i> is very large. Since the point we measure the intensity is between the disks at the distance <i><b>h</b></i> from one of them, variables <i><b>h</b></i> and <i><b>d−h</b></i> can also be assumed as very small.<br />This assumption leads to consider the values <i><b>R²/h²</b></i> and <i><b>R²/(d−h)²</b></i> as infinitely large, which result in the following approximate formula for the intensity of a combined field between the disks:<br /><i><b><span style="text-decoration-line: overline;">E</span> ≅ σ/(2ε<sub>r</sub>·ε<sub>0</sub>) + σ/(2ε<sub>r</sub>·ε<sub>0</sub>) =<br />= σ/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br />So, approximately, the electric field between the large parallel disks on a small distance from each other is uniform and depends only on the density of electric charge on the disks <i><b>σ</b></i> and the dielectric constant of the media between them <i><b>ε<sub>r</sub></b></i>.<br /><br />Let's analyze a process of discharging of electricity between two disks.<br />Since our purpose is to store the charges and to avoid discharge, we should know how much electricity <i><b>Q</b></i> we can store in these two disks before electrons jump from a negatively charged disk to a positively charged because of the force of attraction.<br /><br />The discharge will be more difficult if the force of attraction between the disks is less. The force of attraction is characterized by the intensity of the field between the disks. Knowing intensity of the field between the disks <i><b><span style="text-decoration-line: overline;">E</span></b></i>, which is the force acting on a unit charge, we can calculate the work needed by a charge of <i><b>+1C</b></i> to overcome a distance <i><b>d</b></i> between the disks by multiplying the intensity (force) by the distance.<br /><br />The work needed to discharge <i><b>+1C</b></i>, that is the work needed to move <i><b>+1C</b></i> of charge from one disk to another (the <i>voltage</i> between the disks) is<br /><i><b>V = <span style="text-decoration-line: overline;">E</span>·d = σ·d/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i><br /><br />Let's recall that the purpose of our work is to store as much electric charge in these two disks as possible. But with growing charge <i><b>Q</b></i> proportionally grows the density of electricity <i><b>σ</b></i> and proportionally grows the <i>intensity</i> of the field and the <i>voltage</i> between the disks.<br /><br />Let's introduce the new characteristic that defines the ability of our two disk construction, called <b>capacitor</b>, to hold electric charge - <i>capacity</i> <i><b>C = Q/V</b></i>.<br />Defined as such, the <i>capacity</i> of a <b>capacitor</b> described above equals to<br /><i><b>C = Q/V = σ·A/</b></i>[<i><b>σ·d/(ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b> =<br />= A·(ε<sub>r</sub>·ε<sub>0</sub>)/d</b></i><br /><br />As we see, the <i>capacity</i> of our <b>capacitor</b> depends on three major factors:<br />(a) the area of the disks (<i>capacity</i> is proportional to this area)<br />(b) the distance between disks (<i>capacity</i> is inversely proportional to this area)<br />(c) the dielectric constant of the media between the disks (<i>capacity</i> is proportional to a dielectric constant of the media).<br /><br />To satisfy the purpose of storing as much electricity in the <b>capacitor</b> as possible, we should increase the area of the disks, decrease the distance between the disks and isolate disks from each other with the media with high dielectric constant.<br /><br />The disk shape of a capacitor is not really important, as long as it contains two flat parallel shapes on a small distance from each other. The easiest variation from the practical standpoint is two rectangles. Although different shape leads to slightly different results in the value of <i>capacity</i>, the difference is negligible, and it disappears completely when the distance between planes becomes an infinitesimal value. So, the definition of <i>capacity</i><br /><i><b>C = A·(ε<sub>r</sub>·ε<sub>0</sub>)/d</b></i><br />is quite universal and is widely used as a characteristic of a <b>capacitor</b>.<br /><br />Finally, if we adapt a rectangular form of a capacitor, it's easy to significantly increase the area <i><b>A</b></i> by folding or rolling both surfaces together. This technique is actually utilized in manufacturing capacitors of large capacity and it's called <i>parallel connection of capacitors</i>.<br />In such a way of connecting capacitors the total capacity of an assembly equals to a sum of capacities of individual capacitors<br /><i><b>C = C<sub>1</sub>+C<sub>2</sub>+C<sub>3</sub>+C<sub>4</sub>+...</b></i><br />This picture depicts an idea to stack smaller individual capacitors to enlarge the area, thus allowing to store more electric charge in a smaller volume.<br /><img src="http://www.unizor.com/Pictures/Capacitor.png" style="height: 200px; width: 200px;" />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20035422686338649492020-02-20T12:35:00.001-08:002020-02-20T12:35:49.446-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Voltage<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/IV-x2jnCbLU" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Voltage</u><br /><br />Recall the concept of the electric field <i>potential</i>.<br />By definition, the <b>electric field potential</b> is a quantitative characteristic of an electric field, <u>defined for each position in this field</u>, as the amount of work needed to move a probe - positively electrically charged point-object of <b>+1C</b> (one <i>coulomb</i>) - from infinitely remote point in space, where the field does not exist, to this position in the field.<br /><br />Also recall that in an electric field amount of work to move a charge from one position to another is independent of a trajectory because electrostatic forces are <i>conservative</i>. So, amount of work to move a charge from point A to point B along a straight line between them is the same as if we move along some curve or go from A to an infinitely far point and then return to B.<br /><br />Electric potential for each point of an electric field fully defines this field. If we know the electric potential at each point of a field, we don't have to know what kind of an object is the source of the field, nor its charge, nor shape in order to understand the movement of any probe object in this field.<br /><br />To find the amount of work needed to move a charge <i><b>q</b></i> from a point in the electric field with a potential <i><b>V<sub>1</sub></b></i> to a point with potential <i><b>V<sub>2</sub></b></i> we can simply multiply the electric charge by the difference of electric potentials between these points:<br /><i><b>W = q·(V<sub>2</sub>−V<sub>1</sub>)</b></i><br />The sign of the resulting value of work <i><b>W</b></i> is important. It signifies whether outside force has to perform the work against the forces of the electric field (like forcing a positive charge to go further from the attracting negative charge) or the electric field does the work itself (like forcing a positive charge to go further from the repelling positive charge).<br /><br />The expression in the parenthesis in the above formula for work <i><b>W</b></i>, that signifies the <i>difference in electric potentials</i> between two points in the electric field, is the main component to calculate the work needed to move a charge between these points. This expression has a special name - <b>voltage</b> - in honor of Italian physicist Alessandro Volta.<br /><br />Since the electric potential is the work performed on the unit charge, the convenient unit of measurement of this potential is the unit of work per unit of charge.<br /><br />This unit of measurement is called <i><b>volt</b></i> (symbol <i><b>V</b></i>) and it is defined as such a <i>difference in electric potential</i> between two points in an electric field that one <i>joule</i> of work (<i><b>1J</b></i>) is required to move one <i>coulomb</i> of positive charge (<i><b>+1C</b></i>) between these points:<br /><i><b>1V = 1J/1C</b></i><br /><br />In the electric field produced by an electrically charged point-object with <i><b>Q</b></i> amount of electricity the potential at any point depends only on its distance <i><b>R</b></i> from the source of the field and equals to <i><b>k·Q/R</b></i>. So, the <i>voltage</i> between two points at distances <i><b>R<sub>1</sub></b></i> and <i><b>R<sub>2</sub></b></i> equals to<br />Δ<i><b>V = k·Q·(1/R<sub>1</sub> − 1/R<sub>2</sub>)</b></i><br /><br />In the electric field produced by an electrically charged infinite plane with density of charge <i><b>σ</b></i> the <i>intensity</i> of the field is constant for all points in space and is equal to <i><b>2π·k·σ</b></i> (see Problems 2 of this section). The direction of the force is always perpendicular to the plane. Therefore, moving a probe charge parallel to the plane does not require any work, and the only parameter we have to take into consideration is the height above the plane. The amount of work needed to move a probe charge of <i><b>+1C</b></i> from the height <i><b>H<sub>1</sub></b></i> to height <i><b>H<sub>2</sub></b></i> is the product of the force (<i>intensity</i> of the field) by distance (difference in height). Therefore, it equals to<br />Δ<i><b>V = 2π·k·σ·(H<sub>2</sub>−H<sub>1</sub>)</b></i><br /><br /><i>Important Analogy</i><br /><br />Producing electric charge by separation of electrons from the neutral atoms can be compared with raising the level of water (or any other liquid) in the tall vertical tube with closed valve at the bottom in the gravitational field of Earth.<br />Electric charge causes the existence of electric field that is the source of a force on any electrically charged object.<br />The water raised to a certain height is the source of pressure, which is the force acting on any object at the bottom.<br />The <i>voltage</i>, which is a <i>difference in electric potential</i> between two points in the electric field is analogous to a <i>difference in potential energy</i> between the water at two different heights.<br />Connecting positive and negative charges will cause their mutual neutralization, so no electric field and no voltage will be present anymore.<br />Opening a valve in the vertical tube with water will cause the water to flow down, and no pressure will exist anymore at the bottom of the tube.<br /><br />This analogy might be useful to understand many facts related to electricity, and it goes much deeper than just about voltage. Electric generators, electric motors, conductors, resistors etc. - all can be to an extent compared with corresponding water-based devices. We will address these concepts, as we progress with the course.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-80459200995437257432020-02-10T13:41:00.001-08:002020-02-10T13:41:56.037-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 4<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/Tq8D22axXes" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems 4</u><br /><br /><i>Problem A</i><br /><br />An infinitesimally thin disk <i><b>α</b></i> of radius <i><b>R</b></i> is electrically charged with uniform density of electric charge <i><b>σ</b></i> (coulombs per square meter). What is the <b>intensity</b> of the electric field produced by this disk at point <i><b>P</b></i> positioned at distance <i><b>h</b></i> (meters) from its surface on a perpendicular through its center?<br /><br /><i>Solution</i><br />Please refer to Problems 2 of "Electric Field", as we will use its results. The Problem A from Problems 2 is about intensity of the electric field produced by an infinite plane.<br /><br />Here we will consider a finite electrically charged infinitely thin disk of the radius <i><b>R</b></i> instead of an infinite plane as in Problems 2.<br />Assume, as in Problems 2, the density of electrical charge of this disk is <i><b>σ</b></i> and the point we want to measure the intensity of the field is <i><b>P</b></i> located above the disk on the perpendicular through its center on height <i><b>h</b></i> above it.<br /><br />Going through exactly the same logic as in Problems 2, dividing our disk into concentric rings, we will come to a formula for intensity produced by an infinitely narrow ring of inner radius <i><b>r</b></i> and outer radius <i><b>r+</b>d<b>r</b></i>, where <i><b>r</b></i> is changing from <i><b>0</b></i> to <i><b>R</b></i>:<br /><i>d<b>E(h,r) =<br />= 2π·k·σ·r·</b>d<b>r·h <span style="font-size: medium;">/</span> (h²+r²)<span style="font-size: x-small;"><sup>3/2</sup></span></b></i><br /><br />After substitution<br /><i><b>y = 1 + r²/h²</b></i><br />we obtain<br /><i>d<b>y = 2r·</b>d<b>r/h²</b></i><br /><i><b>2r·</b>d<b>r = h²·</b>d<b>y</b></i><br /><i>d<b>E(h,y) =<br />= π·k·σ·</b>d<b>y·h³ <span style="font-size: medium;">/</span> (h²+r²)<span style="font-size: x-small;"><sup>3/2</sup></span>=<br />= π·k·σ·</b>d<b>y <span style="font-size: medium;">/</span> (1+r²/h²)<span style="font-size: x-small;"><sup>3/2</sup></span> =<br />= π·k·σ·y<span style="font-size: x-small;"><sup>−3/2</sup></span>·</b>d<b>y</b></i><br />which is easy to integrate since it's a plain power function.<br />The limits of integration for <i><b>r</b></i> are from <i><b>0</b></i> to <i><b>R</b></i>.<br />Therefore, the limits of integration for <i><b>y</b></i> are from <i><b>1</b></i> to <i><b>1+R²/h²</b></i>.<br />The indefinite integral of <i><b>y<span style="font-size: x-small;"><sup>−3/2</sup></span></b></i> is <i><b>−2y<span style="font-size: x-small;"><sup>−1/2</sup></span></b></i>. Therefore, the total vector of intensity at point <i><b>P</b></i> equals to<br /><i><b>E(h) =<br />= −2π·k·σ·</b></i>[<i><b>(1+R²/h²)<span style="font-size: x-small;"><sup>−1/2</sup></span>−1</b></i>]<i><b> =<br />= 2π·k·σ·</b></i>[<i><b>1−(1+R²/h²)<span style="font-size: x-small;"><sup>−1/2</sup></span></b></i>]<i><b> =<br />= 2π·k·σ·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />We can rewrite this formula using the <i>permittivity</i> of vacuum <i><b>ε<sub>0</sub>=1/(4π·k)</b></i> as<br /><i><b>E(h) = (σ/2ε<sub>0</sub>)·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />Analyzing this formula, we see that the intensity of the electric field of a uniformly charged disk of radius <i><b>R</b></i> at point above its center on the height <i><b>h</b></i> depends on the ratio <i><b>R/h</b></i>.<br />If the height remains the same, but the radius increases to infinity, the formula transforms into the one we obtained in the Problem 2 for infinite charged plane.<br />If the height <i><b>h</b></i> decreases to zero with a fixed radius <i><b>R</b></i>, the intensity gradually increases to its maximum value <i><b>2π·k·σ</b></i>, which is the same as for an infinite plane in Problems 2. So, for a small height the uniformly charged disk acts like an infinite plane.<br />If the height <i><b>h</b></i> increases to infinity with a fixed radius <i><b>R</b></i>, the intensity gradually decreases to zero.<br />All conclusions are intuitively obvious.<br /><br />Another parameter from which the intensity depends is the medium around a charged object. Knowing from a previous lecture about <i>permittivity</i>, we can consider the space around the charged disk to be not only vacuum, but any media with known <i>dialectic constant</i> <i><b>ε<sub>r</sub></b></i>.<br />In this case, instead of Coulomb's constant <i><b>k</b></i>, we have to use <i><b>1/(4π·ε<sub>r</sub>·ε<sub>0</sub>)</b></i> and the formula for intensity looks like this:<br /><i><b>E(h) =<br />= </b></i>[<i><b>σ/(2ε<sub>r</sub>·ε<sub>0</sub>)</b></i>]<i><b>·</b></i>[<i><b>1−1/√<span style="text-decoration-line: overline;">1+R²/h² </span></b></i>]<br /><br />From the above formula for any media filling the space around a charged disk we see that the greater <i>dielectric constant</i> for this media - the smaller intensity of the field around it.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-10097201066229049392020-02-10T12:01:00.001-08:002020-02-10T12:01:02.026-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field -Permittivity<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/4SXGYEkJShE" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Permittivity</u><br /><br />Let's take a closer look at Coulomb's Law<br /><i><b>F = k·q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i><br />where<br /><i><b>F</b></i> is the magnitude of the force of attraction (in case of opposite charges) or repelling (in case of the same type of charge, positive or negative) in <i>N - newtons</i><br /><i><b>q<sub>A</sub></b></i> is electric charge of point-object A in <i>C - coulombs</i><br /><i><b>q<sub>B</sub></b></i> is electric charge of point-object B in <i>C - coulombs</i><br /><i><b>R</b></i> is the distance between charged objects in <i>m - meters</i><br /><i><b>k</b></i> is a coefficient of proportionality (Coulomb's constant) equals to 9.0·10<sup>9</sup> in <i>N·m²/C²</i><br /><br />The intuitive explanation of the inverse proportionality of this force to a square of a distance between objects A and B was that the force emitting by an electric point-charge is distributed around it in a radial fashion and, at distance <i><b>R</b></i>, should be inversely proportional to an area of a sphere of the radius <i><b>R</b></i>.<br />The area of a sphere of radius <i><b>R</b></i> is <i><b>4πR²</b></i>. Therefore, it's more natural to express Coulomb's with <i><b>4πR²</b></i> in the denominator. Then it will look like this<br /><i><b>F = q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> (4π·ε<sub>0</sub>·R²)</b></i><br />where<br /><i><b>ε<sub>0</sub></b></i> is a constant called <i>permittivity of vacuum</i>.<br />In terms of Coulomb's constant <i><b>k</b></i> it is equal to<br /><i><b>ε<sub>0</sub> = 1/(4π·k) = 8.85419·10<sup>−12</sup></b></i><br />measured in <i>C²/(N·m²)</i><br /><br />In the above definition <i>permittivity</i> is the property of space between the charges to let the force of electric field through. It is analogous to such mechanical properties as <i>resistance</i>, <i>friction</i>, <i>viscosity</i>.<br /><br />As experiments show, the same electric charges at the same distance but in different environments produce electric fields of different intensities.<br /><b>Environment matters</b>. In vacuum a specific point-charge at a specific distance produces the field of one intensity, while, positioned inside a sand box, the same charge at the same distance would produce a field of different intensity.<br /><br />That's why we specifically called <i><b>ε<sub>0</sub></b></i> the <i>permittivity of vacuum</i>, as no other environment was considered. All the experiments described before relate to vacuum as the media where these experiments are conducted. In different environment the force of electric field would differ.<br /><br />This prompts us to introduce an <i>absolute permittivity</i> <i><b>ε<sub>a</sub></b></i> (or simply <i><b>ε</b></i>) of any media and its <i>relative permittivity</i> <i><b>ε<sub>r</sub>=ε<sub>a</sub>/ε<sub>0</sub></b></i>.<br />The <i>relative permittivity</i> of a media is also called its <i>dielectric constant</i> with the value of this <i>dielectric constant</i> for vacuum being equal to 1.<br />For olive oil the <i>dielectric constant</i> is 3, for silicon its about 11-12, for mineral oil its about 2, for marble - 8, for titanium dioxide - between 86 and 173 etc.<br /><br />Now the formula for intensity of electric field in a media with <i>relative permittivity</i> <i><b>ε<sub>r</sub></b></i> looks like this<br /><i><b>F = q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> (4π·ε<sub>r</sub>·ε<sub>0</sub>·R²)</b></i><br />So, generally speaking, when the electric field speads into any media with a <i>dielectric constant</i> <i><b>ε<sub>r</sub></b></i>, we should use the coefficient <i><b>1 <span style="font-size: medium;">/</span> </b></i><b>(<i>4π·ε<sub>r</sub>·ε<sub>0</sub></i>)<i></i></b> instead of Coulomb's constant <i><b>k</b></i>.<br /><br />The greater the value of the <i>dielectric constant</i> - the stronger it resists to penetration of electric field, so the field is weaker than in vacuum for the same charge and distance. Vacuum is the easiest for the electric field to penetrate.<br /><br />Also worth noting that the permittivity of any material depends on its temperature and exact chemical composition. This allows, for example, to measure the temperature or humidity of air by measuring its relative permittivity.<br /><br />Materials with high value of permittivity are used for electrical insulation to prevent the electric field from dissipating around electrical charges.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-14730116643242091412020-01-23T13:33:00.001-08:002020-01-23T13:33:20.234-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 3<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/iJ2Ci7AHyW8" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems 3</u><br /><br /><i>Problem A</i><br /><br />An infinitesimally thin sphere <i><b>α</b></i> of radius <i><b>R</b></i> is electrically charged with uniform density of electric charge <i><b>σ</b></i> (coulombs per square meter).<br />A point <i><b>P</b></i> is inside this sphere at a distance <i><b>h</b></i> (meters) from its center, so <i><b>h</b></i> is less than <i><b>R</b></i>.<br />What is the <b>intensity</b> of the electric field produced by this sphere at point <i><b>P</b></i>?<br /><br /><i>Solution</i><br /><br />The magnitude of the field intensity at point <i><b>P</b></i> produced by any infinitesimal piece of sphere <i><b>α</b></i> is inversely proportional to a square of its distance to point <i><b>P</b></i> and directly proportional to its charge, and the charge, in turn, is proportional to its area with density <i><b>σ</b></i> being a coefficient of proportionality.<br /><br />Let's define a system of Cartesian coordinates in our three-dimensional space with the origin at the center <i><b>O</b></i> of the center of the charged sphere <i><b>α</b></i> with Z-axis along segment <i><b>OP</b></i>.<br /><br />Then the coordinates of point <i><b>P</b></i>, where we have to calculate the vector of electric field intensity, are (<i><b>0,0,h</b></i>).<br /><br />From the considerations of symmetry, the vector of intensity of the field, produced by an entire electrically charged sphere <i><b>α</b></i>, at point <i><b>P</b></i> inside it should be directed along a line <i><b>OP</b></i> from point <i><b>P</b></i> to a center of a sphere, that is along Z-axis. Indeed, for any infinitesimal area of sphere <i><b>α</b></i> near point (<i><b>x,y,z</b></i>) there is an area symmetrical to it relatively to Z-axis near point (<i><b>−x,−y,z</b></i>), which produces the intensity vector of the same magnitude, the same vertical (parallel to <i><b>OP</b></i>) component of it and opposite horizontal component. So, all horizontal components will cancel each other, while vertical ones can be summarized by magnitude.<br />Therefore, we should take into account only projections of all individual intensity vectors from all areas of a sphere onto Z-axis, as all other components will cancel each other.<br /><br />The approach we will choose is to take an infinitesimal area on a sphere in a form of a spherical ring of infinitesimal width, produced by cutting a sphere by two planes parallel to XY-plane at Z-coordinates <i><b>z=r</b></i> and <i><b>z=r+</b>d<b>r</b></i> and calculate the vertical component of the intensity vector produced by it. Then we will integrate the result from <i><b>r=−R</b></i> to <i><b>r=R</b></i>.<br />This choice is based on a simple fact that for every small piece of this spherical ring its distance to point <i><b>P</b></i> is the same, as well as an angle between its vector of intensity and Z-axis is the same, hence the vertical component of the field intensity vector produced by it will be the same as for any other such piece of this spherical ring, if it has the same area, while the horizontal component of the intensity vector will be canceled by a symmetrical piece of this spherical ring lying diametrically across it.<br /><br />Since any infinitesimal part of this ring has exactly the same vertical component of the intensity vector as any other part having the same area, to get the total vertical component of intensity for an entire ring, we can use its total charge that depends on its total area and charge density <i><b>σ</b></i>.<br /><br /><img src="http://www.unizor.com/Pictures/SphericalRing_Side.png" style="height: 200px; width: 200px;" /><br />The area of a spherical ring equals to a difference between areas of two spherical caps.<br />The area of a spherical cap equals to <i><b>2π·R·H</b></i>, where <i><b>H</b></i> is the height of a cap.<br />The spherical cap formed by a plane cutting a sphere at <i><b>z=r</b></i> has height <i><b>R−r</b></i>. The spherical cap formed by a plane cutting a sphere at <i><b>z=r+</b>d<b>r</b></i> has height <i><b>R−r−</b>d<b>r</b></i>.<br />Therefore, the area of a spherical ring between these two cutting planes is<br /><i>area<b>(r,</b>d<b>r) =<br />= 2π·R·</b></i>[<i><b>(R−r)−(R−r−</b>d<b>r)</b></i>] =<i><b><br />= 2π·R·</b>d<b>r</b></i><br /><br />The charge concentrated in this spherical ring is<br /><i>d<b>Q(r) = 2π·σ·R·</b>d<b>r</b></i><br /><br />To find the magnitude of the intensity vector produced by this ring we need to know its charge (calculated above) and the distance to a point, where the intensity is supposed to be calculated. This distance can be calculated using the Pythagorean Theorem:<br /><i><b>L² = (R² − r²) + (h−r)² =<br />= R² + h² −2h·r</b></i><br />The magnitude of the intensity vector produced by this spherical ring is, therefore,<br /><i>d<b>E(h,r) = k·</b>d<b>Q(r)/L² =<br />= 2π·k·σ·R·</b>d<b>r <span style="font-size: medium;">/</span> L² =<br />= 2π·k·σ·R·</b>d<b>r <span style="font-size: medium;">/</span> (R²+h²−2h·r)</b></i><br /><br />We are interested only in vertical component of this vector, which is equal to<br /><i>d<b>E<sub>z</sub>(h,r)= </b>d<b>E(h,r)·sin(</b></i>∠<i><b>PAB) =<br />= </b>d<b>E(h,r)·(h−r)/L =<br />= 2π·k·σ·R·</b>d<b>r·(h−r) <span style="font-size: medium;">/</span> L³ =<br />= 2π·k·σ·R·</b>d<b>r·(h−r) <span style="font-size: medium;">/</span> (R²+h²−2h·r)<span style="font-size: x-small;"><sup>3/2</sup></span></b></i><br /><br />Integrating this by <i><b>r</b></i> from −R to R can be done as follows.<br />First of all, let's substitute<br /><i><b>x = R²+h²−2h·r</b></i><br />Then<br /><i><b>r = (R²+h²−x)/2h</b></i><br /><i>d<b>r = −</b>d<b>x/2h</b></i><br />The limits of integration for <i><b>x</b></i> are from <i>R²+h²+2h·R=(R+h)²</i> to <i>R²+h²−2h·R=(R−h)²</i>.<br />Now the expression to integrate looks like<br /><i>d<b>E<sub>z</sub>(h,x) = C·(h²−R²+x)·x<span style="font-size: x-small;"><sup>−3/2</sup></span>·</b>d<b>x</b></i><br />where constant <i><b>C</b></i> equals to<br /><i><b>C = −π·k·σ·R/(2h²)</b></i><br /><br />Let's integrate the above expression in the limits specified.<br />First, find the indefinite integral.<br /><span style="font-size: large;">∫</span><i><b>C·(h²−R²+x)·x<span style="font-size: x-small;"><sup>−3/2</sup></span>·</b>d<b>x =<br />= −2C·(h²−R²)·x<span style="font-size: x-small;"><sup>−1/2</sup></span> + 2C·x<span style="font-size: x-small;"><sup>1/2</sup></span> =<br />= 2C·[(R²−h²)·x<span style="font-size: x-small;"><sup>−1/2</sup></span> + x<span style="font-size: x-small;"><sup>1/2</sup></span>]</b></i><br />This expression for indefinite integral should be used to calculate the definite integral in limits for <i><b>x</b></i> from <i>(R+h)²</i> to <i>(R−h)²</i>.<br /><br />Assuming that point <i><b>P</b></i> is inside a sphere, that is <i>h</i> is less than radius <i>R</i>,<br /><i><b>((R+h)²)<span style="font-size: x-small;"><sup>1/2</sup></span> = R+h</b></i><br /><i><b>((R−h)²)<span style="font-size: x-small;"><sup>1/2</sup></span> = R−h</b></i><br /><br />Therefore, substituting the upper limit into an expression for an indefinite integral, we get<br /><i><b>2C·[(R²−h²)/(R−h) + (R−h)] = 4C·R</b></i><br />Substituting the lower limit, we get<br /><i><b>2C·[(R²−h²)/(R+h) + (R+h)] = 4C·R</b></i><br />The difference between these two expressions is zero, which means that <u>the intensity of the electric field inside a uniformly charged sphere is zero</u>.<br /><br /><i>Problem B</i><br /><br />An infinitesimally thin sphere <i><b>α</b></i> of radius <i><b>R</b></i> is electrically charged with uniform density of electric charge <i><b>σ</b></i> (coulombs per square meter).<br />A point <i><b>P</b></i> is outside this sphere at a distance <i><b>h</b></i> (meters) from its center, so <i><b>h</b></i> is greater than <i><b>R</b></i>.<br />What is the <b>intensity</b> of the electric field produced by this sphere at point <i><b>P</b></i>?<br /><br /><i>Solution</i><br /><br />Start as in the previous problem up to indefinite integral<br /><i><b>2C·[(R²−h²)·x<span style="font-size: x-small;"><sup>−1/2</sup></span> + x<span style="font-size: x-small;"><sup>1/2</sup></span>]</b></i><br /><br />Assuming that point <i><b>P</b></i> is outside a sphere, that is <i>h</i> is greater than radius <i>R</i>,<br /><i><b>((R+h)²)<span style="font-size: x-small;"><sup>1/2</sup></span> = R+h</b></i><br /><i><b>((R−h)²)<span style="font-size: x-small;"><sup>1/2</sup></span> = h−R</b></i><br /><br />Therefore, substituting the upper limit into an expression for an indefinite integral, we get<br /><i><b>2C·[(R²−h²)/(h−R) + (h−R)] = −4C·R</b></i><br />Substituting the lower limit, we get<br /><i><b>2C·[(R²−h²)/(R+h) + (R+h)] = 4C·R</b></i><br />The difference between them is the intensity of the electric field produced by a sphere at a point outside it:<br /><i><b>E(h) = −8C·R =<br />= 8π·k·σ·R²/(2h²) =<br />= 4π·k·σ·R²/h²</b></i><br /><br />Notice that<br /><i>area(Sphere) = 4π·R²</i><br />Therefore, <i><b>Q = 4π·σ·R²</b></i><br />where <i><b>σ</b></i> is the density of electric charge on a sphere, represents a total charge of a sphere.<br />Hence,<br /><i><b>4π·k·σ·R²/h² = k·Q/h²</b></i><br />and the field intensity of a sphere at a point outside it equals to intensity of a point-object with the same electric charge and located at the center of a sphere.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-82020851862277987702020-01-21T12:00:00.001-08:002020-01-21T12:00:21.699-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 2<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/YAmlNYTA4qs" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems 2</u><br /><br /><i>Problem A</i><br /><br />An infinite infinitesimally thin plane <i><b>α</b></i> is electrically charged with uniform density of electric charge <i><b>σ</b></i> (coulombs per square meter).<br />What is the <b>intensity</b> of the electric field produced by this plane at point <i><b>P</b></i> positioned at a distance <i><b>h</b></i> (meters) from its surface?<br /><br /><i>Solution</i><br /><br />The magnitude of the field intensity at point <i><b>P</b></i> produced by any infinitesimal piece of plane <i><b>α</b></i> is inversely proportional to a square of its distance to point <i><b>P</b></i> and directly proportional to its charge, and the charge, in turn, is proportional to its area with density <i><b>σ</b></i> being a coefficient of proportionality.<br /><br />Let's define a system of cylindrical coordinates in our three-dimensional space with the origin at the projection <i><b>O</b></i> of the point <i><b>P</b></i> onto our electrically charged plane <i><b>α</b></i> with Z-axis along segment <i><b>OP</b></i> and polar coordinates on plane <i><b>α</b></i> with <i><b>r</b></i> for radial distance <i><b>OA</b></i> between any point <i><b>A</b></i> on this plane and the origin of coordinates <i><b>O</b></i> and <i><b>φ</b></i> for a counterclockwise angle from the positive direction of some arbitrarily chosen base ray <i><b>OX</b></i> within plane <i><b>α</b></i>, originated an point <i><b>O</b></i>, to radius <i><b>OA</b></i>.<br /><br />Then the coordinates of point <i><b>P</b></i>, where we have to calculate the vector of electric field intensity, are (<i><b>0,0,h</b></i>). The coordinates of any point <i><b>A</b></i> on plane <i><b>α</b></i> will be (<i><b>r,φ,0</b></i>).<br /><br />From the considerations of symmetry, the vector of intensity of the field, produced by an entire infinite electrically charged plane <i><b>α</b></i>, at point <i><b>P</b></i> outside of it should be directed along a perpendicular <i><b>OP</b></i> from point <i><b>P</b></i> to a plane. Indeed, for any infinitesimal area of plane <i><b>α</b></i> there is an area symmetrical to it relatively to point <i><b>O</b></i>, which produces the intensity vector of the same magnitude, the same vertical (parallel to <i><b>OP</b></i>) component of it and opposite horizontal (within a plane <i><b>α</b></i>) component. So, all horizontal components will cancel each other, while vertical ones can be summarized by magnitude.<br />Therefore, we should take into account only projections of all individual intensity vectors from all areas of a plane onto a perpendicular <i><b>OP</b></i> from point <i><b>P</b></i> onto plane <i><b>α</b></i>, as all other components will cancel each other.<br /><br />The approach we will choose is to take an infinitesimal area on a plane in a form of a ring centered at point <i><b>O</b></i> of infinitesimal width <i>d<b>r</b></i> with inner radius <i><b>r</b></i> and outer radius <i><b>r+</b>d<b>r</b></i> and calculate the vertical component of the intensity vector produced by it. Then we will integrate the result from <i><b>r=0</b></i> to infinity.<br />This choice is based on a simple fact that for every small piece of this ring its distance to point <i><b>P</b></i> is the same, as well as an angle between its vector of intensity and Z-axis is the same, hence the vertical component of the field intensity vector produced by it will be the same as for any other such piece of this ring, if it has the same area, while the horizontal component of the intensity vector will be canceled by a symmetrical piece of this ring lying diametrically across it.<br /><br />Since any infinitesimal part of this ring has exactly the same vertical component of the intensity vector as any other part having the same area, to get the total vertical component of intensity for an entire ring, we can use its total charge that depends on its total area and charge density <i><b>σ</b></i>.<br /><br /><img src="http://www.unizor.com/Pictures/FlatRing.png" style="height: 200px; width: 200px;" /><br />The area of a ring equals to<br /><i>area<b>(r,</b>d<b>r) = π</b></i>[<i><b>(r+</b>d<b>r)²−r²</b></i>]<i><b> =<br />= 2π·r·</b>d<b>r + π·(</b>d<b>r)²</b></i><br />We can drop the infinitesimal of the second order <i><b>π·(</b>d<b>r)²</b></i> and leave only the first component - infinitesimal of the first order, that we plan to integrate by <i><b>r</b></i> from 0 to infinity.<br /><br />The charge concentrated in this ring is<br /><i>d<b>Q(r) = 2π·σ·r·</b>d<b>r</b></i><br /><br />To find the magnitude of the intensity vector produced by this ring we need to know its charge (calculated above) and the distance to a point, where the intensity is supposed to be calculated. This distance can be calculated using the Pythagorean Theorem:<br /><i><b>L² = h² + r²</b></i><br />The magnitude of the intensity vector produced by this ring is, therefore,<br /><i>d<b>E(h,r) = k·</b>d<b>Q(r)/L² =<br />= 2π·k·σ·r·</b>d<b>r <span style="font-size: medium;">/</span> L² =<br />= 2π·k·σ·r·</b>d<b>r <span style="font-size: medium;">/</span> (h² + r²)</b></i><br /><br />We are interested only in vertical component of this vector, which is equal to<br /><i>d<b>E<sub>z</sub>(h,r)= </b>d<b>E(h,r)·sin(</b></i>∠<i><b>PAO) =<br />= </b>d<b>E(h,r)·h/L =<br />= 2π·k·σ·r·</b>d<b>r·h <span style="font-size: medium;">/</span> L³ =<br />= 2π·k·σ·r·</b>d<b>r·h <span style="font-size: medium;">/</span> (h² + r²)<span style="font-size: x-small;"><sup>3/2</sup></span></b></i><br /><br />Integrating this by <i><b>r</b></i> from 0 to ∞ can be done as follows.<br />First of all, let's substitute<br /><i><b>x = r/h</b></i><br />Then<br /><i><b>r = h·x</b></i><br /><i>d<b>r = h·</b>d<b>x</b></i><br />The limits of integration for <i><b>x</b></i> are the same, from 0 to ∞.<br />Now the expression to integrate looks like<br /><i>d<b>E<sub>z</sub>(h,x) =<br />= 2π·k·σ·h³·x·</b>d<b>x <span style="font-size: medium;">/</span> h³(1 + x²)<span style="font-size: x-small;"><sup>3/2</sup></span> =<br />= 2π·k·σ·x·</b>d<b>x <span style="font-size: medium;">/</span> (1 + x²)<span style="font-size: x-small;"><sup>3/2</sup></span></b></i><br /><u>Before going into details of integration, note that this expression does not depend on distance <i><b>h</b></i> from point <i><b>P</b></i> to an electrically charged plane <i><b>α</b></i>. This is quite remarkable!</u><br />No matter how far point <i><b>P</b></i> is from plane <i><b>α</b></i>, the intensity of electric field at this point is the same.<br /><br />To integrate the last expression for a projection onto Z-axis of the intensity of electric field produced by an infinitesimal area of plane <i><b>α</b></i>, introduce another substitution:<br /><i><b>y = x² + 1</b></i><br />Then<br /><i><b>x·</b>d<b>x = </b>d<b>y/2</b></i><br />The limits of integration for <i><b>y</b></i> are from 1 to ∞.<br />The expression to integrate becomes<br /><i>d<b>E<sub>z</sub>(y) = π·k·σ·</b>d<b>y <span style="font-size: medium;">/</span> y<span style="font-size: x-small;"><sup>3/2</sup></span> =<br />= π·k·σ·y<span style="font-size: x-small;"><sup>−3/2</sup></span>·</b>d<b>y</b></i><br /><br />This is easy to integrate. The indefinite integral of <i><b>y<span style="font-size: x-small;"><sup>n</sup></span></b></i> is <i><b>y<span style="font-size: x-small;"><sup>n+1</sup></span>/(n+1)</b></i>. Using this for <i><b>n=−3/2</b></i>, we get an indefinite integral of our function<br /><i><b>−2·π·k·σ·y<span style="font-size: x-small;"><sup>−1/2</sup></span> + C</b></i><br />Using the Newton-Leibniz formula for limits from 1 to ∞, this gives the value of the magnitude of the total intensity of a charged plane <i><b>α</b></i>:<br /><i><b>E = </b></i><span style="font-size: large;">∫</span><sub>[1,∞]</sub><i><b>π·k·σ·y<span style="font-size: x-small;"><sup>−3/2</sup></span>·</b>d<b>y = 2π·k·σ</b></i><br /><br />Let's note again that this value is independent of the distance <i><b>h</b></i> of point <i><b>P</b></i>, where we measure the intensity of the electric field, from an electrically charged plane <i><b>α</b></i>. It only depends on the density <i><b>σ</b></i> of electric charge on this plane.<br /><br />As for direction of the intensity vector, as we suggested above, it's always perpendicular to the plane <i><b>α</b></i>.<br />Hence, we can say that the electric field produced by a uniformly charged plane is <b>uniform</b>, at each point in space it is directed along a perpendicular to a plane and has a magnitude <i><b>E=2π·k·σ</b></i>, where <i><b>σ</b></i> represents the density of electric charge on a plane and <i><b>k</b></i> is a Coulomb's constant.<br /><br /><i>Problem B</i><br /><br />An infinite infinitesimally thin plane <i><b>α</b></i> is electrically charged with uniform density of electric charge <i><b>σ</b></i> (coulombs per square meter).<br />What is the <b>work</b> needed to move a charge <i><b>q</b></i> (coulombs) from point <i><b>M</b></i> positioned at a distance <i><b>m</b></i> (meters) from its surface to point <i><b>N</b></i> positioned at a distance <i><b>n</b></i> (meters) from its surface?<br /><br /><i>Solution</i><br /><br />Notice that positions of points <i><b>M</b></i> and <i><b>N</b></i> are given only in terms of their distance to a charged plane <i><b>α</b></i>, that is in terms of vertical displacement. Distance between them in the horizontal direction is irrelevant since any horizontal movement will be perpendicular to the vectors of field intensity and, therefore, require no work to be done.<br /><br />So, our work only depends on the distance along the vertical and can be calculated as<br /><i><b>W<sub>MN</sub> = E·(n−m)·q =<br />= 2π·k·σ·q·(n−m)</b></i><br /><br /><i><b><br /></b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-18067019695944394052020-01-17T11:51:00.001-08:002020-01-17T11:51:49.738-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Field Poten...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/5nfeF3xd-k8" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Field Potential</u><br /><br /><i>Coulomb's Force</i><br />The general form of the Coulomb's Law, when two electrically charged point-objects, <b><i>A</i></b> and <b><i>B</i></b>, are involved, is<br /><i><b>F = k·q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i><br />where<br /><i><b>F</b></i> is the magnitude of the force of attraction (in case of opposite charges) or repelling (in case of the same type of charge, positive or negative) in <i>newtons(N)</i><br /><i><b>q<sub>A</sub></b></i> is electric charge of point-object A in <i>coulombs(C)</i><br /><i><b>q<sub>B</sub></b></i> is electric charge of point-object B in <i>coulombs(C)</i><br /><i><b>R</b></i> is the distance between charged objects in <i>meters(m)</i><br /><i><b>k</b></i> is a coefficient of proportionality, the Coulomb's constant, equals to 9.0·10<sup>9</sup> in <i>N·m²/C²</i><br /><br />We have introduced a concept of <i>electric field intensity</i> as a <b>force</b> acting on a probe point-object <b><i>B</i></b>, charged with +1C of electricity, from a field produced by the main object <b><i>A</i></b>. This force is a characteristic of a field at a point where a probe object is located and is equal to<br /><i><b>E = k·q<sub>A</sub> <span style="font-size: medium;">/</span> R²</b></i><br /><br />If we want to move a probe point-object from one point in the field to another in uniform (without acceleration) motion, we have to take into account this force. It can help us to do the move, if this force acts in the direction of a motion, or prevent this motion, if it acts against it. In a way, the electric field becomes our partner in motion, helping or preventing us to do the move.<br /><br /><i>Work of Coulomb's Force</i><br />Of obvious interest is the amount of work needed to accomplish the move. If we act against the force of electric field intensity, we have to spend certain amount of energy to do the work. If the field force helps us, we do not spend any energy because the field does it for us. Similar considerations were presented in the Gravitation part of this course.<br /><br />Recall from the Mechanics part of this course that the work of the force <i>F</i>, acting at an angle <i>φ</i> to a trajectory on the distance <i>S</i>, is<br /><i><b>W = F·S·cos(φ)</b></i><br /><br />For a non-uniform motion and variable force all components of this formula are dependent on some parameter <i>x</i>, like time or distance:<br /><i>d<b>W(x) = F(x)·</b>d<b>S(x)·cos(φ(x))</b></i><br />where we have to use infinitesimal increments of work <i>d<b>W(x)</b></i> done by force <i><b>F(x)</b></i> on infinitesimal distance <i>d<b>S(x)</b></i>.<br />The angle <i><b>φ(x)</b></i> is the angle between a vector of force <i><b>F(x)</b></i> and a tangential to a trajectory at point <i><b>S(x)</b></i>.<br /><br />Using a concept of <i>scalar product</i> of vectors and considering force and interval of trajectory as vectors, the same definition can be written as<br /><i>d<b>W(x) = (<span style="text-decoration-line: overline;">F(x)</span>·</b>d<b><span style="text-decoration-line: overline;">S(x)</span>)</b></i><br /><br />The latter represents the most rigorous definition of <b>work</b>.<br />Integration by parameter <b><i>x</i></b> from <b><i>x=x<sub>start</sub></i></b> to <b><i>x=x<sub>end</sub></i></b> can be used to calculate the total work<br /><br /><b><i>W<sub>[x<sub>start</sub> , x<sub>end</sub> ]</sub></i></b> performed by a variable force <b><i>F(x)</i></b>, acting on an object in a non-uniform motion, on certain distance <b><i>S(x)</i></b> along its trajectory, as the parameter <i><b>x</b></i> changes from <i><b>x<sub>start</sub></b></i> to <i><b>x<sub>end</sub></b></i>.<br /><br />In case of a motion of an electrically charged object in an electrical field the force is the Coulomb's force.<br />Let's analyze the work needed to move such an object in the field of electrically charged point-object from one position to another.<br /><br /><i>Case 1. Radial Motion</i><br />Let the charge of the main point-object in the center of the electrical field be <i><b>Q</b></i>. We move a probe object of charge <i><b>q</b></i> along a radius from it to the center of a field from distance <i><b>r<sub>1</sub></b></i> to <i><b>r<sub>2</sub></b></i>.<br />The Coulomb's force on a distance <i><b>x</b></i> from the center equals to<br /><i><b>F(x) = k·Q·q <span style="font-size: medium;">/</span> x²</b></i><br />The direction of this force is along the radial trajectory and the sign of the Coulomb's force properly describes whether the resulting work will be positive (in case of similarly charged main and probe objects, + and + or − and −) or negative (in case of opposite charges, + and − or − and +).<br /><br />Since the force is variable and depends on the distance between the main object and the probe object, to calculate the work needed to move the probe object, we have to integrate the product of this force by an infinitesimal increment of the distance <i>d<b>x</b></i> on a segment from <i><b>r<sub>1</sub></b></i> to <i><b>r<sub>2</sub></b></i>.<br /><b><i>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> = <span style="font-size: large;">∫</span><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub>k·Q·q·</i></b><i>d<b>x <span style="font-size: medium;">/</span> x²</b></i><br />Since the indefinite integral (anti-derivative) of <i><b>1<span style="font-size: medium;">/</span>x²</b></i> is <i><b>−1<span style="font-size: medium;">/</span>x</b></i>, the amount of work is<br /><b><i>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> = k·Q·q·</i></b>(<i><b>1<span style="font-size: medium;">/</span>r<sub>1</sub>−1<span style="font-size: medium;">/</span>r<sub>2</sub></b></i>)<br />This work is additive. If we move from a distance of <i><b>r<sub>1</sub></b></i> to a distance <i><b>r<sub>2</sub></b></i> and then from a distance <i><b>r<sub>2</sub></b></i> to a distance <i><b>r<sub>3</sub></b></i>, the total work will be equal to a sum of works, which, in turn, would be the same as if we move directly to distance <i><b>r<sub>3</sub></b></i> without stopping at <i><b>r<sub>2</sub></b></i><br /><b><i>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> + W<sub>[r<sub>2</sub>,r<sub>3</sub>]</sub> = W<sub>[r<sub>1</sub>,r<sub>3</sub>]</sub></i></b><br /><br /><i>Case 2. Circular Motion</i><br />Consider now that we move a probe object circularly, not changing the distance from the center of the electric field.<br />In this case the vector of force (radial) is always perpendicular to the vector of trajectory (tangential). As a result, this motion can be performed without any work done by us or the field. So, for a circular motion the work performed is always zero.<br />Obviously, this work, as we move the probe object circularly, is also additive.<br /><br /><i>Case 3. General</i><br />Any vector of force in a central electric field can be represented as a sum of two vectors - radial, that changes the distance to a center of a field, and tangential (along a circle), which is perpendicular to a radius. That means that any infinitesimal increment of work can be represented as a sum of two increments - radial and tangential. Since the latter is always zero, the amount of work performed to facilitate this motion depends only on the distances to the center at the beginning and at the end of the motion.<br /><br /><i>Conservative Forces</i><br />The immediate consequence from this consideration is that the work needed to move a charged point-object from one point in the radial electric field to another is independent of the trajectory and only depends on starting and ending position in the field. Even more, for radial electric field it depends only on starting and ending distances to a center of the field.<br /><br />This independence of work from trajectory is a characteristic not only of radial electric field, but of the whole class of the fields - those produced by <i>conservative forces</i>, and <u>electrostatic forces are <i>conservative</i></u>. <i>Gravitational forces</i> are also of the same type.<br />As an example of <i>non-conservative</i> forces, consider an object moving inside the water from one point to another. Since the water always resists the movement, the longer the trajectory that connects two points - the more work is needed to travel along this trajectory.<br /><br /><i>Electric Field Potential</i><br />The <d>electric field potential is a quantitative characteristic of an electric field, <u>defined for each position in this field</u>, as the amount of work needed to move a probe point-object of +1C from infinity (where the field does not exist) to this position in the field.<br /><br />In the radial field produced by the point-object charged with <i><b>Q</b></i> amount of electricity the <d>electric field potential at any point in the field depends only on a distance of this point to a center of the field.<br />Using the formula above for <i><b>r<sub>1</sub>=∞</b></i> and <i><b>r<sub>2</sub>=r</b></i> for a probe object charged with <i><b>q=+1C</b></i> of electricity, we obtain the formula for an electric potential at distance <i><b>r</b></i> from a center of the field, where a point-object charged with <i><b>Q</b></i> amount of electricity is located<br /><i><b>V(r) = −k·Q <span style="font-size: medium;">/</span> r</b></i><br /><br />Notice that the derivative of potential <i><b>V(r)</b></i> by distance <i><b>r</b></i> from a center gives the field intensity:<br /><i><b>V'(r) = k·Q <span style="font-size: medium;">/</span> r²</b></i><br />So, knowing the potential at each point of the radial field, we can determine the intensity at each point.<br /><br />Since, as we stated above, the work performed to move a probe object in the electric field does not depend on trajectory, we can accomplish moving a probe object charged with +1C of electricity from distance <i><b>r<sub>1</sub></b></i> to distance <i><b>r<sub>2</sub></b></i> by, first, moving it to infinity, which results in amount of work <nobr><i><b>W<sub>1</sub> = −V(r<sub>1</sub>)</b></i></nobr>, then from infinity to distance <i><b>r<sub>2</sub></b></i>, which results in amount of work <nobr><i><b>W<sub>2</sub> = V(r<sub>2</sub>)</b></i></nobr>. The sum of them for a probe object charged with +1C of electricity gives the same amount of work as to move it directly from distance <i><b>r<sub>1</sub></b></i> to <i><b>r<sub>2</sub></b></i> as calculated above:<br /><i><b>W = W<sub>1</sub> + W<sub>2</sub> = −V(r<sub>1</sub>)+V(r<sub>2</sub>) =<br />= k·Q·q·</b></i>(<i><b>1<span style="font-size: medium;">/</span>r<sub>1</sub>−1<span style="font-size: medium;">/</span>r<sub>2</sub></b></i>)<i><b> = W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i><br /><br />Electric potential for each point of an electric field fully defines this field. If we know the electric potential in each point of a field, we don't have to know what kind of an object is the source of the field, nor its charge, nor shape.<br /><br />To find the amount of work needed to move a charge <i><b>q</b></i> from a point in the electric field with a potential <i><b>V<sub>1</sub></b></i> to a point with potential <i><b>V<sub>2</sub></b></i> we use the formula <i><b>W = q·(V<sub>2</sub>−V<sub>1</sub>)</b></i></d></d>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-86061614071845400962020-01-13T12:33:00.001-08:002020-01-13T12:33:34.873-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Intensity<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/LnU8ue0NIXw" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Field Intensity</u><br /><br /><b>Electric field intensity</b> is the force (based on the <b>Coulomb's Law</b>) of an electric field of some electrically charged object <i><b>A</b></i>, exhorted on a probe point-object <i><b>B</b></i>, charged with one coulomb of positive electricity (+1C), positioned at some point in the electric field around a main object <i><b>A</b></i>.<br /><br />In other words, it's the force experienced by a probe point-object, charged with +1C of electricity, positioned at some point in space around a main electrically charged object <i><b>A</b></i>.<br /><br />This is a <b>measure of the intensity of the electric field</b> of some charged object at a specific point in space. So, it's a function of two parameters: the electric charge in the main object <i><b>A</b></i> and a position in space relatively to this object.<br /><br />In many cases the word "intensity" is replaced with a word "strength" or just dropped from the conversation. So, terms <b>electric field intensity</b>, <b>electric field strength</b> or in some cases simply <b>electric field</b> are synonymous.<br /><br />First of all, <b>electric field intensity</b> is a force and, therefore, a <b>vector</b>. Since electrically charged objects attract or repel each other, depending on the type of their charges (positive with deficiency of electrons or negative with excess of electrons), this force is directed along the line connecting a main object, whose electric field intensity we measure, and a probe point-object, charged with +1C of electricity, positioned somewhere in space around the main object.<br /><br />The magnitude of the vector of electric field intensity can be calculated based on the Coulomb's Law.<br />The general form of the Coulomb's Law, when two electrically charged point objects, <b><i>A</i></b> and <b><i>B</i></b>, are involved, is<br /><i><b>E = k·q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i><br />where<br /><i><b>E</b></i> is the magnitude of the force of attraction (in case of opposite charges) or repelling (in case of the same type of charge, positive or negative) in <i>N - newtons</i><br /><i><b>q<sub>A</sub></b></i> is electric charge of point-object A in <i>C - coulombs</i><br /><i><b>q<sub>B</sub></b></i> is electric charge of point-object B in <i>C - coulombs</i><br /><i><b>R</b></i> is the distance between charged objects in <i>m - meters</i><br /><i><b>k</b></i> is a coefficient of proportionality, equals to 9.0·10<sup>9</sup> in <i>N·m²/C²</i><br /><br />Since the electric charge of a probe point-object <b><i>B</i></b>, that we use to measure the intensity of an electric field of some charged point-object <b><i>A</i></b>, is <i><b>q<sub>B</sub>=+1C</b></i>, the magnitude of the electric field intensity of point-object <b><i>A</i></b> with electric charge <i><b>q<sub>A</sub></b></i> at a distance <i><b>R</b></i> from it is<br /><i><b>E = k·q<sub>A</sub> <span style="font-size: medium;">/</span> R²</b></i><br /><br />The direction of the vector of this force of electric field intensity is along the line connecting the main and the probe point-objects towards the main object, if its charge is negative and attracting a positively charged probe object, or away from the main object, if its charge is positive and repelling a positively charged probe object.<br /><br />If two or more main point-objects charged with electricity are positioned in some configuration, a probe point-object will experience some force from each of them. All these forces will be combined, according to the rules of vector addition, and the resulting force is the <b>intensity</b> of the combined electric field of all main point-objects.<br /><br /><i>Example 1</i><br />Consider a pair of point-objects at points A and B, charged with positive charge <i>+q</i> each, at distance <i>2d</i> from each other. What would be an <i>intensity E(x)</i> of their combined electric field at a point <i>P</i> on a perpendicular bisector of a segment <i>AB</i> at distance <i>x</i> from a midpoint <i>M</i> of this segment?<br /><br />Magnitude of the intensity of electric field from each object is<br /><i><b>E<sub>A</sub> = E<sub>B</sub> = k·q/(d²+x²)</b></i><br />One of them is directed from <i>A</i> to <i>P</i>, another - from <i>B</i> to <i>P</i>. They are at angle to each other, so we need the rule of parallelogram to find a resulting force. Let ∠<i>APM</i> be <i>φ</i>.<br /><i><b>tan(φ) = d/x</b></i><br />Representing each field intensity vector as the sum of two vectors - one along the line <i>MP</i> and another along the line <i>AB</i>, and taking into consideration only the components along <i>MP</i> (the other two nullify each other), we can calculate the resulting intensity<br /><i><b>E = E<sub>A</sub>·cos(φ) + E<sub>B</sub>·cos(φ) =<br />= 2k·q·cos(φ) <span style="font-size: medium;">/</span> (d²+x²) =<br />= 2k·q·x <span style="font-size: medium;">/</span> </b></i>[<i><b>(d²+x²)<sup>3/2</sup></b></i>]<br /><br /><i>Example 2</i><br />Consider an infinitely thin rod of a length <i>2d</i> charged with electricity such that the density of electrical charge (amount of charge per unit of length) equals to <i>λ</i>.<br />Our task is to determine the electrical field intensity at any point outside this rod.<br /><br />Let's establish the frame of reference with the origin of coordinates at the midpoint of our rod and the X-axis along a rod. So, the rod is positioned on the X-axis from <i>x=−d</i> to <i>x=+d</i>.<br /><br />From the consideration of symmetry it is obvious that the two main parameters of the position in space are important: how far a point is from the X-axis and how far the projection of the point on the X-axis is from the center of the rod. This allows us to establish XY-plane as going through the rod and a point in space where the electric field intensity is supposed to be determined and ignore the Z-axis, so the position of the rod and a point, where intensity is to be established, can be represented on XY-plane as below.<br /><img src="http://www.unizor.com/Pictures/ElectricalIntensityRod.png" style="height: 200px; width: 200px;" /><br />Point <i>P(a,b)</i> is the one, where the field intensity is to be established. Y-coordinate <i>y=b</i> represents the distance from point <i>P</i> to the rod along a perpendicular to the rod and X-coordinate represents the distance from the projection of point P on the X-axis to the midpoint of the rod.<br /><br />The probe charge of <i>+1C</i> is at point <i>P(a,b)</i> and the field intensity at this point is the sum of all forces exhorted by the pieces of rod onto this probe charge.<br /><br />Consider infinitesimal piece of the rod of the length <i>dx</i> positioned at X-coordinate <i>x</i>. The plan is to determine the force it exhort onto the probe charge of <i>+1C</i> at point <i>P(a,b)</i> and integrate the result from <i>x=−d</i> to <i>x=d</i>.<br /><br />The electric charge of the piece of the rod of the length <i>dx</i> is <i><b>λ·</b>d<b>x</b></i>, where <i><b>λ</b></i> is the given density of electric charge in the rod.<br />The square of the distance <i><b>r</b></i> from this piece of the rod to point <i>P(a,b)</i> is<br /><i><b>r² = (a−x)² + b²</b></i><br /><br />Now we can apply the <b>Coulomb's Law</b> to determine the infinitesimal electric force <i>d<b>F</b></i> between this piece of the rod of the length <i>d<b>x</b></i> and a probe charge of <i>+1C</i> at point <i>P(a,b)</i><br /><i>d<b>E = k·λ·</b>d<b>x <span style="font-size: medium;">/</span> </b></i>[<i><b>(a−x)²+b²</b></i>]<br />The above is the magnitude of the electrical force. Its direction is along the line connecting a piece of the rod with point <i>P(a,b)</i>.<br /><br />We cannot integrate this expression directly since the forces from different pieces of the rod have different direction. We have to represent this force as a sum of two forces - horizontal force <i>F<sub>x</sub></i> along the X-axis and vertical force <i>F<sub>y</sub></i> along the Y-axis. Then we can separately integrate each component to get two components of the final force.<br /><br />Simple math gives us the following expressions for component of the force <i>F</i><br /><i>d<b>E<sub>x</sub> = </b>d<b>E·(a−x) <span style="font-size: medium;">/</span> r =<br />= k·λ·</b>d<b>x·(a−x) <span style="font-size: medium;">/</span> </b></i>[<i><b>(a−x)²+b²</b></i>]<i><b><sup>3/2</sup></b></i><br /><i>d<b>E<sub>y</sub> = </b>d<b>E·b <span style="font-size: medium;">/</span> r =<br />= k·λ·</b>d<b>x·b <span style="font-size: medium;">/</span> </b></i>[<i><b>(a−x)²+b²</b></i>]<i><b><sup>3/2</sup></b></i><br /><br />Now you see how important is Mathematics to succeed in Physics!<br /><br />Let's integrate each force, horizontal and vertical, on <i>x</i>∈[<i>−d,d</i>] interval.<br /><br />First, let's find indefinite integral for <i><b>E<sub>x</sub></b></i><br /><i><b><span style="font-size: large;">∫</span>k·λ·</b>d<b>x·(a−x)<span style="font-size: medium;">/</span></b></i>[<i><b>(a−x)²+b²</b></i>]<i><b><sup>3/2</sup> =</b></i><br />...substitute <i>y=(a−x)²+b²</i><br />= <i><b>−0.5·k·λ·<span style="font-size: large;">∫</span>y<sup>−3/2</sup></b>d<b>y =<br />= k·λ·y<sup>−1/2</sup></b></i><br />Definite integral for <i>y</i> should be taken in limits from <i>(a+d)²+b²</i> to <i>(a−d)²+b²</i>, which results in the following expression for <i><b>E<sub>x</sub></b></i>: <i><b>E<sub>x</sub> = k·λ·</b></i>{[<i><b>(a−d)²+b²</b></i>]<i><b><sup>−1/2</sup>−</b></i>[<i><b>(a+d)²+b²</b></i>]<i><b><sup>−1/2</sup></b></i>}<br /><br />Incidentally,<br /><i>r<sub>right</sub></i> = [<i>(a−d)²+b²</i>]<i><sup>1/2</sup></i><br />is the distance from point <i>P(a,b)</i> to the right end of the rod and<br /><i>r<sub>left</sub></i> = [<i>(a+d)²+b²</i>]<i><sup>1/2</sup></i><br />is the distance to the left end.<br />So, the formula for horizontal component of the resulting field force is<br /><i><b>E<sub>x</sub> = k·λ·</b></i>[<i><b>1/r<sub>right</sub> − 1/r<sub>left</sub></b></i>]<i><b></b></i><br />Interestingly, as the length of the rod increases to infinity, the horizontal component of the field strength diminishes to zero, as both <i>1/r<sub>right</sub></i> and <i>1/r<sub>left</sub></i> diminish to zero. The obvious reason is that with an infinitely long rod the horizontal forces directed to the left are balanced by horizontal forces directed to the right.<br /><br />Now let's address the vertical component of the electric field intensity <i><b>E<sub>y</sub></b></i>.<br />First, let's calculate the indefinite integral<br /><i><b><span style="font-size: large;">∫</span>k·λ·</b>d<b>x·b <span style="font-size: medium;">/</span> </b></i>[<i><b>(a−x)²+b²</b></i>]<i><b><sup>3/2</sup> =</b></i><br />...substitute <i>(a−x)/b=tan(y)</i><br />...<i>tan(y)=(a−x)/b</i><br />...<i>(a−x)²+b² = b²·</i>[<i>tan²(y)+1</i>]<i> =<br />...= b²/cos²(y)</i><br />...[<i>(a−x)²+b²</i>]<i><sup>−3/2</sup> = b<sup>−3</sup>·cos<sup>3</sup>(y)</i><br />...<i>dx=−b/cos²(y)·dy</i><br /><i><b>= −(k·λ/b)·<span style="font-size: large;">∫</span>cos(y)·</b>d<b>y =<br />= (k·λ/b)·sin(y) + C</b></i><br /><br />As far as limits of integration, if <i>x</i>∈[<i>−d,d</i>] then <i>y</i>∈[<i>y<sub>1</sub>,y<sub>2</sub></i>], where<br /><i>y<sub>1</sub> = arctan((a+d)/b)</i> and<br /><i>y<sub>2</sub> = arctan((a−d)/b)</i><br />To find the definite integral in the limits above, we will use a trigonometric identity<br /><i>sin(arctan(z)) = z <span style="font-size: medium;">/</span> √<span style="text-decoration-line: overline;">(1+z²)</span></i><br /><br />Now we can express the vertical component of the field intensity as<br /><i><b>E<sub>y</sub> = (k·λ/b)·</b></i>[<i><b>sin(y<sub>2</sub>)−sin(y<sub>1</sub>)</b></i>]<br />where <i>y<sub>1</sub></i> and <i>y<sub>2</sub></i> are defined above.<br /><i><b>E<sub>y</sub> = (k·λ/b)·</b></i>{<i><b>(a−d)·</b></i>[<i><b>(a−d)²+b²</b></i>]<i><b><sup>−1/2</sup> − (a+d)·</b></i>[<i><b>(a+d)²+b²</b></i>]<i><b><sup>−1/2</sup></b></i>}<br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-57219708546048174482020-01-10T19:21:00.001-08:002020-01-10T19:21:34.140-08:00Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 1<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/-uwRDvJDihg" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems 1</u><br /><br /><i>Problem A</i><br />At three vertices <b><i>P</i></b>, <b><i>Q</i></b> and <b><i>R</i></b> of an equilateral triangle with the length of a side <i><b>d</b></i> there are three electrical charges <b><i>q<sub>P</sub></i></b>, <b><i>q<sub>Q</sub></i></b> and <b><i>q<sub>R</sub></i></b>. All charges are equal in magnitude to <i><b>q</b></i>, but the first two are positive, while the third is negative:<br /><b><i>q<sub>P</sub> = q<sub>Q</sub> = −q<sub>R</sub> = q</i></b><br />Determine the magnitude and direction of the combined electrical attraction force acting on charge <i><b>q<sub>R</sub></b></i> from <i><b>q<sub>P</sub></b></i> and <i><b>q<sub>Q</sub></b></i>.<br /><br /><i>Solution</i><br /><i><b>F<sub>P</sub> = −k·q²<span style="font-size: medium;">/</span>d²</b></i> (<i>from q<sub>P</sub> on q<sub>R</sub></i>)<br /><i><b>F<sub>Q</sub> = −k·q²<span style="font-size: medium;">/</span>d²</b></i> (<i>from q<sub>Q</sub> on q<sub>R</sub></i>)<br />These forces are at angle π/3 to each other. Adding them as vectors. The direction of a combined force is from point <i><b>R</b></i> to a midpoint between <i><b>P</b></i> and <i><b>Q</b></i>.<br />Its magnitude is<br /><i><b>F<sub>P+Q</sub> = −k·q²·√<span style="text-decoration-line: overline;">3</span><span style="font-size: medium;">/</span>d²</b></i><br /><br /><i>Problem B</i><br />Two point-objects of mass <i><b>m</b></i> each are hanging at the same level above the ground on two parallel vertical threads in the gravitational field with a free fall acceleration <i><b>g</b></i>.<br />When they are charged with the same amount of electricity <i><b>q</b></i>, they move away from each other to a distance <i><b>d</b></i> from each other, and each thread will make some angle with a vertical.<br />Determine this angle as a function of known parameters.<br /><br /><i>Solution</i><br /><i><b>F<sub>e</sub></b></i> is an electrical repelling force between objects.<br /><i><b>T</b></i> is a thread tension.<br /><i><b>φ</b></i> is an angle that each thread makes with a vertical.<br /><i><b>F<sub>e</sub> = k·q²/d²</b></i> (<i>Coulomb's Law</i>)<br /><i><b>T·cos(φ) = m·g</b></i> (<i>= weight</i>)<br /><i><b>T·sin(φ) = F<sub>e</sub></b></i> (<i>= repelling)</i><br /><i><b>tan(φ) = F<sub>e</sub> <span style="font-size: medium;">/</span>(m·g)</b></i><br /><i><b>tan(φ) = k·q² <span style="font-size: medium;">/</span>(m·g·d²)</b></i><br /><br /><i>Problem C</i><br />We assume the Bohr's classical model of an atom.<br />The atom of hydrogen has 1 proton and 1 electron rotating around it on a distance <i><b>R</b></i>.<br />The electric charge of a proton is positive <i><b>+q</b></i>, for an electron it is <i><b>−q</b></i>.<br />The mass of an electron is <i><b>m</b></i>.<br />What is the angular speed and frequency of rotation of electron?<br /><br /><i>Solution</i><br /><i><b>F<sub>e</sub></b></i> is an electrical attracting force of proton to electron that keeps electron on its orbit.<br /><i><b>F<sub>e</sub> = k·q²/R²</b></i> (<i>Coulomb's Law</i>)<br /><i><b>ω</b></i> is angular speed of an electron.<br /><i><b>m·R·ω² = F<sub>e</sub></b></i> (<i>Newton's Law</i>)<br /><i><b>ω = √<span style="text-decoration-line: overline;">k·q²/(R³·m)</span></b></i><br /><i><b>ν = ω/(2π)</b></i><br />Let's substitute real data (all numbers are in SI units):<br /><i><b>k = 9.0·10<sup>9</sup></b> N·m²/C²</i><br /><i><b>q = 1.602·10<sup>−19</sup></b> C</i><br /><i><b>m = 9.109·10<sup>−31</sup></b> kg</i><br /><i><b>R = 25·10<sup>−12</sup></b> m</i> (empirical)<br /><i><b>ω ≅ 1.274·10<sup>17</sup></b> rad/sec</i><br /><i><b>ν ≅ 2.028·10<sup>16</sup></b> rev/sec</i><br /><br /><i>Problem D</i><br />A proton of mass <nobr><i><b>m=1.673·10<sup>−27</sup></b>kg</i></nobr> and charge of <nobr><i><b>q=1.602·10<sup>−19</sup></b>C</i></nobr> is launched from a distance <nobr><i><b>a=0.5</b>m</i></nobr> towards a nucleus of Uranium that has <i><b>N=92</b></i> protons.<br />What should be the minimal initial speed <i><b>v</b></i> of bombarding proton to overcome the repelling force of all protons inside the nucleus of Uranium and get closer to it on a distance <i><b>b=0.001</b>m</i> from a nucleus?<br /><br /><i>Solution</i><br />Initial kinetic energy of a bombarding proton <i><b>E=m·v²/2</b></i> should be equal to a work <i><b>A</b></i> of electric repelling force between a bombarding proton and the protons inside a nucleus. This work is done by a variable repelling force that depends on a distance between bombarding proton and a nucleus.<br />Since the force depends on the distance, we have to integrate the work done by this force along the distance covered by a bombarding proton.<br />Let <i><b>x</b></i> be a variable distance of a proton to a nucleus.<br /><i><b>F(x) = k·q·N·q/x²=k·N·q²/x²</b></i><br />Then the work done by this force on a segment from <i><b>x</b></i> to <i><b>x+</b>d<b>x</b></i> is<br /><i>d<b>A = k·N·q²·</b>d<b>x/x²</b></i><br />The total amount of work done by electrical repelling force is<br /><span style="font-size: large;">∫</span><sub>[a,b]</sub><i><b>k·N·q²·</b>d<b>x/x² =<br />= k·N·q²·</b></i>[<i><b>(1/b)−(1/a)</b></i>]<br />This is supposed to be equal to kinetic energy of a bombarding proton at the beginning of its motion<br /><i><b>m·v²/2 = k·N·q²·</b></i>[<i><b>(1/b)−(1/a)</b></i>]<br />This gives the value of speed <i><b>v</b></i><br /><i><b>v = √<span style="text-decoration-line: overline;">2·k·N·q²·[(1/b)−(1/a)]/m</span></b></i><br />Substituting the values listed above:<br /><i><b>v ≅ 68.64</b> m/sec</i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-42142558182095885752020-01-07T11:55:00.001-08:002020-01-07T11:55:50.215-08:00Unizor - Physics4Teens - Electromagnetism - Electrical Field - Coulomb's...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/pmrbIy8Eick" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Coulomb's Law</u><br /><br />As we know, excess of electrons above the number of protons in an object is what we call <i>negative charge</i>. Deficiency of electrons is called <i>positive charge</i>.<br />We also know that two positively charged objects or two negatively charged objects repel each other, if positioned close to each other, while oppositely charged (one positive and another negative) attract each other.<br /><br />It means that there are some forces around electrically charged objects that act in space around them. That is precisely what the term <b>force field</b> means. So, there is an <i>electrical force field</i> that surrounds each electrically charged object.<br /><br />The next task is to determine the strength of the electrical forces in the <i>electrical field</i>.<br />Intuitively, the force of attraction or repelling between two electrically charged objects must depend on the number of excessive or deficient electrons in each. The most obvious hypothesis is that the force must be proportional to the number of excess or deficient electrons in each object.<br /><br />Just as a thought experiment, imagine two point objects <i><b>A</b></i> and <i><b>B</b></i> with one excess electron in each <i><b>e1<sub>A</sub></b></i> and <i><b>e1<sub>B</sub></b></i>. The object <i><b>A</b></i> will repel object <i><b>B</b></i> with some strength <i><b>F</b></i>. If the number of excess electrons in object <i><b>A</b></i> is increased to two (<i><b>e1<sub>A</sub></b></i> and <i><b>e2<sub>A</sub></b></i>), the forces of repelling <i><b>B</b></i> must be added: one part from <i><b>e1<sub>A</sub></b></i> to <i><b>e1<sub>B</sub></b></i> and another from <i><b>e2<sub>A</sub></b></i> to <i><b>e1<sub>B</sub></b></i>. So, the repelling force acting on <i><b>B</b></i> will be doubled.<br /><br />Similar arguments used <i><b>M</b></i> times for <i><b>M</b></i> excess electrons in object <i><b>A</b></i> will lead to multiplication of the initial force by <i><b>M</b></i>. If we increase the number of excess electrons in object <i><b>B</b></i> to <i><b>N</b></i>, we will have to multiply our force by <i><b>N</b></i>.<br />As a result, the total force will be proportional to a product <i><b>M·N</b></i>.<br /><br />We know that the unit of electric charge <i>coulomb</i> is proportional to a charge of one electron. More precisely, the charge of 1 electron is 1.602176634·10<sup>−19</sup>C.<br />Therefore, if objects <i><b>A</b></i> and <i><b>B</b></i> have electric charge <i><b>q<sub>A</sub></b></i> and <i><b>q<sub>B</sub></b></i> in <i>coulombs</i>, the force of attraction or repelling between them is proportional to <i><b>q<sub>A</sub>·q<sub>B</sub></b></i>.<br /><br />The next variable that should be taken into consideration when examining the force of electric field is the distance between objects. The logic we will use to analyze the dependency of the force on distance is similar to the one we used in case of gravitational field.<br /><br />Consider a set of tiny springs attached to each electron of the charged object <i><b>A</b></i>. Each such spring represents a force developed by one electron.<br />Obviously, the greater the distance between a probe object <i><b>B</b></i> and object <i><b>A</b></i> - the smaller is the density of springs in the space. It is intuitively obvious that the force of attraction or repelling acting on object <i><b>B</b></i> is proportional to a density of springs in the area where <i><b>B</b></i> is located, the less springs are observed where <i><b>B</b></i> is - the smaller the force will be and vice versa.<br /><br />In turn, the density of springs is inversely proportional to a square of a distance from object <i><b>A</b></i> because the area of a sphere equals to 4πR<sup>2</sup>, where R is a radius.<br /><br />Therefore, the force of an electrical field at a distance <i><b>R</b></i> from its source (a charged object) is inversely proportional to a square of <i><b>R</b></i>.<br /><br />Summarizing all the above, we suspect that the force of attraction or repelling between two charged objects with charges <i><b>q<sub>A</sub></b></i> and <i><b>q<sub>B</sub></b></i> at a distance <i><b>R</b></i> from each other should be proportional to<br /><i><b>q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i><br /><br />Experimentally, this was confirmed and the only detail we need is to adjust the units of measurement, so the resulting force will be in the units we usually use.<br />In the SI system, where the force is expressed in <i>newtons</i>, electric charge in <i>coulombs</i> and the distance in <i>meters</i> the force of attraction or repelling is<br /><i><b>F = k·q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i><br />where<br /><i><b>F</b></i> is the magnitude of the force of attraction (in case of opposite charges) or repelling (in case of the same type of charge, positive or negative) in <i>N - newtons</i><br /><i><b>q<sub>A</sub></b></i> is electric charge of object A in <i>C - coulombs</i><br /><i><b>q<sub>B</sub></b></i> is electric charge of object B in <i>C - coulombs</i><br /><i><b>R</b></i> is the distance between charged objects in <i>m - meters</i><br /><i><b>k</b></i> is a coefficient of proportionality, equals to 9.0·10<sup>9</sup> in <i>N·m²/C²</i><br /><br />The above is the <b>Coulomb's Law</b>, discovered by French physicist Charles-Augustin de Coulomb in 1785.<br /><br />The <b>Coulomb's Law</b> describes the force between two charged objects. If both have the same "sign", both are positively charged with deficiency of electrons or both are negatively charged with excess of electrons, the sign of the force is <b>positive</b>, it's a <b>repulsive</b> force. If the charges are of opposite "sign", one positive with deficiency of electrons and another negative with excess of electrons, the force is <b>negative</b>, it's <b>attractive</b> force.<br />The word "sign" we took in quotes because it's just an artificial way of designating different types of charges that physicists use for convenience.<br /><br />As we see, the <b>Coulomb's Law</b> for electric field looks very similar to the <b>Newton's Law</b> for gravitational field.<br />The fundamental difference between these two fields is that gravity always attracts, while electrically charged objects can attract or repel each other, depending on what kind of electrical charge they have. This is the asymmetry of gravitation and a strong argument to consider gravitational field as something fundamentally different from electrical field. Indeed, the General Theory of Relativity by Einstein suggests that the gravitational field is the result of curvature of the space we live in.<br /><br />Another purely quantitative difference between these two fields is the magnitude of the force.<br />Consider, hypothetically, two electrons at a distance of one millimeter from each other. They are repelling each other because of electric force, that depends on their charge, and attract each other because of gravity, that depends on their masses.<br />Let's compare these two forces using the <b>Coulomb's Law</b> and the <b>Newton's Law</b>, using standard SI units.<br />Electric charge of an electron is 1.602176634·10<sup>−19</sup> C.<br />Mass of an electron is 9.1093837015·10<sup>−31</sup> kg.<br /><br /><u><i>Electrical force</i><br />(repelling)</u><br /><i><b>F<sub>e</sub> = k·q<sub>A</sub>·q<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i>, where<br /><i>k=9.0·10<sup>9</sup></i><br /><i>q<sub>A</sub>=q<sub>B</sub>=1.602176634·10<sup>−19</sup></i><br /><i>R=0.001</i><br />Resulting electrical repelling force is<br /><i><b>F<sub>e</sub> ≅ 2.31·10<sup>−22</sup> N</b></i><br /><br /><u><i>Gravitational force</i><br />(attracting)</u><br /><i><b>F<sub>g</sub> = G·m<sub>A</sub>·m<sub>B</sub> <span style="font-size: medium;">/</span> R²</b></i>, where<br /><i>G=6.674·10<sup>−11</sup></i><br /><i>m<sub>A</sub>=m<sub>B</sub>=9.1093837015·10<sup>−31</sup></i><br /><i>R=0.001</i><br />Resulting gravitational attracting force is<br /><i><b>F<sub>g</sub> ≅ 5.53·10<sup>−65</sup> N</b></i><br /><br />As you see, the difference is huge. Electrical force between two electrons is significantly stronger that the gravitational force. On a subatomic level the gravitational forces can be ignored. On a planetary level the electrical charges are often small, planets are, generally speaking, electrically neutral or very close to neutral, so the gravitational forces play the major role.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-25747260753784142862019-12-26T11:51:00.001-08:002019-12-26T11:51:02.080-08:00Unizor - Physics4Teens - Electromagnetism - Unit of Electric Charge<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/F0i5KN74bvA" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Electric Charge Unit - Coulomb</u><br /><br />Since electrons are the carriers of electricity, we can measure the amount of <i>negative</i> electricity in any object as the number of excess electrons in it, that is the number of electrons that do not have a proton to pair with.<br /><br />In case of deficiency of electrons, we can measure the amount of <i>positive</i> electricity in any object as the number of excess protons, that is the number of protons that do not have an electron to pair with.<br /><br />The problem with this way of measuring the amount of electricity is that this unit (amount of electricity in one electron or one proton) is very small and inconvenient for practical matters.<br />For this reason physicists use a larger unit of electric charge - <b>coulomb</b>, abbreviated as <b>C</b>.<br />In this unit of measurement (and this is the contemporary definition of this unit) the electric charge of an electron, used as negative value, or of a proton, used as positive value, is<br /><i><b>e=1.602176634·10<sup>−19</sup>C</b></i><br /><br />Therefore, <b>1 Coulomb</b> is approximately equal to the amount of electricity in <nobr><b>1 <span style="font-size: medium;">/</span> (1.602176634·10<sup>−19</sup>)</b></nobr> electrons.<br />The word "approximately" is used because the above quantity must be an integer (since we are talking about the <u>number</u> of electrons), which it is not. So, the true definition of a <b>coulomb</b> is as stated above.<br /><br />Historically, <b>coulomb</b> was defined differently. When electrons are moving along some conductive material, they form an <i>electric current</i>, which physicists have measured in <i>amperes</i> - units of electric current, which they have defined separately, using electromagnetic properties of a current. Knowing the <i>electric current</i>, they have defined 1 coulomb of electric charge as an amount of electricity, transferred by an <i>electric current</i> of 1 ampere during 1 second.<br />This different, more complicated approach to define <b>coulomb</b> was recently changed to a simpler one described above.<br /><br />To have an understanding of the amount of electric charge of <nobr><i>1 coulomb</i></nobr>, let's get back to a previous lecture that stated that the total number of electrons in one cubic centimeter of iron is 2.2·10<sup>24</sup>. Reducing the size to one cubic millimeter, it makes the number of electrons in it equal to 2.2·10<sup>21</sup>.<br />If each electron has a charge of 1.602176634·10<sup>−19</sup>C, as is defined above, we can calculate the total electric charge of all electrons in a cubic millimeter of iron:<br /><i><b>2.2·1.6·10<sup>21−19</sup>C ≅ 3.5·10<sup>2</sup></b></i><br /><br />So, in one cubic millimeter of electrically neutral iron we have about 350C of negative electric charge in all its electrons and 350C of positive electric charge in all its protons. Their mutual attraction hold the structure of the atoms inside. There is no excess nor deficiency of electrons, which makes the whole object electrically neutral.<br /><br /><i>Coulomb</i> is a large unit. Getting back to an experiment with two pieces of iron presented in the previous lecture, let's reduce the size of pieces to a cubic millimeter, position them at a distance of 100 meters and magically transform all electrons from one piece to another. Then the charge of one piece will be positive +350C and the charge of another will be negative −350C.<br />Under these conditions the attraction force between them will be about 1.1·10<sup>11</sup>N (<i>newtons</i>), which is huge and is about twice the weight of a great pyramid of Giza.<br /><br />As another example, the amount of electricity passing through a regular incandescent lamp of 60W is about 0.5C per second.<br /><br />We present these examples without any proof, just for demonstration of the concept of <i>amount of electricity</i>. In the future lectures we will learn the laws of electricity and all these calculations will be presented in details.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-87529918761306806652019-12-23T19:44:00.001-08:002019-12-23T19:44:07.527-08:00Unizor - Physics4Teens - Electromagnetism - Carrier of Electricity<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/o1BWXzln1mQ" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Carrier of Electricity</u><br /><br /><i>Electricity</i> is a form of energy.<br /><i>Mechanical</i> energy is carried by moving objects. <i>Thermal</i> energy is carried by oscillating molecules. <i>Electricity</i>, as a form of energy, must have its carrier too.<br /><br />As we know, any object consists of molecules that retain the object's properties. There are thousands of different molecules, corresponding to thousands of different in their characteristics objects.<br /><br />Molecules are comprised from atoms, which have completely different characteristics. There are just a little more than a hundred different atoms, from their compositions all molecules are built.<br /><br />Each atom, according to a classical model that we will use when talking about electricity, consists of a nucleus, where certain number of protons and neutrons are bundled together, and electrons that rotate on different orbits around a nucleus.<br /><br />What force keeps electrons on their orbits?<br />Similarly to gravitational force that keeps planets on their orbits around a sun, there is a force between a nucleus and electrons circling on their orbits.<br />This force is called <i>electric force</i>. As in case of gravity, we can talk about <i>electric field</i> as a domain of space, where electric forces are acting.<br /><br />Numerous experiments show that the attracting <i>electric force</i> exists between protons and electrons. Neutrons do not play any role in keeping electrons on their orbits. Further experiments showed that, while protons and electrons are attracted to each other, two protons repulse each other, and so are two electrons. That prompted the designation of terms <i>positive</i> and <i>negative</i> to describe the electric energy carried, correspondingly, by protons and electrons and the term <i>electric charge</i> to name and quantify this form of energy.<br /><br />So, we say that <b>protons</b> carry <i>positive electric charge</i>, while <b>electrons</b> carry <i>negative electric charge</i>. Positive and negative electric charges attract and neutralize each other, while two positive or two negative charges repulse each other.<br /><br />In a normal state atoms have equal number of protons and electrons, thus are <i>electrically neutral</i>, which implies that the amount of electric charge in one proton and one electron are equal, though opposite in "sign". Obviously, our designation of "positive" and "negative" is just for a convenience of description and ease of calculations.<br /><br />Under some circumstances the number of electrons in an object might be greater or less than the number of protons, which causes the entire object to be <i>negatively</i> (when there is an excess of electrons) or <i>positively</i> (when there is a deficiency of electrons) charged.<br /><br />In case of excess of electrons, some electrons are no longer attached to a nucleus, but travel freely inside the object. Because of the excess of electrons, the entire object becomes <i>negatively charged</i>.<br />When there is a deficiency in number of electrons, certain spots on certain orbits around certain nuclei remain empty. The entire object becomes <i>positively charged</i>.<br />If a negatively charged object A comes to contact with a positively charged object B, excess electrons from A might migrate to B, thus diminishing negative charge of A by diminishing excess of electrons and diminishing positive charge of B by filling empty spaces on the orbits of those atoms that had deficiency of electrons.<br /><br />Electric force is really strong. As an example, let us consider two small cubes of iron of 1 cm³ each positioned at the distance of 1 meter from each other.<br />There are 8.4·10<sup>22</sup> atoms in each cube. Each atom contains 26 protons, 30 neutrons and 26 electrons.<br />If we magically strip all atoms in one cube of iron of all their electrons and transfer them to another cube, the one with deficiency of electrons will be positively charged and the one with excess electrons will be negatively charged. So, they will attract each other.<br />The total number of excess electrons in one cube, which is equal to the total number of deficient electrons in another one is<br /><i><b>26·8.4·10<sup>22</sup> ≅ 2.2·10<sup>24</sup></b></i><br />The distance between them is 1 meter. According to Coulomb Law, that we will study later, the force of attraction between these two small cubes of iron will be, approximately<br /><i><b>F = 1.11·10<sup>21</sup> N (newtons)</b></i><br />THIS IS HUGE!<br />Just as a comparison, the force of gravity that keeps the planet Mars on its orbit around the Sun is equal to <i><b>1.6·10<sup>21</sup> N</b></i>, which is quite comparable.<br />It also shows how much stronger <i>electrical forces</i> are, compared to <i>gravitational</i>.<br /><br />Fortunately, we never have to deal with such strong forces, all atoms that comprise an object are never stripped of all their electrons.<br /><br />It's easy to see electricity in action.<br />Lightning is the flow of electrons from the negatively charged cloud to some object on the ground.<br />Getting off some synthetic clothes is accompanied by small visible in the dark and audible sparks, which are also the flows of electrons from an object with excess of electrons to an object with their deficiency.<br /><br />An <i>electroscope</i> is a device that can demonstrate the excess or deficiency of electrons, that is whether an object is electrically charged.<br /><img src="http://www.unizor.com/Pictures/Electroscope.png" style="height: 150px; width: 200px;" /><br />Initially, the electroscope is electrically neutral and aluminum foils are in vertical position, as there is no electrical force between them.<br />If electrically charged object touched the metal disc of an initially neutral electroscope, excess or deficiency of electrons in an object will be shared with metal disc and, further, with aluminum foils. This will negatively or positively charge both aluminum foils, and they will split apart because the same electrical charge (positive or negative) repulse objects (aluminum foils in this case).<br />With proper calibration this effect can be used to measure the amount of electricity by the angle of deviation of aluminum foils from vertical, as more electrons are in excess or deficiency - the stronger the repulsive force will be between the foils, and the greater angle of deviation of their position from the vertical will be.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-36938997835457674002019-12-03T10:16:00.001-08:002019-12-03T10:16:20.607-08:00Unizor - Physics4Teens - Energy - Light as Quants<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/pYSCh1p3Zh0" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Quants of Light</u><br /><br />By the end of 19th century the <i>electrons</i>, as carriers of electricity, were discovered by Sir Joseph Thomson in 1897, and many scientists experimented with electricity.<br /><br />At that time the <i>wave theory of light</i> was dominant. It could explain most of experimental facts and was shared by most physicists.<br />However, there were some new interesting experimental facts that could not be easily explained in the framework of the <b>classical</b> <i>wave theory of light</i> as oscillation of electromagnetic field.<br /><br />The <i>photoelectric effect</i> was one of such experimental facts that physicists could not fit into classical wave theory.<br /><br />Consider a simple electrical experiment when two poles, positive and negative, positioned close to each other, are gradually charged. After a charge reaches certain value, a spark between these poles causes the discharge of electricity.<br /><br />The electric charge was attributed to electrons with a negatively charged pole having more electrons than in an electrically neutral state and a positively charged one having less electrons than in an electrically neutral state.<br />The electric spark was the flow of excess electrons from a negative pole to a positive one, thus bringing them both to an electrically neutral state.<br /><br />It was observed that, if the negative pole is lit by light, the discharge occurs earlier, with less amount of charge accumulated in the poles.<br />This was so-called <i>photoelectric effect</i>.<br /><br />The experimental characteristics of the <i>photoelectric effect</i> were:<br />(a) if <i>photoelectric effect</i> is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;<br />(b) the speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;<br />(c) for each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.<br /><br />The explanation coming to mind within a framework of the <i>wave theory of light</i> would be as follows.<br />Light is oscillations of the electromagnetic field. Electrons, accumulated during the charging process, are vibrating more intensely as a result of the oscillations of the electromagnetic field of the light, whose energy is transformed to electrons, so photo-electrons leave the negative pole easier, thus facilitating an earlier discharge.<br /><br />This would be a great explanation if not for a couple of contradictory facts.<br />The first contradiction is the property (b) of the <i>photoelectric effect</i>. According to the classical wave theory, the speed of photo-electrons must be dependent on the intensity of the light (amplitude of electromagnetic waves), which was not observed. And the (c) property is also unexplainable by classical wave theory, because, again, within a framework of the classical wave theory for any frequency we can find an intensity of light sufficient to "knock" out the electrons from the negative pole or keep the light of lesser intensity long enough time to transfer to electrons sufficient amount of energy to fly off the surface of the pole, which was not observed.<br /><br />The explanation of these phenomena came with introduction of <i>quants of light</i> - a hypothesis offered by Planck and used by Einstein to explain the properties of photoelectric effect.<br /><br />According to the explanation of photoelectric effect offered by Einstein, light propagates in space not as a continuous stream of waves of electromagnetic oscillations, but in small indivisible packets (<i>quants of light</i> or <i>photons</i>), separated in space and traveling along the same path, thus resurrecting the <i>corpuscular theory</i>, but without rejecting the electromagnetic origin of light.<br /><br />The energy of each <i>photon</i> proportionally depends on the frequency of oscillation of electromagnetic field that carries the light, not on intensity of light, with intensity of light being just a measure of the number of <i>photons</i> passing through a point in space in a unit of time.<br /><br />The energy of light is absorbed by electrons also in these <i>photons</i>. To break away the electron needs certain minimal energy.<br />If the energy of a single photon is sufficient to overcome the atomic forces that keep the electron inside the negative pole, this electron becomes a photo-electron and flies away to a positive pole.<br />If the energy of a photon is less than this minimal amount necessary to overcome the atomic forces, this energy is dissipated as heat, and no photo-electrons are produced.<br /><br />Within the framework of this new <i><b>quantum theory of light</b></i> all the characteristics of the <i>photoelectric effect</i> can find their explanation.<br />Let's analyze them.<br /><br />(a) If <i>photoelectric effect</i> is observed with specific frequency of the light, the number of electrons leaving the negative pole in a unit of time is proportional to intensity of the light;<br /><b>Explanation</b>: each photon has sufficient amount of energy to "knock" out an electron, and intensity is the number of such <i>photons</i> per unit of time.<br /><br />(b) The speed of photo-electrons and, therefore, their kinetic energy do not depend on intensity of light, but on its frequency; higher intensity light produces more electrons, but their speed remains the same, while higher frequency light produces faster photo-electrons;<br /><b>Explanation</b>: speed of electrons and, therefore, their kinetic energy depend on the energy of a <i>photon</i> that "knocked" these electrons out, which, in turn, is proportional to the frequency of light, while intensity of light (the number of <i>photons</i> per unit of time) affects the number of photo-electrons produced by light, not their individual energy.<br /><br />(c) For each material, used as a negative pole, there is minimal frequency of light necessary to initiate the photoelectric effect; high intensity or prolong time exposure to light of a lesser frequency do not produce photoelectric effect.<br /><b>Explanation</b>: the energy, needed by an electron to break away from atomic forces that keep it inside the material, obviously depend on the material; as soon as the light frequency is sufficient for one <i>photon</i> to carry that amount of energy, the photoelectric effect can start; <i>photons</i> of lesser level of frequency cannot "knock" out the electrons from the surrounding material, and the energy of the light is just dissipated as heat.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-47607749201597744192019-11-14T19:10:00.001-08:002019-11-14T19:10:37.857-08:00Unizor - Physics4Teens - Energy of Light - Light as Waves<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/ZPbN9tLUjXY" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Waves</u><br /><br /><i>Corpuscular theory</i> of light, suggested by Newton, while adequately explaining such properties of light as traveling along a straight line and reflection off the mirror, has not been able to explain certain other properties of light observed by experimentators. In particular, <i>interference</i>, <i>diffraction</i> and <i>polarization</i> of light remained unexplainable in the framework of <i>corpuscular theory</i>.<br /><br />Attempts to model the light as oscillation of waves were made by several scientists, but the main formulation and explanation of wave-like properties of light is mostly related to a brilliant English physician and physicist Thomas Young.<br />His famous double-slit experiment, that showed the interference of light passing through two small close to each other slits, convincingly proved that light has wave-like properties.<br /><br />This experiment is extremely simple, anybody can reproduce it at home, its schema is below.<br /><img src="http://www.unizor.com/Pictures/DoubleSlit.jpg" style="height: 200px; width: 200px;" /><br /><br />The light from a single monochromatic source equidistant from two slits goes through these slits and is projected on a screen. The picture on the screen is a sequence of light and dark stripes parallel to the slits. This is a picture of <i>interference</i> that can only be explained within a framework of waves.<br />Two light waves coming through two slits to a surface of a screen in phase (both at its top wave state or both at the bottom) increase each other, making bright stripes. Those that come in opposite phases (one at its top state, while another at the bottom) nullify each other, making dark stripes.<br /><br /><img src="http://www.unizor.com/Pictures/DoubleSlitStripes.png" style="height: 200px; width: 200px;" /><br /><br />Let's interpret the <i>interference</i> of light from the wave theory viewpoint.<br /><br />First of all, if light is a wave, we have to determine the carrier of these waves. Until late 19th century physicists accepted a hypothesis that certain substance called <i>ether</i> (or <i>aether</i>) fills all the space and is a material carrier of waves of light and other waves of electromagnetic oscillations.<br />This hypothesis of <i>ether</i> was later on rejected, but for understanding of wave-like properties of light we can still think about <i>ether</i> as a substance, through which light travels as oscillations of this substance.<br />It might be considered similar to sound spreading inside a metal object, when you hit this metal object with another one. The crystal structure of the metal will start vibrating at the point of impact (the source of the sound) and these vibrations will spread throughout the whole object in all directions in a form of compression waves.<br />Notice that in this model the amplitude of vibrations is a characteristic of the strength of the sound and speed of propagation of sound waves depends on properties of material the object is made of.<br /><br />If light is the waves of a carrier substance, we can talk about this wave's length (distance between two adjacent crests or two adjacent troughs), amplitude (maximum deviation from the neutral position) and speed of propagation (how far a crest moves in a unit of time).<br /><br />If the speed of light propagation in the <i>ether</i> is <i><b>c</b>(m/sec)</i> and the wave length is <i><b>λ</b></i>, it means that in a <i><b>1</b> sec</i> the crest of a wave moves by a distance <i><b>c</b> meters</i>. With a wave length <i><b>λ</b> meters</i> there will be <i><b>f=c/λ</b> crests</i> on this distance. So, the frequency of oscillation <i><b>f</b></i>, that is the number of oscillations per second, the wave length <i><b>λ</b></i> and the speed of propagation <i><b>c</b></i> are related by the equation<br /><i><b>f = c/λ</b></i><br />or<br /><i><b>λ = c/f</b></i><br />or<br /><i><b>f·λ = c</b></i><br /><br />Let's consider the double slit experiment and analyze what will be observed on the screen at the point equidistant from both slits. Two slits can be considered as two independent sources of light and, considering they are letting through the light emitted by a single source equidistant from them, we can assume that the waves of light are coming <i>in-phase</i> to these slits and go through them also <i>in-phase</i>. Then the two beams of light travel the same distance to a point on a screen that we have chosen as equidistant from the slits, arriving to this point also <i>in-phase</i>. At that point they are combined and intensity of the light is increasing at there. That's why the stripe on a screen in the middle of a picture is bright, it consists of points equidistant from both slits.<br /><br />If, however, the distance these two beams of light have to travel towards the screen is different, the compounded effect of them, observed on a screen, will be dependent on whether the beams come to a screen <i>in-phase</i> or not.<br />More precisely, if the difference in the distance these beams travel towards a screen is a multiple of the length of a wave, they will come <i>in-phase</i>, and the overlapping beams will strengthen each other, there will be a bright spot on a screen.<br />If the difference in the distance equals to a multiple of wave length plus half a wave length, they will come <i>out-of-phase</i>, and the overlapping beams will nullify each other, there will be a dark spot on a screen.<br />In other cases the effect depends on how close the difference in distance is to a multiple of wave length or a multiple of wave length plus half a wave length with intermediary results.<br /><br />So, the wave theory can explain the effect of interference. Other wave-related properties of light that can be explained within the framework of the wave theory will be presented in the Optics chapter of this course.<br /><br />While the corpuscular properties of light do exist, they also can be explained within a framework of the wave theory and the energy of waves. That makes the wave theory of light more universal and more acceptable among physicists, though the auxiliary concept of <i>ether</i> has been totally rejected.<br /><br />The concept of <i>ether</i> was rejected as there was no experiment that could prove its existence. Moreover, a famous experiment of Michelson-Morley has proven that the speed of light does not depend on the speed of the source of light, which cannot be explained from the position of <i>ether</i> as a medium, through which the light waves propagate.<br /><br />There were some interesting observations of electric and magnetic properties, which related to an experimental fact that the speed of propagation of electric signals along wires is practically the same as the speed of light. This and other experiments prompted physicists to seek the explanation of the nature of light in the domain of electricity and magnetism.<br /><br />James Maxwell came up with his famous equations that describe the <i>electromagnetic field</i> and concluded that light is the oscillations of <i>electromagnetic field</i>. There is no special medium, like <i>ether</i>, that carries the light as the sound is carried by the waves of compression of the medium. Instead, the <i>electromagnetic field</i> propagates through space, thus carrying the light.<br />The properties of <i>electromagnetic field</i> will be presented in the parts of this course dedicated to <i>Electricity</i> and <i>Magnetism</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-16735381918438972062019-11-05T11:32:00.001-08:002019-11-05T11:32:43.538-08:00Unizor - Physics4Teens - Energy - Energy of Light - Light as Corpuscles<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/v3PrPxQBJ1o" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Corpuscles</u><br /><br />The nature of light was one of the topics of Newton's research. He experimented with light, made instruments to research the nature of light and formulated the first theory of light - <i>corpuscular theory</i>.<br /><br />According to <i>corpuscular theory</i>, light is a bunch of very small particles emitted by a source of light and flying with a very high speed. When they enter our eye, we see the light as a result of bombarding of inner surface of an eye with these tiny particles of light that Newton called <i>corpuscles</i>.<br />Each <i>light corpuscle</i> has mass, speed, kinetic energy and trajectory, like any other material object.<br /><br />Furthermore, Newton decided that there are different kinds of <i>light corpuscles</i> that cause our perception as different colors. He assumed that different colors of <i>light corpuscles</i> are due to their different sizes.<br /><br />He also realized that the white color is a combination of different kinds of <i>light corpuscles</i> that can be separated into different individual colors. Thus, white light after going through a green glass becomes green, because a green glass separates different kinds of <i>light corpuscles</i>, letting through only those that are perceived by our eye as green.<br /><br />According to Newton, <i>light corpuscles</i> are elastic, which explains perfectly their reflection from the mirror.<br /><br />The effect of refraction of light, when it changes the direction going from one medium, like air, into another, like water, was explained by Newton as a result of changing the speed of propagation of <i>light corpuscles</i>, when they go from one medium into another.<br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>diffraction</i>, when the light seems bending around an edge of an obstacles or aperture.<br />Here is the picture formed on a screen by red light going through a small round hole.<br /><img src="http://www.unizor.com/Pictures/Diffraction.png" style="height: 200px; width: 200px;" /><br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>interference</i>, when the light going through two small holes positioned near each other forms a complex wave-like picture on the screen.<br /><img src="http://www.unizor.com/Pictures/Interference.png" style="height: 150px; width: 200px;" /><br /><br />The <i>corpuscular theory</i> of light was unable to explain the effect of <i>polarization</i>, when the light consecutively going through two crystals of tourmaline changes its intensity from maximum to zero, depending on orientation of these crystals relatively to each other, when we rotate them around the axis coinciding with the direction of light.<br /><img src="http://www.unizor.com/Pictures/Polarization.png" style="height: 130px; width: 200px;" /><br /><br />These and some other difficulties in explanation of observed properties of light led to another theory - the <i>wave theory of light</i>.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-26592601450851128222019-10-31T10:41:00.001-07:002019-10-31T10:41:09.678-07:00Unizor - Physics4Teens - Energy - Light as Energy Carrier<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/lWQLQFzZw24" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Light as Energy Carrier</u><br /><br />Sun is a source of light and a source of heat. Just looking at light, we don't see energy it carries, but, touching a surface that was under Sun's light for some time, we feel its temperature, which is a measure of inner energy of the molecules inside the object. The object warms up under the Sun's light, which can only be explained by energy the light carries.<br /><br />Let's conduct an experiment with light to prove that it carries energy.<br />On YouTube there are a few videos that show it, one of them demonstrates a <a href="https://www.youtube.com/watch?v=j7UtjEjh7k4">radiometer</a>.<br /><br />The radiometer ("solar mill") looks like this:<br /><img src="http://www.unizor.com/Pictures/Radiometer.png" style="height: 240px; width: 200px;" /><br />Four small plates are arranged on a spinning wheel in the vacuum to avoid interference of air motion. Each plate has two surfaces, one silver and another black.<br />When light is directed on this "solar mill", black sides absorb light and are heated more than silver ones that reflect light.<br />As a result, the molecules near the black surface are moving more intensely than those near the surface of the silver side, thus pushing the silver side more and causing the rotation of the "solar mill".<br /><br />So, there is no doubt that light carries energy. An important question is, how it does it.<br />We used to think about heat as the intensity of molecular motion. In case of light there is no such motion, light travels from Sun to Earth through vacuum.<br />Apparently, the situation is similar to gravity in a sense that gravity carries energy, but does not require any medium, like molecules, to carry it. Recall that we have introduced a concept of a <i>field</i> as a certain domain of space were forces exist and energy is present without any material substance. Somewhat similar situation is with light. Its nature is the waves of <i>electromagnetic field</i>.<br /><br /><i>Electromagnetic field</i> is a completely different substance than <i>gravitational field</i>, but both are capable to carry energy without any material presence.<br /><br />Classifying light as the waves of an <i>electromagnetic field</i>, we are opening the door to using this wave model for explanation of different kinds of light.<br />First of all, let's consider the light we see with a naked eye. The vision itself is possible only if the light carries some energy, that agitates some cells inside our eye, that, in turn, send an electric signal to a brain - one more argument toward a light as a carrier of energy.<br /><br />Light that can be seen by a naked eye is called <i>visible</i>. But what about different colors that we can differentiate by an eye and different intensity of light that we view as "bright" or "faint"? The only explanation within a wave theory of light is that different intensities and colors that our eye sees are attributable to different kinds of waves of an electromagnetic field.<br /><br />Any wave has two major characteristics: amplitude and frequency. This is similar to a pendulum, where amplitude is the maximum angle of deviation from a vertical and frequency is measured as a number of oscillations per unit of time. Light, as a wave, also has these two characteristics. The <i>amplitude</i> is an intensity of light, while <i>frequency</i> of the visible light is viewed as its color.<br /><br />We all know that the photo laboratories, developing old fashioned films, are using rather faint red light during the developing process in order not to overexpose the film to light. The obvious reason is that red light carries less energy than white one, that is known to be a combination of many differently colored kinds of light. So, the color-defining frequency, as well as an intensity-defining amplitude of electromagnetic waves, determine the amount of energy carried by light.<br /><br />Not only visible light is a manifestation of electromagnetic waves. From cosmic radiation called <i>gamma rays</i> to <i>x-rays</i> to <i>ultra-violet light</i> to <i>visible light</i> to <i>infra-red light</i> to <i>microwave</i> to <i>radio waves</i> - all are electromagnetic waves of different frequencies.<br /><br />The unit of measurement of frequency is called <i>Hertz</i>, abbreviated as <i><b>Hz</b></i> with <i><b>1 Hz</b></i> meaning 1 oscillation per second.<br /><br />The range of frequencies of electromagnetic waves is from a few oscillations per second for very low frequency radio waves to 10<sup>24</sup> oscillations per second for very high frequency gamma rays. The visible spectrum of frequencies is close to 10<sup>15</sup> oscillations per second with the light perceived as red having a smaller frequency around 0.4·10<sup>15</sup> Hz, followed in increasing frequency order by orange, yellow, green, blue and violet with a frequency around 0.7·10<sup>15</sup> Hz.<br /><br />Amount of energy carried by light depends on the amplitude and frequency of electromagnetic waves that constitute this light. Generally speaking, the higher the amplitude - the higher the energy is carried by light in a unit of time and, similarly, the higher the frequency - the higher the energy is carried by light in a unit of time.<br /><br />Dependency of the energy carried by light on the frequency of electromagnetic waves that constitute this light is a more complex problem, that was solved in the framework of the Quantum Theory of light. According to Quantum Theory, electromagnetic waves propagate in packets called <i>photons</i>. Each <i>photon</i> carries an energy proportional to the frequency of waves that constitute this <i>photon</i>, and the amplitude of electromagnetic waves is, simply, a measure of the number of photons participating in these electromagnetic waves. That's why the picture of a light as a sinusoidal wave is a very simplified view on the nature of light as it is understood by contemporary physics.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-85391035758300597142019-10-15T20:18:00.001-07:002019-10-15T20:18:42.738-07:00Unizor - Physics4Teens - Energy - Gravitational Potential<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/fWXogAfvR30" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 4 -<br />Solid Sphere</u><br /><br /><i>1. Determine the potential of the gravitational field of a uniform solid sphere at any point outside it.</i><br /><br />Let's establish a system of coordinates with a sphere's center at the origin of coordinates and X-axis going through a point of interest <i><b>P</b></i>, where we have to determine the gravitational potential.<br /><br />Assume that the sphere's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>. Then its volume is <i><b>4πR³/3;</b></i> and the mass density per unit of volume is <i><b>ρ=3M<span style="font-size: medium;">/</span>(4πR³)</b></i>.<br />Assume further that X-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>, which is greater than the radius of a sphere <i><b>R</b></i>.<br />If, instead of a sphere, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>O(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub>(H) = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br /><br />To calculate a gravitational potential of a solid sphere at point <i><b>P</b></i> on the X-axis, let's divide it into infinite number of infinitesimally thin concentric spherical shells, all centered at the origin of coordinates and use the results presented in the previous lecture about spherical shell.<br /><br />As the variable of integration we will chose a radius of a spherical shell <i><b>r</b></i> that varies from <i>0</i> to <i>R</i>. Its outside surface area is <i><b>4πr²</b></i>, its thickness is <i>d<b>r</b></i> and, therefore, its volume is <i><b>4πr²·</b>d<b>r</b></i>.<br />This allows us to calculate the mass of this spherical shell using the volume and mass density calculated above.<br /><i>d<b>m = ρ·4πr²·</b>d<b>r =<br />= 3M·4πr²·</b>d<b>r<span style="font-size: medium;">/</span>(4πR³) =<br />= 3M·r²·</b>d<b>r<span style="font-size: medium;">/</span>R³</b></i><br />The formula for gravitational potential of a spherical shell, derived in the previous lecture was <i>V=−G·M<span style="font-size: medium;">/</span>H</i>, where <i>G</i> is a gravitational constant, <i>M</i> is a mass of a spherical shell and <i>H</i> is a distance from a center of a shell to a point of interest.<br />In case of a solid sphere divided into infinite number of infinitesimally thin concentric spherical shells the distance <i>H</i> remains the same. So, all we have to do is to substitute the mass in the formula for a shell with the variable mass of a shells we divided our solid sphere and to integrate by variable radius <i>r</i>:<br /><i><b>V(H) = −(G/H)<span style="font-size: large;">∫</span><sub>[0;R]</sub></b>d<b>m =<br />= −(G/H)<span style="font-size: large;">∫</span><sub>[0;R]</sub>3M·r²·</b>d<b>r<span style="font-size: medium;">/</span>R³ =<br />= −</b></i>[<i><b>3M·G/(H·R³)</b></i>]<i><b><span style="font-size: large;">∫</span><sub>[0;R]</sub>r²·</b>d<b>r</b></i><br />The indefinite integral of <i>r²</i> is <i>r³/3</i>, which gives the value of the integral<br /><i><b><span style="font-size: large;">∫</span><sub>[0;R]</sub>r²·</b>d<b>r = R³/3 − 0 = R³/3</b></i><br />Therefore, finally,<br /><i><b>V(H) = −G·M/H</b></i><br /><br /><b>Remarkably, the formula is exactly the same as if the whole mass was concentrated in the center of a sphere, the same as in case of a spherical shell</b>.<br /><br />It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.<br /><br />In theory, this result was easily predictable. The gravitational potential of each spherical shell is the same as if its mass is concentrated at its center. All shells are concentric, therefore the masses of all of them are concentrated in the origin of coordinates and can be added together, since the gravitational potential is additive. Thus, we come to the same value of gravitational potential of a sphere, as if its total mass is concentrated in one point - its center.<br /><br />Let's analyze the force of gravity, acting on a probe object of a mass <i><b>m</b></i> at a point of interest on the distance <i><b>H</b></i> from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:<br /><i><b>F(H) = m·</b>d<b>V(H)/</b>d<b>H =<br />= G·m·M/H²</b></i><br />which is a well known Newton's Law of Gravitation.<br /><br /><i>2. Determine the potential of the gravitational field of a uniform solid sphere of radius <i><b>R</b></i> at a point inside it at distance <i><b>r</b></i> from a center.</i><br /><br />For this problem, as in the <i>Problem 1</i> above, we will need a mass density per unit of volume <i><b>ρ=3M<span style="font-size: medium;">/</span>(4πR³)</b></i>.<br /><br />Assume that a probe object is a distance <i><b>r</b></i> from a center of sphere, which is less than the radius of a sphere <i><b>R</b></i>. Let's calculate the gravitational potential <i><b>V(r)</b></i> of the combination of two separate sources - the solid sphere of radius <i><b>r</b></i> with a probe object on its surface and a thick empty spherical object between a surface of a sphere of the radius <i><b>r</b></i> and surface of a sphere of the radius <i><b>R</b></i>.<br /><br />The gravitational potential of a uniform solid sphere of radius <i><b>r</b></i> on its surface is discussed above as a <i>Problem 1</i>. To use the results of this problem, we need a mass <i><b>M<sub>1</sub>(r)</b></i> of a source of gravity and the distance of a point of interest from a center <i><b>H</b></i>.<br />The mass is<br /><i><b>M<sub>1</sub>(r) = ρ·4πr³<span style="font-size: medium;">/</span>3 = M·r³<span style="font-size: medium;">/</span>R³</b></i><br />The distance form a center is<br /><i><b>H = r</b></i><br />The gravitational potential on the surface of this solid sphere of the radius <i><b>r</b></i> equals to<br /><i><b>V<sub>1</sub>(r) = −G·M·r²/R³</b></i><br /><br />Consider now the second source of gravity - a thick empty sphere between the radiuses <i><b>r</b></i> and <i><b>R</b></i>.<br />As in <i>Problem 1</i> above, we will divide a thick empty sphere into an infinite number of concentric infinitesimally thin spherical shells of a variable radius <i><b>x</b></i> and thickness <i>d<b>x</b></i>.<br />The mass of each shell is<br /><i>d<b>m(x) = ρ·4πx²·</b>d<b>x =<br />= 3M·4πx²·</b>d<b>x<span style="font-size: medium;">/</span>(4πR³) =<br />= 3M·x²·</b>d<b>x<span style="font-size: medium;">/</span>R³</b></i><br />The potential inside such an infinitesimally thin spherical shell of radius <i><b>x</b></i> is, as we know from a previous lecture, constant and equals to<br /><i>d<b>V(x) = −G·</b>d<b>m(x)<span style="font-size: medium;">/</span>x =<br />= −3G·M·x·</b>d<b>x<span style="font-size: medium;">/</span>R³</b></i><br />To get a full potential inside such a thick empty sphere we have to perform integration of this expression from <i>x=r</i> to <i>x=R</i>.<br /><i><b>V<sub>2</sub>(r) = (−3G·M<span style="font-size: medium;">/</span>R³)<span style="font-size: large;">∫</span><sub>[r;R]</sub>x·</b>d<b>x =<br />= −3G·M·(R²−r²)/(2R³)</b></i><br /><br />The total potential at distance <i><b>r</b></i> from a center equal to sum of two potentials calculated above<br /><i><b>V(r) = V<sub>1</sub>(r) + V<sub>2</sub>(r) =<br />= −G·M·r²/R³ −<br />−3G·M·(R²−r²)/(2R³) =<br />= −G·M·(3R²−r²)/(2R³)</b></i><br /><br />On the outer surface of this sphere, when <i><b>r=R</b></i>, the above formula converts into the one derived in <i>Problem 1</i>:<br /><i><b>V(R) = −G·M·/R</b></i><br /><br />In the center of a solid sphere, when <i><b>r=0</b></i>, the potential is<br /><i><b>V(R) = −(3/2)·G·M·/R</b></i><br /><br />Let's analyze the force of gravity, acting on a probe object of a mass <i><b>m</b></i> at a point of interest on the distance <i><b>r</b></i> from a center of a solid sphere. This is a mass of a prove object multiplied by a derivative of the gravitational potential:<br /><i>Case inside a solid sphere</i>:<br /><i><b>F(r) = m·</b>d<b>V(r)/</b>d<b>r =<br />= G·m·M·r/R³</b></i><br />So, as we move from a center of a solid sphere (<i><b>r=0</b></i>) towards its outer surface (<i><b>r=R</b></i>), the force is linearly growing from zero at the center to <i><b>G·m·M/R²</b></i> at the end on the surface.<br /><i>Case outside a solid sphere (using the results of Problem 1 above for V(H)=−G·M/H, where H=r is greater than R)</i>:<br /><i><b>F(r) = m·</b>d<b>V(r)/</b>d<b>r =<br />= G·m·M/r²</b></i><br />So, as we move from a surface of a solid sphere (<i><b>r=R</b></i>) outwards to infinity, increasing <i><b>r</b></i>, the force is decreasing inversely to a square of a distance from the center from <i><b>G·m·M/R²</b></i> to zero at infinity.<br /><br />It's quite interesting to graph the force of gravitation as a function of a distance of a probe object from a center of a solid sphere. We have two different functions that represent this force, one inside and one outside the surface of a sphere.<br />The graph looks like this:<br /><img src="http://www.unizor.com/Pictures/GravitySolidSphere.png" style="height: 150px; width: 200px;" /><br /><br /><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-78215585584398073532019-10-07T12:37:00.001-07:002019-10-07T12:37:05.587-07:00Unizor - Physics4Teens - Energy - Gravitational Potential - Thin Spheric...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/xI86jwVxmQk" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 3 -<br />Thin Spherical Shell</u><br /><br /><i>1. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point outside it.</i><br /><br />Let's establish a system of coordinates with a spherical shell's center at the origin of coordinates and X-axis going through a point of interest <i><b>P</b></i>, where we have to determine the gravitational potential.<br /><br />Assume that the sphere's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>. Then its surface is <i><b>4πR²</b></i> and the mass density per unit of surface area is <i><b>ρ=M<span style="font-size: medium;">/(4πR²)</span></b></i><span style="font-size: medium;">.<br />Assume further that X-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>, which is greater than the radius of a spherical shell <i><b>R</b></i>.<br /><img src="http://www.unizor.com/Pictures/SphericalShell.png" style="height: 100px; width: 200px;" /><br /><br />If, instead of a spherical shell, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>O(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub> = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br /><br />To calculate a gravitational potential of an infinitesimally thin spherical shell at point <i><b>P</b></i> on the X-axis, let's divide a spherical shell into infinite number of infinitesimally thin rings that are parallel to the YZ-plane and, therefore, perpendicular to X-axis, that goes through a center of each ring.<br /><br />The angle <i><b>φ</b></i> from X-axis (that is, from <i><b>OP</b></i>) to a radius from an origin of coordinates to any point on a ring will be our variable of integration.<br />Then the radius of a ring will be<br /><i><b>r(φ) = R·sin(φ)</b></i><br />The distance from the origin of coordinates to a center of a ring is <i><b>R·cos(φ)</b></i>.<br />The area of a ring between angles <i><b>φ</b></i> and <i><b>φ+</b>d<b>φ</b></i> will be equal to the product of infinitesimal width of a ring <i><b>R·</b>d<b>φ</b></i> and its circumference <i><b>2πR·sin(φ)</b></i><br />Therefore, the mass of a ring will be<br /><i>d<b>m(φ) = ρ·2πR²·sin(φ)·</b>d<b>φ =<br />= M·2πsin(φ)·</b>d<b>φ<span style="font-size: medium;">/</span>(4π) =<br />= M·sin(φ)·</b>d<b>φ<span style="font-size: medium;">/</span>2</b></i><br /><br />Knowing the mass of a ring <i>d<b>m(φ)</b></i>, its radius <i><b>r(φ)</b></i> and the distance from the ring's center to point of interest <i><b>P</b></i>, that is equal to <i><b>H−R·cos(φ)</b></i>, we can use the formula of the ring's potential from a previous lecture<br /><i>V = −G·M <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²</span></i><br />substituting<br /><i>d<b>V(φ)</b></i> instead of <i>V</i><br /><i>d<b>m(φ)</b></i> instead of <i>M</i><br /><i><b>H−R·cos(φ)</b></i> instead of <i>H</i><br /><i><b>r(φ)</b></i> instead of <i>R</i><br /><br />Therefore,<br /><i>d<b>V(φ) = −G·</b>d<b>m(φ) <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">r²(φ)+[H−R·cos(φ)]²</span> =<br />= −G·M·sin(φ)·</b>d<b>φ <span style="font-size: medium;">/</span>2√<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br /><br />Now all we need is to integrate this by <i><b>φ</b></i> in limits from <i>0</i> to <i>π</i>.<br />Substitute<br /><i><b>y = √<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br />Incidentally, the geometric meaning of this value is the distance from point of interest <i><b>P</b></i> to any point on a ring for a particular angle <i><b>φ</b></i>.<br />Then<br /><i>d<b>y = R·H·sin(φ)·</b>d<b>φ <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br /><br />The limits of integration for <i><b>φ</b></i> from <i>0</i> to <i>π</i> in terms of <i><b>y</b></i> are from |<i><b>H−R</b></i>| (which, for our case of point <i><b>P</b></i> being outside the sphere, equals to <i><b>H−R</b></i>) to <i><b>H+R</b></i>.<br /><br />In terms of <i><b>y</b></i><br /><i>d<b>V(y) = −G·M·</b>d<b>y <span style="font-size: medium;">/</span>(2R·H)</b></i><br />which we have to integrate by <i><b>y</b></i> from <i><b>H−R</b></i> to <i><b>H+R</b></i>.<br /><br />Simple integration of this function by <i><b>y</b></i> on a segment [<i><b>H−R;H+R</b></i>] produces <i><b>−G·M·y<span style="font-size: medium;">/</span>(2R·H)</b></i> in limits from <i><b>H−R</b></i> to <i><b>H+R</b></i>:<br /><i><b>V = −G·M·(H+R)<span style="font-size: medium;">/</span>(2R·H) +<br />+ G·M·(H−R)<span style="font-size: medium;">/</span>(2R·H) =<br />= −G·M<span style="font-size: medium;">/</span>H</b></i><br /><br /><b>Remarkably, it's exactly the same gravitational potential, as if the whole mass was concentrated in a center of a spherical shell, as noted above as <i><b>V<sub>0</sub></b></i></b>.<br /><br />It's a justification for considering gravitational potential of space objects in many cases to be approximately equivalent to a potential of point-masses of the same mass concentrated in one point, the center of mass of an object.<br /><br /><i>2. Determine the potential of the gravitational field of a uniform infinitesimally thin spherical shell at any point inside it.</i><br /><br />Using the same notation as in the previous case, this problem requires the distance from a point of interest <i><b>P</b></i> to a center of a spherical shell <i><b>O</b></i> to be less than the radius <i><b>R</b></i> of a spherical shell.<br />Doing exactly the same manipulation and substitution<br /><i><b>y = √<span style="text-decoration-line: overline;">R²+H²−2R·H·cos(φ)</span></b></i><br />we see that the only difference from the previous case is in the limits of integration in terms of <i><b>y</b></i>.<br />The limits of integration for <i><b>φ</b></i> from <i>0</i> to <i>π</i> in terms of <i><b>y</b></i> are from |<i><b>H−R</b></i>| (which, in this case of point <i><b>P</b></i> being inside the sphere, equals to <i><b>R−H</b></i>) to <i><b>H+R</b></i>.<br /><br />Integration by <i><b>y</b></i> on a segment [<i><b>R−H;H+R</b></i>] produces <i><b>−G·M·y<span style="font-size: medium;">/</span>(2R·H)</b></i> in limits from <i><b>R−H</b></i> to <i><b>H+R</b></i>:<br /><i><b>V = −G·M·(H+R)<span style="font-size: medium;">/</span>(2R·H) + G·M·(R−H)<span style="font-size: medium;">/</span>(2R·H) =<br />= −G·M<span style="font-size: medium;">/</span>R</b></i><br /><br />Remarkably, it's constant and is independent of the position of point <i><b>P</b></i> inside a spherical shell.<br /><br />We have mentioned in the earlier lecture on gravitational field that in one dimensional case the gravitational force is a derivative of gravitational potential by distance from the source of gravity times mass of a probe object:<br /><b><i>F(r)=G·M·m <span style="font-size: medium;">/</span>r²=m·</i></b><i>d<b>V(r)/</b>d<b>r</b></i><br /><br />The fact that the gravitational potential is constant and, therefore, its derivative is zero, signifies that there is no force of gravity inside a spherical shell. The forces of gravity from all directions nullify each other.<br /><br />An intuitive explanation of this is in the fact that, if you consider any conical surface with a vertex at point <i><b>P</b></i> inside a sphere, cutting pieces of spherical shell's surface in both directions, the areas of the pieces will be proportional to a square of a distance from point <i><b>P</b></i>, while the gravitational forces produced by these pieces of surface are inversely proportional to a square of a distance from point <i><b>P</b></i>, thus both forces from opposite ends of a cone are equal in magnitude and opposite in direction, thus nullify each other.</span><br /><br /><span style="font-size: medium;"><br /></span>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-2146035884629930812019-10-04T19:26:00.001-07:002019-10-04T19:26:02.124-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/nbQAFewy5Y4" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 2</u><br /><br /><i>1. Determine the potential of the gravitational field of an infinitely thin uniform solid ring at any point on the line perpendicular to a plane of the ring and going through its center.</i><br /><br />Let's establish a system of coordinates with a ring in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.<br />Assume that the ring's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of length is <i><b>ρ=M/(2πR)</b></i>.<br />Assume further that the Z-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>.<br /><br />If, instead of a ring, we had a point mass <i><b>M</b></i> concentrated in its center at point <i><b>(0,0,0)</b></i>, its gravitational potential at a point <i><b>P</b></i> would be<br /><i><b>V<sub>0</sub> = −G·M/H</b></i><br />(remember that the gravitational potential is negative, as it equals to amount of work needed to bring a unit of mass from infinity to point <i><b>P</b></i>, and the field performs this work for us, so we perform negative work).<br /><br />Since the mass in our case is distributed along the circumference of a ring, and every point on a ring is on a distance <i><b>r=√<span style="text-decoration-line: overline;">R²+H²</span></b></i> from point <i><b>P</b></i>, which is further from this point than the center of a ring, the gravitational potential of a ring at point <i><b>P</b></i> will be smaller.<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br />Therefore, to calculate a gravitational potential of a ring, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest <i><b>P</b></i> and integrate all these potentials.<br /><br />Let's choose an angle from the positive direction of the X-axis to a point on a ring as the main integration variable <i><b>φ</b></i>∈[<i><b>0;2π</b></i>]. Its increment <i>d<b>φ</b></i> gives an increment of the circumference of a ring<br /><i>d<b>l = R·</b>d<b>φ</b></i><br />The mass of this infinitesimal segment of a ring is<br /><i>d<b>m = ρ·</b>d<b>l = M·R·</b>d<b>φ <span style="font-size: medium;">/</span>(2πR) = M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π)</b></i><br /><br />The distance from this infinitecimal segment of a ring to a point of interest <i><b>P</b></i> is independent of variable <i><b>φ</b></i> and is equal to constant <i><b>r=√<span style="text-decoration-line: overline;">R²+H²</span></b></i>.<br /><br />Therefore, gravitational potential of an infinitecimal segment of a ring is<br /><i>d<b>V = −G·</b>d<b>m <span style="font-size: medium;">/</span>r = −G·M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π√<span style="text-decoration-line: overline;">R²+H²</span>)</b></i><br /><br />Integrating this by variable <i><b>φ</b></i> on [<i><b>0;2φ</b></i>], we obtain the total gravitational potential of a ring at point <i><b>P</b></i>:<br /><i><b>V = <span style="font-size: large;">∫</span><sub>[0;2π]</sub></b>d<b>V = −<span style="font-size: large;">∫</span><sub>[0;2π]</sub>G·M·</b>d<b>φ <span style="font-size: medium;">/</span>(2π√<span style="text-decoration-line: overline;">R²+H²</span>)</b></i><br />Finally,<br /><i><b>V = −G·M <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">R²+H²</span></b></i><br /><br /><i>2. Determine the potential of the gravitational field of an infinitely thin uniform solid disc at any point on the line perpendicular to a plane of the disc and going through its center.</i><br /><br />Let's establish a system of coordinates with a disc in the XY-plane with its center coincided with the origin of coordinates. Then the Z-axis is the line, on which we have to determine the gravitational potential of the ring.<br />Assume that the ring's radius is <i><b>R</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of surface is <i><b>ρ=M/(πR²)</b></i>.<br />Assume further that the Z-coordinate of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, is <i><b>H</b></i>.<br /><br />Let's split our disc into infinite number of infinitely thin concentric rings of radius from <i><b>x=0</b></i> to <i><b>x=R</b></i> of width <i>d<b>x</b></i> each and use the previous problem to determine the potential of each ring.<br /><br />The mass of each ring is<br /><i>d<b>m(x) = ρ·2πx·</b>d<b>x</b></i><br />This gravitational potential of this ring at point <i><b>P</b></i>, according to the previous problem, is<br /><i>d<b>V(x) = −G·</b>d<b>m(x) <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span> =<br />= −G·ρ·2πx·</b>d<b>x <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span> =<br />= −G·M·2πx·</b>d<b>x <span style="font-size: medium;">/</span>(πR²√<span style="text-decoration-line: overline;">x²+H²</span>) =<br />= −G·M·2x·</b>d<b>x <span style="font-size: medium;">/</span>(R²√<span style="text-decoration-line: overline;">x²+H²</span>)</b></i><br /><br />To determine gravitational potential of an entire disc, we have to integrate this expression in limits from <i><b>x=0</b></i> to <i><b>x=R</b></i>.<br /><i><b>V = <span style="font-size: large;">∫</span><sub>[0;R]</sub></b>d<b>V(x) = −k·<span style="font-size: large;">∫</span><sub>[0;R]</sub>2x·</b>d<b>x <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">x²+H²</span></b></i><br />where <i><b>k = G·M <span style="font-size: medium;">/</span>R²</b></i><br /><br />Substituting <i><b>y=x²+H²</b></i> and noticing the <i>d<b>y=2x·</b>d<b>x</b></i>, we get<br /><i><b>V = −k·<span style="font-size: large;">∫</span><sub>[H²;H²+R²]</sub> </b>d<b>y <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">y</span></b></i><br />The derivative of <i><b>√<span style="text-decoration-line: overline;">y</span></b></i> is <i><b>1 <span style="font-size: medium;">/</span>(2√<span style="text-decoration-line: overline;">y</span>)</b></i> Therefore, the indefinite integral of <i><b>1 <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">y</span></b></i> is<br /><i><b>2√<span style="text-decoration-line: overline;">y</span> + C</b></i><br /><br />Finally,<br /><i><b>V = −k·(2√<span style="text-decoration-line: overline;">H²+R²</span>−2H) = −2G·M·(√<span style="text-decoration-line: overline;">H²+R²</span>−H) <span style="font-size: medium;">/</span>R²</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-15361333087752802492019-09-24T19:31:00.001-07:002019-09-24T19:31:01.396-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravit...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/siZCP7D-Huo" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravity Integration 1</u><br /><br /><i>Determine the potential of the gravitational field of an infinitely thin solid rod at any point outside of it.</i><br /><br />Let's establish a system of coordinates with a rod and a point mass lying in the XY-plane with the rod on the X-axis with one end at point <i><b>A(a,0)</b></i> and another at point <i><b>B(b,0)</b></i>.<br />Assume that the rod's length is <i><b>L=b−a</b></i> and the mass is <i><b>M</b></i>, so the density of mass per unit of length is <i><b>ρ=M/L</b></i>.<br />Assume further that the coordinates of a point <i><b>P</b></i>, where we want to calculate the gravitational potential, are <i><b>(p,q)</b></i>.<br /><br />If, instead of a rod, we had a point mass <i><b>M</b></i> concentrated in the midpoint of a rod at point <i><b>((a+b)/2,0)</b></i>, its gravitational potential at a point <i><b>(p,q)</b></i> would be<br /><i><b>V<sub>0</sub>=G·M/r</b></i><br />where <i><b>r</b></i> is the distance between the midpoint of a rod and a point of measurement of gravitational potential <i><b>P</b></i>:<br /><i><b>r = </b></i>{<i><b></b></i>[<i><b>(p−(a+b)/2</b></i>]<i><b><sup>2</sup> + q<sup>2</sup></b></i>}<i><b><sup>1/2</sup></b></i><br /><br />Since the mass in our case is distributed along the rod, the gravitational potential will be different.<br /><br />As we know, gravitational potential is additive, that is the potential of a combined gravitational field produced by two or more sources of gravitation at some point equals to sum of their individual gravitational potentials at this point.<br />Therefore, to calculate a gravitational potential of a rod, we can divide it into infinite number of infinitesimal pieces, calculate the gravitational potential of each piece at the point of interest <i><b>P</b></i> and integrate all these potentials.<br /><br />Consider a picture below (we recommend to save it locally to see in the bigger format).<br /><img src="http://www.unizor.com/Pictures/GravityRod.png" style="height: 120px; width: 200px;" /><br />As a variable, we will use an X-coordinate of a point on a rod <i><b>Q</b></i> and calculate the gravitational potential at point of interest <i><b>P(p,q)</b></i> from an infinitesimal segment of a rod of the length <i>d<b>x</b></i> around point <i><b>Q(x,0)</b></i>.<br />Knowing that, we will integrate the result by <i><b>x</b></i> on a segment [<i><b>a;b</b></i>] to get the gravitational potential of the rod.<br /><br />The infinitesimal segment of a rod <i>d<b>x</b></i>, positioned around a point <i><b>Q(x,0)</b></i>, has an infinitesimal mass <i>d<b>m</b></i> that can be calculated based on the total mass of a rod <i><b>M</b></i> and its length <i><b>L=b−a</b></i> as<br /><i>d<b>m = M·</b>d<b>x <span style="font-size: medium;">/</span>L</b></i><br /><br />The gravitational potential of this segment depends on its mass <i>d<b>m</b></i> and its distance <i><b>r(x)</b></i> to a point of interest <i><b>P(p,q)</b></i>.<br /><i>d<b>V = G·</b>d<b>m <span style="font-size: medium;">/</span>r(x)</b></i><br />Obviously,<br /><i><b>r(x) = </b></i>[<i><b>(p−x)<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i><br />Combining all this, the full gravitational potential of a rod [<i><b>a;b</b></i>] of mass <i><b>M</b></i> at point <i><b>P(p,q)</b></i> will then be<br /><i><b>V(p,q) = <span style="font-size: large;">∫</span><sub>a</sub><sup><sup>b</sup></sup>G·</b>d<b>m <span style="font-size: medium;">/</span>r(x) = <span style="font-size: large;">∫</span><sub>a</sub><sup><sup>b</sup></sup>G·M·</b>d<b>x<span style="font-size: medium;">/</span></b></i>{<i><b>L·</b></i>[<i><b>(p−x)<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i>}<br /><br />We can use the known indefinite integral<br /><i><b><span style="font-size: large;">∫</span></b>d<b>t <span style="font-size: medium;">/</span>√<span style="text-decoration-line: overline;">(t²+c²)</span> = ln|t+√<span style="text-decoration-line: overline;">(t²+c²)</span>|</b></i><br /><br />Let's substitute in the integral for gravitational potential <i><b>t=x−p</b></i>.<br />Then<br /><i><b>V(p,q) = <span style="font-size: large;">∫</span>G·M·</b>d<b>t <span style="font-size: medium;">/</span>L·</b></i>[<i><b>t<sup>2</sup>+q<sup>2</sup></b></i>]<i><b><sup>1/2</sup></b></i><br />where integration is from <i><b>t=a−p</b></i> to <i><b>t=b−p</b></i>.<br /><i><b>V(p,q) = (G·M/L)·</b></i>[<i><b>ln|b−p+√<span style="text-decoration-line: overline;">(b−p)²+q²</span>| − ln|a−p+√<span style="text-decoration-line: overline;">(a−p)²+q²</span>|</b></i>]<br />where <i><b>L = b−a</b></i><br /><br />Since the difference of logarithms is a logarithm of the result of division,<br /><i><b>V(p,q) = G·M·ln(R) <span style="font-size: medium;">/</span>L</b></i><br />where<br /><i><b>L = b−a</b></i> and<br /><i><b>R = |b−p+√<span style="text-decoration-line: overline;">(b−p)²+q²</span>| <span style="font-size: medium;">/</span> |a−p+√<span style="text-decoration-line: overline;">(a−p)²+q²</span>|</b></i>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-69288424898840422132019-09-20T13:21:00.001-07:002019-09-20T13:21:27.523-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Energy...<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/HGOK0nYEe2w" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Gravitational Energy Conservation</u><br /><br />While moving an object from a distance <i><b>r<sub>1</sub></b></i> to a distance <i><b>r<sub>2</sub></b></i> from the center of gravity, the gravitational field has performed certain work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i>, spending certain amount of energy. Since energy must be conserved, it should materialize in some other way.<br /><br />Indeed, the kinetic energy of a probe object at the end of its movement from point {<i>r<sub>1</sub>,0,0</i>} to {<i>r<sub>2</sub>,0,0</i>} must be equal to the work performed by the field.<br /><br />We have positioned our probe object at point {<i>r<sub>1</sub>,0,0</i>} without any initial speed, that is <i><b>V<sub>r<sub>1</sub></sub>=0</b></i>. Therefore, the kinetic energy <i><b>K<sub>r<sub>1</sub></sub></b></i> at this initial point is zero.<br />At the end of a motion at point {<i>r<sub>2</sub>,0,0</i>} the speed <i><b>V<sub>r<sub>2</sub></sub></b></i> must have such a value that the kinetic energy <i><b>K<sub>r<sub>2</sub></sub></b></i> would be equal to work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i> performed by the field.<br /><i><b>K<sub>r<sub>2</sub></sub> = m·V²<sub>r<sub>2</sub></sub> <span style="font-size: medium;">/</span>2 = W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i><br /><br />From this equation, knowing how to calculate the work performed by a gravitational field (see the previous lecture), we can find a speed of a probe object at the end of its motion from point {<i>r<sub>1</sub>,0,0</i>}, where it was at rest, to point {<i>r<sub>2</sub>,0,0</i>}:<br /><i><b>m·V²<sub>r<sub>2</sub></sub> <span style="font-size: medium;">/</span>2 =<br />= <i>(</i>G·M <span style="font-size: medium;">/</span>r<sub>2</sub> − G·M <span style="font-size: medium;">/</span>r<sub>1</sub></b></i><b>)<i>·m</i></b><br /><i><b>V²<sub>r<sub>2</sub></sub> = 2·<i></i>G·M·(1<span style="font-size: medium;">/</span>r<sub>2</sub> − 1<span style="font-size: medium;">/</span>r<sub>1</sub></b></i><b>)<i></i></b><br /><br />In a particular case, when <i><b>r<sub>1</sub>=∞</b></i> and <i><b>r<sub>2</sub>=r</b></i>, that is a probe object falls with no initial speed from the infinitely long distance from a source of gravity to a point at distance <i><b>r</b></i> from it, the formula is simplified:<br /><i><b>V²<sub>r</sub> = 2·<i></i>G·M <span style="font-size: medium;">/</span>r</b></i><br /><br />We would like to warn against falling into a point-mass that is a source of gravity, when the final distance from it is zero, that is <i><b>r=0</b></i> in the above equation. It obviously produces infinite speed and infinite kinetic energy, which does not correspond to reality. The most important reason for this deviation from the reality is our assumption about a source of gravity to be a point-mass. Real objects have certain non-zero dimensions. For example, in case of a gravitational field around our planet should not be analyzed by this formula for values of <i><b>r</b></i> less than the radius of Earth.<br /><br />Back to <i>energy conservation</i>.<br />The <i>potential energy</i> of an object is a measure of work that it can do, if left alone, that depends on a position of an object relative to other objects and such properties as its mass. Actually, these two parameters are the only ones needed to calculate the potential energy of a probe object in a gravitational field, provided we know everything about the field.<br /><br />As we know (see the previous lecture), amount of work we need to move a probe object of mass <i><b>m</b></i> from an infinite distance to a distance <i><b>r</b></i> from a source of gravity equals to<br /><i><b>W<sub>r</sub> = −G·M·m <span style="font-size: medium;">/</span>r</b></i><br />It's negative from our external to the gravitational field viewpoint, because we don't actually perform work, the field performs it for us. So, from the external viewpoint, the field gives certain energy to external object by performing some work on it, similar to a person, pushing the cart, spends energy, transferring it to a cart.<br /><br />In this expression, skipping over the universal gravitational constant <i><b>G</b></i>, components <i><b>M</b></i> (mass of a source of gravitational field) and <i><b>r</b></i> (distance from the center of the gravitational field) characterize the gravitational field, while <i><b>m</b></i> (mass of a probe object) characterizes the object, whose potential energy we measure.<br /><br />This energy is transferred to a probe object as its <i>potential energy</i>. If an object is not moving from this position, because some force holds it there, it retains this potential energy. As soon as there is no force holding it there, it will start moving towards the source of gravity, losing its <i>potential energy</i> and gaining the <i>kinetic energy</i> because it will move faster and faster.<br /><br />As an example, let's calculate the kinetic energy and final speed of a small asteroid, free falling on the surface of the Moon, assuming the Moon is the only source of gravity in the Universe.<br />The Universal Gravitational Constant is<br /><i><b>G=6.67408·10<sup>−11</sup></b></i>,<br />its units are <i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup></i>.<br />The mass of the Moon is <i><b>M=7.34767309·10<sup>22</sup></b> kg</i>.<br />The radius of the Moon is <i><b>r=1.7371·10<sup>6</sup></b> m</i>.<br />Let's assume that an asteroid falling on the Moon is relatively small one, say, <i><b>m=50</b> kg</i>.<br /><br />According to the formula above, the gravitational field of the Moon did the work that equals to<br /><i><b>W ≅ 6.67408·10<sup>−11</sup>·7.34767309·10<sup>22</sup>·50 <span style="font-size: medium;">/</span>(1.7371·10<sup>6</sup>) ≅ 141,151,800 </b>(joules)</i><br />let's check the units to make sure we get joules, the units of work<br /><i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup>·kg·kg·m<sup>−1</sup> = kg·m<sup>2</sup>·sec<sup>−2</sup> = N·m = J</i><br /><br />The final speed <i><b>V</b></i> can be calculated by equating this amount of work and kinetic energy of an asteroid:<br /><i><b>V<sup>2</sup> ≅ 2·6.67408·10<sup>−11</sup>·7.34767309·10<sup>22</sup> <span style="font-size: medium;">/</span>(1.7371·10<sup>6</sup>) ≅ 5646072</b></i><br />let's check the units to make sure we get the square of speed units<br /><i>m<sup>3</sup>·kg<sup>−1</sup>·sec<sup>−2</sup>·kg·m<sup>−1</sup> = m<sup>2</sup>·sec<sup>−2</sup> = (m/sec)<sup>2</sup></i><br /><br />From this the speed of an asteroid falling from infinity onto Moon's surface is<br /><i><b>V ≅ √<span style="text-decoration-line: overline;">5646072</span> ≅ 2376 </b>(m/sec)</i><br />or about <i><b>2.4 km/sec</b></i>.<br /><br />Incidentally, this is the so-called <i>escape speed</i> from the Moon, the initial speed needed for an object to leave the gravitational field of the Moon. A stone, thrown perpendicularly to the surface of the Moon with an initial speed less than that will go for certain distance away from the Moon, but then it will be brought back by the Moon's gravitation. Only if the initial speed is equal or exceeds the one above, the distance an object will go will be infinite, that is the object will leave the gravitational field of the Moon.<br /><br />Let's do similar calculations for the Earth, using the same assumptions, the same asteroid and the same units of measurement.<br /><i><b>M = 5.972·10<sup>24</sup></b> kg</i><br /><i><b>r = 6.371·10<sup>6</sup></b> m</i><br /><i><b>W ≅ 6.67408·10<sup>−11</sup>·5.972·10<sup>24</sup>·50 <span style="font-size: medium;">/</span>(6.371·10<sup>6</sup>) ≅ 3,128,049,424 </b>(joules)</i><br /><br /><i><b>V<sup>2</sup> ≅ 2·6.67408·10<sup>−11</sup>·5.972·10<sup>24</sup> <span style="font-size: medium;">/</span>(6.371·10<sup>6</sup>) ≅ 125121977</b> (m/sec)<sup>2</sup></i><br />From this the speed of an asteroid falling from infinity onto Earth's surface is<br /><i><b>V≅√<span style="text-decoration-line: overline;">125121977</span>≅11186</b> m/sec</i><br />or about <i><b>11.2</b> km/sec</i>.<br /><br />This is also the <i>escape speed</i> needed to fly away from Earth's gravitational field.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0