tag:blogger.com,1999:blog-37414104180967168272020-01-23T13:33:20.537-08:00Unizor - 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.comBlogger395125tag: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.com0tag:blogger.com,1999:blog-3741410418096716827.post-64974552626984366942019-09-16T19:28:00.001-07:002019-09-16T19:28:28.295-07:00Unizor - Physics4Teens - Energy - Gravitational Field - Problems<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/37caTtBahno" width="480"></iframe><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><u>Problems on Gravity</u><br /><br /><i>Problem 1</i><br />Gravitational potential of a spherical gravitational field around a point-mass <i><b>M</b></i> at a distance <i><b>r</b></i> from it is defined as the work performed by gravity to bring a probe object of a unit mass from infinity to this point and is expressed as<br /><i><b>V<sub>r</sub> = −G·M <span style="font-size: medium;">/</span>r</b></i><br />Why is this formula independent of trajectory of a probe object or its exact final position relative to the point-mass <i><b>M</b></i>, but only on a distance itself from the source of gravity?<br /><br /><i>Solution</i><br />Any movement can be represented as infinitely many infinitesimal displacements, combined together into a trajectory.<br />In our three-dimensional world the force and an infinitesimal displacement of a probe object are vectors, so the infinitesimal work <i>d<b>W</b></i> performed by the force of gravity <i><b><span style="text-decoration-line: overline;">F</span></b></i> during the movement of a probe object, described by the infinitesimal displacement <i><span style="text-decoration-line: overline;">d<b>S</b></span></i>, is a scalar product of these two vectors:<br /><i>d<b>W = <span style="text-decoration-line: overline;">F</span>·</b><span style="text-decoration-line: overline;">d<b>S</b></span></i><br />Note that the vector of gravitational force <i><b><span style="text-decoration-line: overline;">F</span></b></i> is always directed towards the source of gravity.<br />Since a displacement vector <i><span style="text-decoration-line: overline;">d<b>S</b></span></i> can be represented as a sum of radial (towards the source of gravity) <i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub></b></i> and tangential (perpendicular to radius) <i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i> components, the above expression for a differential of work can be written as<br /><i>d<b>W = <span style="text-decoration-line: overline;">F</span>·</b></i><b>(<i></i></b><i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub> + </b><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i><b>)<i> =<br />= <span style="text-decoration-line: overline;">F</span>·</i></b><i><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>r</sub> + <span style="text-decoration-line: overline;">F</span>·</b><span style="text-decoration-line: overline;">d<b>S</b></span><b><sub>t</sub></b></i><br />The second component in the above expression is a scalar product of two perpendicular vectors and is equal to zero. That's why we can completely ignore tangential movements, when calculating the work done by a central gravitational field, as not contributing to the amount of work. The total amount of work will be the same as if our probe object moved along a straight line towards the source of gravity and stopped at a distance <i><b>r</b></i> from it.<br /><br /><i>Problem 2</i><br />Given two point-masses of mass <i><b>M</b></i> each, fixed at a distance <i><b>2R</b></i> from each other.<br />Prove that the gravitational potential of a gravitational field produced by both of them at each point on a perpendicular bisector between them equals to a sum of individual gravitational potentials of these point-masses at this point, as if they were the only source of gravitation. In other words, prove that gravitational potential is additive in this case.<br /><br /><i>Solution</i><br />Let's draw a diagram of this problem (you can download it to display in a bigger format).<br /><img src="http://www.unizor.com/Pictures/Gravity2Masses.png" style="height: 100px; width: 200px;" /><br />Our two point-masses are at points <i>A</i> and <i>B</i>, the probe object is at point <i>D</i> on a perpendicular bisector of a segment <i>AB</i> going through point <i>C</i>.<br />The force of gravity towards point <i>A</i> is a segment <i>DE</i>, the force of gravity towards point <i>B</i> is a segment <i>DF</i>.<br />We will calculate the potential of a combined gravitational field of two point-masses at point <i>D</i>, where the probe object is located.<br />Let's assume that the segment <i>CD</i> equals to <i><b>h</b></i>.<br />The magnitude of each gravitational force equals to<br /><i><b>F = G·M·m <span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)</b></i><br />Represent each of these forces as a sum of two vectors, one (green on a drawing) going vertically along the bisector <i>CD</i>, another (red) going horizontally parallel to <i>AB</i>.<br />Vertical components of these two forces will add to each other, as equal in magnitude and similarly directed downwards on a drawing, while horizontal ones will cancel each other, as equal in magnitude and opposite in direction to each other. So, the combined force acting on a probe object is a sum of vertical components of gravitational forces with a magnitude<br /><i><b>F<sub>tot</sub> = 2·G·M·m·sin(φ)<span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)</b></i><br />Since <i><b>sin(φ) = CD/AD</b></i>,<br /><i><b>sin(φ) = h <span style="font-size: medium;">/</span></b></i>[(<i><b>h<sup>2</sup>+r<sup>2</sup></b></i>)<i><b><sup>1/2</sup></b></i>]<br /><i><b>F<sub>tot</sub> = 2·G·M·m·h <span style="font-size: medium;">/</span>(h<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />If the gravitational field pulls a probe object along the perpendicular bisector of a segment <i>AB</i> from infinity to a distance <i><b>h</b></i> from the segment, the magnitude of a combined force of gravity, as a function of a distance from the segment <i><b>x</b></i> is changing, according to a similar formula:<br /><i><b>F<sub>tot</sub>(x) = 2·G·M·m·x <span style="font-size: medium;">/</span>(x<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />To calculate work performed by a gravitational field pulling a probe object from infinity to height <i><b>h</b></i> above the segment <i>AB</i>, we have to integrate<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span><sub>[∞;h]</sub>F<sub>tot</sub>(x)·</b>d<b>x</b></i><br />It's supposed to be negative, since the direction of a force is opposite to a positive direction of the coordinate axis, we will take it into account later.<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span>2GMm·x·</b>d<b>x <span style="font-size: medium;">/</span>(x<sup>2</sup>+r<sup>2</sup>)<sup>3/2</sup></b></i><br />(within the same limits of integration [<i><b>∞;h</b></i>])<br />This integral can be easily calculated by substituting<br /><i><b>y=x<sup>2</sup>+r<sup>2</sup></b></i>,<br /><i><b>2·x·</b>d<b>x = </b>d<b>y</b></i>,<br />infinite limit of integration remaining infinite and the <i><b>x=h</b></i> limit transforming into <i><b>y=h<sup>2</sup>+r<sup>2</sup></b></i>. Now the work expression is<br /><i><b>W<sub>tot</sub> = <span style="font-size: large;">∫</span>G·M·m·y<sup>−3/2</sup>·</b>d<b>y</b></i><br />with limits from <i><b>y=∞</b></i> to <i><b>y=h<sup>2</sup>+r<sup>2</sup></b></i>.<br />The indefinite integral (anti-derivative) of <i><b>y<sup>−3/2</sup></b></i> is <i><b>−2·y<sup>−1/2</sup></b></i>.<br />Therefore, the value of integral and the work are<br /><i><b>W<sub>tot</sub> = −2·G·M·m·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />For a unit mass <i><b>m=1</b></i> this work is a gravitational potential of a combined gravitational field produced by two point-masses on a distance <i><b>h</b></i> from a midpoint between them along a perpendicular bisector<br /><i><b>V<sub>tot</sub> = −2·G·M·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />At the same time, the gravitational potential of a field produced by each one of the point-masses, considered separately, equals to<br /><i><b>V<sub>single</sub> = −G·M·(h<sup>2</sup>+r<sup>2</sup>)<sup>−1/2</sup></b></i><br />As we see, the gravitational potential of two point-masses equals to a sum of gravitational potential of each of them, considered separately.<br /><b>IMPORTANT NOTE</b><br />With more cumbersome calculations this principle can be proven for any two (not necessarily equal) point-masses at any point in space (not necessarily along the perpendicular bisector). This principle means that <b>gravitational potential is additive</b>, that is the <u>gravitational potential of any set of objects at any point in space equals to sum of their individual gravitational potentials</u>.<br /><br /><i>Problem 3</i><br />Express mass <i><b>M</b></i> of a spherical planet in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface.<br /><br /><i>Solution</i><br />Let <i><b>m</b></i> be a mass of a probe object lying on a planet's surface.<br />According to the Newton's 2nd Law, its weight is<br /><i><b>P = m·g</b></i><br />According to the Universal Law of Gravitation, the force of gravitation between a planet and a probe object is<br /><i><b>F<sub>gravity</sub> = G·M·m <span style="font-size: medium;">/</span>R<sup>2</sup></b></i><br />Since the force of gravitation is the weight <i><b>F<sub>gravity</sub> = P</b></i>,<br /><i><b>m·g = G·M·m <span style="font-size: medium;">/</span>R<sup>2</sup></b></i><br />from which<br /><i><b>M = g·R<sup>2</sup> <span style="font-size: medium;">/</span>G</b></i><br /><br /><i>Problem 4</i><br />Express gravitational potential <i><b>V<sub>R</sub></b></i> of a spherical planet on its surface in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface.<br /><br /><i>Solution</i><br />From the definition of a gravitational potential on a distance <i><b>R</b></i> from a source of gravity<br /><i><b>V<sub>R</sub> = −G·M <span style="font-size: medium;">/</span>R</b></i><br />Using the expression of the planet's mass in terms of its radius <i><b>R</b></i> and a free fall acceleration <i><b>g</b></i> on its surface (see above),<br /><i><b>M = g·R<sup>2</sup> <span style="font-size: medium;">/</span>G</b></i><br />Substituting this mass into a formula for potential,<br /><i><b>V<sub>R</sub> = −G·g·R<sup>2</sup> <span style="font-size: medium;">/</span>(G·R) = −g·R</b></i><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-68786613128947835842019-09-10T04:52:00.002-07:002019-09-10T04:53:55.892-07:00Unizor - Physics4Teens - Energy - Energy of Gravitational Field - Gravitational Field<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/XFuuNzOnO_Y" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Gravitational Field</u><br /><br />Studying <i>forces</i>, we have paid attention to a force of attraction, that exists between any material objects, the <i>force of gravity</i>.<br/>For example, if a comet from outer space flies not far from a Sun, it is attracted by Sun and changes its straight line trajectory.<br/><br/>In Mechanics we used to see the force as something between the objects touching each other, like a man pushing a wagon. In case of gravity the force obviously exists, but it acts on a distance, in "empty" space.<br/>In Physics this concept of force acting on a distance is described by a term <i>field</i>. Basically, <i>field</i> is the area in space where some force acts on all objects or only on objects that have specific property. The force in this case depends on a point in space and an object that experiences this force and, as a result of the action of force, changes its movement.<br/><br/><i>Gravitational field</i> exists around any material object (the source object of a field) and acts as an attraction towards this source object, experienced by any other material object (probe object) positioned in this field.<br/>As described in the "Gravity, Weight" chapter of "Mechanics" part of this course, the magnitude of the <i>gravitational force</i> <i><b>F</b></i> is proportional to a product of masses of a source object and a probe object, <b><i>M</i></b> and <b><i>m</i></b>, and it is inversely proportional to a square of a distance <b><i>r</i></b> between these objects:<br/><b><i>F = G·M·m <font size=4>/</font>r²</i></b><br/>where <b><i>G</i></b> - a constant of proportionality, since the units of force (N - newtons) have been defined already, and we want to measure the gravitational force in the same units as any other force.<br/><br/>The direction of the <i>gravitational force</i> acting on a probe object is towards the source object.<br/><br/>Let's return to our example of a comet flying not far from the Sun and, being attracted to the Sun, changing its trajectory. Obviously, to change the trajectory, some energy must be spent. So, we conclude that <i>gravitational field</i> has certain amount of energy at each point that it spends by applying the <i>force</i> onto a probe object.<br/><br/>To quantify this, assume that the source of gravity is a point mass <i><b>M</b></i> fixed at the origin of coordinates. Position a probe object of mass <i><b>m</b></i> at coordinates {<i>r<sub>1</sub>,0,0</i>} and let it go. The force of gravity will cause the motion of this probe object towards the center of gravity, the origin of coordinates, so the movement will be along the X-axis. Let the ending position of the probe object be {<i>r<sub>2</sub>,0,0</i>}, where <i>r<sub>2</sub></i> is smaller then <i>r<sub>1</sub></i>. Let <i><b>x</b></i> be a variable X-coordinate (distance to the origin).<br/><br/>According to the Universal Law of Gravitation, the force of attraction of a probe object towards the source of a gravitational field at distance <i><b>x</b></i> from the origin equals to<br/><i><b>F = −G·M·m <font size=4>/</font>x²</b></i><br/>where minus in front of it signifies that this force is directed opposite to increasing the X-coordinate.<br/>This force causes the motion and, therefore, does some work, moving a probe object from point {<i>r<sub>1</sub>,0,0</i>} to point {<i>r<sub>2</sub>,0,0</i>} along the X-axis. To calculate the work done by this variable force, we can integrate <i><b>F·</b>d<b>x</b></i> from <i><b>x=r<sub>1</sub></b></i> to <i><b>x=r<sub>2</sub></b></i>:<br/><i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> = <font size=5>∫</font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub>F·</b>d<b>x =<br/>= −<font size=5>∫</font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub>G·M·m·</b>d<b>x <font size=4>/</font>x² =<br/>= G·M·m <font size=4>/</font>x<font size=5></b>|<b></font><sub>[r<sub>1</sub>,r<sub>2</sub>]</sub> =<br/>= G·M·m <font size=4>/</font>r<sub>2</sub> − G·M·m <font size=4>/</font>r<sub>1</sub> =<br/>= <i>(</i>G·M <font size=4>/</font>r<sub>2</sub> − G·M <font size=4>/</font>r<sub>1</sub></i>)<i>·m</b></i><br/><br/>The expression<br/><i><b>V(r) = −G·M<font size=4>/</font>r</b></i><br/>is called <i>gravitational potential</i>.<br/>It's a characteristic of a <i>gravitational field</i> sourced by a point mass <i><b>M</b></i> at a distance <i><b>r</b></i> from a source.<br/>It equals to work needed by <b>external forces</b> to bring a probe object of mass <i><b>m=1</b></i> to a point at distance <i><b>r</b></i> from a source of the field <b>from infinity</b>.<br/>Indeed, set <i><b>m=1</b></i>, <i><b>r<sub>1</sub>=∞</b></i> and <i><b>r<sub>1</sub>=r</b></i> in the above formula for work <i><b>W<sub>[r<sub>1</sub>,r<sub>2</sub>]</sub></b></i> and take into consideration that gravitational field "helps" external forces to move a probe object, so the external forces spend negative amount of energy.<br/><br/>Using this concept of <i>gravitational potential</i> <i><b>V(r)</b></i>, we can state that, to move a probe object of a unit mass from distance <i><b>r<sub>1</sub></b></i> relative to a source of gravitational field to a distance <i><b>r<sub>2</sub></b></i> relative to its source in the gravitational field with <i>gravitational potential</i> <i><b>V(r)</b></i>, we have to spend the amount of energy equal to <i><b>V(r<sub>1</sub>)−V(r<sub>2</sub>)</b></i>.<br/>For a probe object of any mass <i><b>m</b></i> this amount should be multiplied by <i><b>m</b></i>.<br/>If <i><b>r<sub>2</sub></b></i> is greater than <i><b>r<sub>1</sub></b></i>, that is we move a probe object further from the source of gravity, working against the gravitational force, this expression is positive, we have to apply effort against the force of gravity. In an opposite case, when <i><b>r<sub>2</sub></b></i> is smaller than <i><b>r<sub>1</sub></b></i>, that is we move closer to a source of gravity, the gravitational force "helps" us, we don't have to apply any efforts, and our work is negative.<br/><br/>Therefore, an expression <i><b>E<sub>P</sub>=m·V(r)</b></i> represents <i>potential energy</i> of a probe object of mass <i><b>m</b></i> at a distance <i><b>r</b></i> from a source of a gravitational field with <i>gravitational potential</i> <i><b>V(r)</b></i>.<br/><br/>A useful consequence from a concept of a <i>gravitational potential</i> is that the <i>force of gravity</i> can be expressed as the derivative of the <i>gravitational potential</i>.<br/><b><i>F = G·M·m <font size=4>/</font>r² = m·</b>d<b>V(r)/</b>d<b>r</i></b><br/>which emphasizes the statement that the <i>gravitational potential</i> is a characteristic of a field itself, not its source.<br/>We, therefore, can discuss <i>gravitational field</i> as an abstract concept defined only by the function called <i>gravitational potential</i>.<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-54619554922461790012019-08-18T06:19:00.001-07:002019-09-10T04:42:09.654-07:00Unizor - Physics4Teens - Energy - Energy of Nucleus - Fusion<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/XcmHR2eISpc" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Nucleus Fusion</u><br /><br /><br /><br /><i>Fusion</i> is a <i>nuclear reaction</i>, when light nuclei are brought together and combined into a heavier ones.<br /><br />The reason for this reaction to release the energy is the difference <br />between amount of energy needed to overcome the repulsion between nuclei<br /> because they have the same positive electric charge (this energy is <br />consumed by <i>fusion</i>) and the potential energy released by <i>strong forces</i>, when the formation of a combined nucleus occurs (this energy is released by <i>fusion</i>).<br /><br /><u>The former is less than the latter</u>.<br /><br /><br /><br />When the light nuclei are fused into a heavier one, the excess of potential energy of <i>strong forces</i>, released in the process of <i>fusion</i>,<br /> over the energy needed to squeeze together protons against their <br />repulsion is converted into thermal and electromagnetic field energy.<br /><br /><br /><br />Analogy to this process can be two magnets separated by a spring.<br /><img src="http://www.unizor.com/Pictures/FusionAnalogy.png" style="height: 400px; width: 200px;" /><br /><br />The magnets represent two separate protons, the magnetic force of attraction between them represents the <i>strong force</i><br /> that is supposed to hold the nucleus together, when these particles are<br /> close to each other, the spring represents the electrical repulsive <br />force between them, acting on a larger distance, as both are positively <br />charged.<br /><br />It's known that magnetic force is inversely proportional to a square of a<br /> distance between objects, while the resistance of a spring against <br />contraction obeys the Hooke's Law and is proportional to the length of <br />contraction.<br /><br />On the picture magnets are separated. To bring them together, we have to<br /> spend certain amount of energy to move against a spring that resists <br />contraction. But the magnetic attraction grows faster then the <br />resistance of the spring, so, at some moment this attraction will be <br />greater than the resistance of a spring. At this moment nothing would <br />prevent magnets to fuse.<br /><br /><br /><br />As is in the above analogy, if we want to fuse two protons, we have to bring them together sufficiently close for <i>strong forces</i> to overtake the repulsion of their positive charges.<br /><br /><br /><br />Consider the following nuclear reaction of <i>fusion</i>.<br /><br />One nucleus of hydrogen isotope <i>deuterium</i> <b><sup>1</sup>H<sup>2</sup></b> with atomic mass 2 contains one proton and one neutron.<br /><br />One nucleus of hydrogen isotope <i>tritium</i> <b><sup>1</sup>H<sup>3</sup></b> with atomic mass 3 contains one proton and two neutrons.<br /><br />If we force these two nuclei to fuse, they will form a nucleus of <i>helium</i> <b><sup>2</sup>He<sup>4</sup></b> and releasing certain amount of energy:<br /><br /><b><sup>1</sup>H<sup>2</sup> + <sup>1</sup>H<sup>3</sup> = <sup>2</sup>He<sup>4</sup> + <sup>0</sup>n<sup>1</sup></b><br /><br /><br /><br />It's not easy to overcome the repulsion of protons. High temperature and<br /> pressure, like in the core of our Sun, are conditions where it happens.<br /> On Earth these conditions are created in the nuclear bomb, using the <br />atomic bomd to achieve proper amount of heat and pressure, thus creating<br /> an uncontrlled <i>fusion</i>.<br /><br />Controlled nuclear reaction of <i>fusion</i> is what scientists are working on right now. So far, it's still in the experimental stage.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-43415492098950650422019-08-12T18:50:00.001-07:002019-08-12T18:50:19.438-07:00Unizor - Physics4Teens - Energy - Energy of a Nucleus - Fission<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/6WdOzCFKxxw" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br/><br/><u>Nucleus Fission</u><br /><br /><i>Fission</i>, first of all, is a <i>nuclear reaction</i>, when heavier nuclei are split into lighter ones.<br/>The reason for this reaction to release the energy is the difference between amount of energy needed to break <i>strong forces</i> that hold the nucleus together (this energy is consumed by <i>fission</i>) and amount of potential energy in positively charged and repelling protons inside nucleus (this energy is released by <i>fission</i>).<br/><u>The former is less than the latter</u>.<br/><br/>When the heavy nucleus is broken into parts, the excess of potential energy of squeezed together protons against their repelling force over the energy of strong forces that keep nucleus together is converted into thermal and electromagnetic field energy.<br/><br/>Analogy of this is a spring squeezed tightly and held in this position by a thread. A thread plays the role of <i>strong forces</i>, while a potential energy of a squeezed spring plays the role of protons kept close to each other by a this force. When you cut a thread, the spring will release the potential energy, similarly to protons repelling from each other.<br/><br/>Electrically positively charged protons repel each other and, at the same time, are bonded together by <i>strong forces</i> inside a nucleus. At the same time neutrons are also bonded by <i>strong forces</i> among themselves and with protons without any repulsion.<br/>So, the more neutrons the nucleus has - the stronger it is. Neutrons only add "bonding material" to a nucleus without adding any repelling forces that work against the nucleus' stability.<br/><br/>Uranium-238 with 92 protons and 146 neutrons (<sup>92</sup>U<sup>238</sup>) naturally occurs on Earth and is relatively stable.<br/>Uranium-235 with the same 92 protons and 143 neutrons (<sup>92</sup>U<sup>235</sup>) has less "bonding material" (less neutrons) and is more susceptible to fission.<br/><br/>All it takes to break the nucleus of <sup>92</sup>U<sup>235</sup> is a little "push" from outside, which can be accomplished by bombarding it with neutrons. In the process of <i>fission</i>, caused by hitting a nucleus of <sup>92</sup>U<sup>235</sup> with a neutron, it can transforms into Barium-141 with 56 protons and 85 neutrons <sup>56</sup>Ba<sup>141</sup>, Krypton-92 with 36 protons and 56 neutrons <sup>36</sup>Kr<sup>92</sup> and 3 free neutrons.<br/>As we see, the numbers of protons is balanced (input: 92, output: 56 and 36), as well as a number of neutrons (input: 1 free hitting neutron and 143 in a nucleus of <sup>92</sup>U<sup>235</sup> total 144, output: 85 in a nucleus <sup>56</sup>Ba<sup>141</sup>, 56 in a nucleus of <sup>36</sup>Kr<sup>92</sup> and 3 new free neutrons total 144).<br/><br/>Let's express this reaction in a formula (letter <i><b>n</b></i> denotes a neutron):<br/><i><b><sup>0</sup>n<sup>1</sup> + <sup>92</sup>U<sup>235</sup> =<br/><sup>56</sup>Ba<sup>141</sup> + <sup>36</sup>Kr<sup>92</sup> + 3·<sup>0</sup>n<sup>1</sup></b></i><br/><br/>What's interesting in this reaction is that it not only produces energy because we break a heavy nucleus into lighter ones, but also that it produces 2 new neutrons that can bombard other atoms, causing a <b>chain reaction</b> and, potentially, an explosion (atomic bomb). However, if we absorb extra neutrons, it will allow to slowly release of nuclear energy (nuclear power stations).<br/><br/>Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-69734146366792166512019-08-05T18:46:00.001-07:002019-08-05T18:46:22.756-07:00Unizor - Physics4Teens - Energy - Energy of a Nucleus<iframe allowfullscreen="" frameborder="0" height="270" src="https://www.youtube.com/embed/zi6gTkhDIds" width="480"></iframe><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Energy of Nucleus</u><br /><br /><br /><br />In this lecture we will analyze the energy aspect of nucleus - the central part of an atom.<br /><br /><br /><br />By now we have built a pyramid of energy types, related to the depth of our view inside the matter.<br /><br /><br /><br />First, we analyzed the <i>mechanical energy</i> - the energy of moving macro-objects.<br /><br /><br /><br />Our next view deep into the world of macro-objects uncovered the <i>molecules</i> - the smallest parts of macro-objects that retain their characteristics. The movement of these molecules was the source of <i>thermal energy</i>, which we often call the <i>heat</i>.<br /><br /><br /><br />Next step inside the molecules uncovered <i>atoms</i>, as the molecules'<br /> components. There are about 100 types of atoms and their composition <br />inside the molecules creates all the thousands of different molecules. <i>Chemical reactions</i><br /> change the composition of atoms in molecules, thereby creating new <br />molecules from the atoms of old molecules. This process broke some <br />inter-atomic bonds and created the new ones and is the source of <i>chemical energy</i>.<br /><br /><br /><br />Now we look deep inside the atoms and find there 3 major <b>elementary particles</b> - electrically positively charged <i>protons</i> and electrically neutral <i>neutrons</i> inside a small but heavy <i>nucleus</i> and electrically negatively charged <i>electrons</i>,<br /> circulating around nucleus on different orbits. For electrically <br />neutral atoms the numbers of protons and electrons are equal. <i>Nuclear energy</i> is hidden inside the nucleus and is the subject of this lecture.<br /><br /><br /><br />The first question we would like to answer is "What holds nucleus, its <br />protons and neutrons, together, considering protons, as electrically <br />positively charged particles must repel each other?"<br /><br /><br /><br />The answer is simple. There are other forces in the Universe, not only <br />electrostatic ones, that act in this case. These intra-nucleus forces <br />that hold the nucleus together are called <i><b>strong forces</b></i>. They are <b>strong</b><br /> because they are the source of attraction between the protons that is <br />stronger than electrostatic repelling. However, these strong forces act <br />only on a very small distance, comparable to the size of a nucleus <br />inside an atom. For example, at a distance <i>10<sup>−15</sup>m</i> the strong force is more than 100 times stronger than electrostatic one.<br /><br /><br /><br />If, by regrouping <i>protons</i> and <i>neutrons</i>, we will be able to create different <i>atoms</i> (inasmuch as regrouping <i>atoms</i> in <b>chemical reaction</b> we create new <i>molecules</i>), a new source of energy, based on <b>strong forces</b>, the <i>nuclear energy</i>, can be uncovered in the course of <b>nuclear reaction</b>.<br /><br /><br /><br />There is another form of <i>nuclear reaction</i> related to <br />transformation of elementary particles. Under certain circumstance a <br />neutron inside a nucleus can transform into proton and, to keep the <br />total electrical charge in balance, it emits an electron. This reaction <br />is called <i>beta-decay</i> and it also produces energy in the form of electromagnetic waves of very high frequency (<i>gamma-rays</i>).<br /><br /><br /><br />Nuclear reactions are a very powerful source of nuclear energy, which is<br /> so much more powerful than other types of energy, that, if misused, it <br />might represent a danger for life on our planet.<br /><br /><br /><br />There is a clear analogy between nuclear and chemical reactions.<br /><br />What happens with atoms in the chemical reaction, happens with protons <br />and neutrons in nuclear reaction. Some atomic bonds break in a chemical <br />reaction, some are created. Some nuclear bonds between protons and <br />neutrons break in a nuclear reaction, some are created.<br /><br /><br /><br />Sometimes the chemical reaction happens by itself, as long as <br />participating substances are close together, but sometimes we have to <br />initiate it, like lighting methane gas with a spark or a flame of a <br />match to initiate continuous burning.<br /><br />Similar approach is valid for nuclear reaction. Sometimes it happens by <br />itself, but sometimes it should be started, like bombarding the nucleus <br />with neutrons, after which it continues by itself.<br /><br /><br /><br />Here is an interesting fact.<br /><br />Physicists have measured the masses of protons, neutrons and many <br />different nuclei that contain these protons and neutrons and have <br />discovered that the sum of masses of individual protons and neutrons is <br />greater than the mass of a nucleus that contain these exact particles.<br /><br />For example,<br /><br />mass of proton is 1.0072766 atomic mass units or 1.6726·10<sup>-27</sup>kg,<br /><br />mass of neutron is 1.0086654 atomic mass units or 1.6749·10<sup>-27</sup>kg.<br /><br />At the same time, mass of deuterium nucleus, that contains 1 proton and 1<br /> neutron is 2.0135532 atomic mass units, which is smaller than the sum <br />of masses of proton and neutron (1.0072766 + 1.0086654 = 2.015942).<br /><br />This so-called "mass defect" is directly related to nuclear energy - the energy of <i>strong forces</i> that hold the nucleus together.<br /><br /><br /><br />A simplified explanation of this effect is based on the law of energy <br />conservation. Consider the force of gravity between a planet and an <br />object above its surface. The object has certain potential energy and, <br />if dropped to the ground, this potential energy transforms into other <br />forms, like kinetic, thermal etc.<br /><br /><br /><br />Similarly, if we consider two independent neutrons (or neutron and <br />proton, or two protons) on a very small distance from each other, but <br />not forming a nucleus, there is a potential energy of the <i>strong forces</i><br /> acting between them. If we let these two particles to form a nucleus, <br />analogously to an object falling towards the surface of a planet, this <br />potential energy should be transformed into other forms, like thermal.<br /><br /><br /><br />Now the Theory of Relativity comes to play, that has established the equivalence of <b>mass</b> and <b>energy</b> by a famous formula <i><b>E=m·c²</b></i>.<br /> According to this equivalence, if some energy is released during the <br />formation of a nucleus from individual protons and neutrons, there must <br />be certain amount of mass released associated with this energy. That is <br />the explanation of "mass defect".<br /><br /><br /><br />It should be noted that to form a nucleus of deuterium from 1 proton and<br /> 1 neutron is easier than to form a nucleus that contains more than one <br />proton, because electrostatic repulsion between positively charged <br />protons prevents their bonding. So, to bring protons sufficiently close <br />to each other for <i>strong forces</i> to overcome the electrostatic <br />repulsion, we have to spend some energy. The net energy released by <br />forming a nucleus from protons and neutrons is the difference between <br />the energy released from <i>strong forces</i> taking hold of these particles inside a nucleus and the energy consumed to overcome repulsion of protons.<br /><br /><br /><br />Actually, as we attempt to form bigger nuclei, the energy we have to <br />spend to overcome electrostatic repulsion forces become greater than <br />amount of energy released by forming a nucleus. This border line is <br />approximately around the nucleus of iron <i><b>Fe</b></i>. Forming iron <br />and heavier elements from protons and electrons is a process that <br />consumes more energy than releases. These heavier nuclei will produce <br />energy, if we reverse the procedure, breaking them into individual <br />protons and neutrons.<br /><br /><br /><br />The mechanisms described above are used in nuclear reactors and atomic <br />bomb, where heavier elements are broken into lighter ones (fission), <br />releasing energy, and in hydrogen bomb, where lighter elements are <br />bonded together to release the energy (fusion).Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-20467534891672103982019-07-29T14:09:00.001-07:002019-07-29T14:09:21.377-07:00Unizor - Physics4Teens - Energy - Atoms and Chemical Reactions - Interat...<br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Interatomic Bonds</u><br /><br /><br /><br />Atoms in a molecule are bonded together to form a stable chemical substance or compound.<br /><br />The mechanism of bonding is quite complex and different for different <br />molecules. In fact, the complexity of these bonds is outside of the <br />scope of this course. However, certain basic knowledge about molecular <br />bonding and molecular structure is necessary to understand the following<br /> lecture, where we will make certain calculations related to energy <br />produced or consumed in chemical reactions.<br /><br /><br /><br />The key to a mechanism of bonding atoms into molecules lies in an internal structure of atoms.<br /><br />For our purposes we can consider the orbital model of atom as consisting<br /> of electrically positive nucleus and electrically negative electrons <br />circulating on different orbits around a nucleus. <b>This is only a model</b>,<br /> not an exact representation of what's really happening inside the atom,<br /> but this model gives relatively good results that correspond to some <br />simple experiments.<br /><br /><br /><br />Two different particles can be found in a nucleus - positively charged <br />protons and electrically neutral neutrons. The number of protons inside a<br /> nucleus and electrons circulating on different orbits around a nucleus <br />should be the same for electrically neutral atoms in their most common <br />state.<br /><br /><br /><br />For reasons not well understood by many physicists, each orbit can have <br />certain maximum number of electrons that can circulate on it without <br />"bumping" into each other. The higher the orbit - the more electrons it <br />can hold. The lowest orbit can hold no more than 2 electrons, the next -<br /> no more than 8, the next - no more than 14 etc.<br /><br /><br /><br />Consider a few examples.<br /><br /><br /><br />1. Let's consider the structure of a simplest molecule - the molecule of<br /> hydrogen, formed by two atoms of hydrogen. Each hydrogen atom has one <br />electron on the lowest orbit around a nucleus. The maximum number of <br />electrons on this orbit is two, in which case the compound becomes much <br />more stable. So, two atoms of hydrogen grab each other and the two <br />electrons, each from its own atom, are shared by a couple of atoms, thus<br /> creating a stable molecule of hydrogen with symbol <i><b>H<sub>2</sub></b></i>. The bond between two atoms of hydrogen is formed by one pair of shared electrons, so structurally the molecule of hydrogen <i><b>H<sub>2</sub></b></i> can be pictured as<br /><br /><i><b>H−H</b></i>.<br /><br /><br /><br />2. Atom of oxygen has 8 electrons - 2 on the lowest orbit and 6 on the <br />next higher one. The next higher orbit is stable when it has 8 <br />electrons. So, two atoms of oxygen are grabbing each other and share 2 <br />out of 6 electrons on the outer orbit with another atom. So, each atom <br />has 4 "personal" electrons, 2 electrons that it shares with another atom<br /> and 2 electrons that the other atom shares with it. Thus, the orbit <br />becomes full, all 8 spots are filled. The bond between two atoms of <br />oxygen is formed by two pairs of shared electrons, so structurally the <br />molecule of oxygen <i><b>O<sub>2</sub></b></i> can be pictured as<br /><br /><i><b>O=O</b></i><br /><br />(notice double link between the atoms).<br /><br /><br /><br />3. Our next example is gas methane. Its molecule consists of one atom of<br /> carbon (6 electrons, 2 of them on the lowest orbit, 4 - on the next <br />one) and 4 atoms of hydrogen (1 electron on the lowest orbit of each <br />atom). Obviously, having only 4 electrons on the second orbit, carbon is<br /> actively looking for electrons to fill the orbit. It needs 4 of them to<br /> complete an orbit of 8 electrons. Exactly this it finds in 4 atoms of <br />hydrogen that need to complete their own lowest orbit. Sharing <br />electrons, one atom of carbon and 4 atoms of hydrogen fill their <br />corresponding orbits, thus creating a molecule of methane <i><b>CH<sub>4</sub></b></i> with can be pictured as<br /><br /><i><b> H<br /><br /> |<br /><br /> H−C−H<br /><br /> |<br /><br /> H</b></i><br /><br /><br /><br />4. Carbon dioxide molecule contains 1 atom of carbon, that needs 4 <br />electrons to complete its orbit, and 2 atoms of oxygen, each needs 2 <br />electrons to complete its orbit: <i><b>CO<sub>2</sub></b></i>. By <br />sharing 2 electrons from each atom of oxygen with 4 electrons from atom <br />of carbon they all fill up their outer orbit of electrons and become a <br />stable molecule, pictured as<br /><br /><i><b>O=C=O</b></i><br /><br />(notice double link between the atoms).<br /><br /><br /><br />5. Ethanol molecule contains 2 atoms of carbon, 1 atom of oxygen and 6 atoms of hydrogen connected as follows<br /><br /><i><b> H H<br /><br /> | |<br /><br /> H−C−C−O−H<br /><br /> | |<br /><br /> H H</b></i><br /><br />(notice single bond between atoms of carbon and oxygen in ethanol, while<br /> the bond between them in carbon dioxide has double link)<br /><br /><br /><br />6. Hydrogen peroxide molecule contains 2 atoms of hydrogen and 2 atoms of oxygen connected as follows<br /><br /><i><b>H−O−O−H</b></i><br /><br />(notice single bond between atoms of oxygen, not like in a molecule of oxygen)<br /><br /><br /><br />Numerous examples above illustrate that bonds between atoms can be <br />different, even between the same atoms in different molecules. That's <br />why it is important to understand the structure of molecules, how <br />exactly the atoms are linked and what kind of links exist between them. <br />This is the basis for calculation of the amount of energy produced or <br />consumed by chemical reactions that rearrange the atoms from one set of <br />molecules to another.<br /><br /><br /><br />Obviously, bonds <i><b>O−O</b></i> and <i><b>O=O</b></i> are different. <br />The first one is facilitated by one shared electron, the second one - by<br /> two. The amounts of energy, needed to break these bonds, are different <br />too. Therefore, when calculating the energy of chemical reaction, it's <br />important to understand the kind of bond between atoms in each separate <br />case.Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-66298229959949839832019-07-23T12:21:00.001-07:002019-07-23T12:21:16.716-07:00Unizor - Physics4Teens - Energy - Chemical Energy of Atomic Bonds<br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Energy of Atomic Bonds<br /><br />in Molecules</u><br /><br /><br /><br />In this lecture we will analyze the energy aspect of chemical reactions.<br /><br />Consider the reaction of burning of methane. This gas is used in regular<br /> gas stoves, so the reaction happens every time we cook something.<br /><br />A molecule of methane consists of one atom of carbon <i><b>C</b></i> and four atoms of hydrogen <i><b>H</b></i>, the chemical formula of methane is <i><b>CH<sub>4</sub></b></i>.<br /><br />You can imagine a molecule of methane as a tetrahedron, in its center is<br /> an atom of carbon and on each of its four vertices is an atom of <br />hydrogen.<br /><br />A molecule of oxygen, as we know, consists of two atoms of oxygen and has a chemical formula <i><b>O<sub>2</sub></b></i>.<br /><br /><br /><br />As a result of the reaction of burning of methane, water and carbon dioxide are produced, according to the following equation:<br /><br /><i><b>CH<sub>4</sub> + 2O<sub>2</sub> = 2H<sub>2</sub>O + CO<sub>2</sub></b></i><br /><br />So, during this reaction<br /><br />(a) four atomic bonds between carbon and hydrogen in one molecule of methane are broken,<br /><br />(b) one atomic bond in each molecule of oxygen (out of two) are broken,<br /><br />(c) two atomic bonds between hydrogen and oxygen in each molecule of water (out of two) are created,<br /><br />(d) two atomic bonds between carbon and oxygen in a molecule of carbon dioxide are created.<br /><br /><br /><br />Amounts of potential energy of the different atomic bonds are <br />experimentally determined, which would lead to calculation of the amount<br /> of chemical energy released (for exothermic) or consumed (by <br />endothermic) reaction.<br /><br /><br /><br />To make experiments to determine potential energy of the bonds inside a <br />molecule, we have to make experiments with known amounts of components <br />in chemical reaction. The reaction above includes one molecule of <br />methane and two molecules of oxygen. Obviously, we cannot experiment <br />with one or two molecules. The solution is to experiment with <b>proportional</b> amounts of components, say, 1 million of molecules of methane and 2 million of molecules of oxygen.<br /><br /><br /><br />To explain how to do this, we have to get deeper into atoms. Physics <br />models atoms as consisting of three kinds of elementary particles - <br />protons (electrically positively charged), neutrons (electrically <br />neutral) and electrons (electrically negatively charged). This is a <br />relatively simple model, that corresponds to most of experiments, though<br /> the reality is more complex than this. For our purposes we can view <br />this model of atom as a nucleus, that contains certain number of protons<br /> and neutrons, and a number of electrons circulating the nucleus on <br />different orbits.<br /><br /><br /><br />Electrons are very light relatively to protons and neutrons, so the mass<br /> of an atom is concentrated, mostly, in its nucleus. Protons and neutron<br /> have approximately the same mass, which is called <i>atomic mass unit</i>. So, the mass of an atom in <i>atomic mass units</i><br /> ("atomic weight") is equal to the number of protons and neutrons in its<br /> nucleus. This mass is known for each element of the Periodic Table of <br />Mendeleev, that is for each known atom.<br /><br />For example, it is determined that atom of hydrogen <i><b>H</b></i> has atomic weight of 1 atomic unit, atom of carbon <i><b>C</b></i> has atomic weight of 12, atom of oxygen <i><b>O</b></i> has atomic weight 16.<br /><br /><br /><br />Knowing atomic weights of atoms, we can calculate atomic weight of molecules. Thus, the atomic weight of a molecule of methane <i><b>CH<sub>4</sub></b></i> is 12+4=16. Atomic weight of a molecule of oxygen <i><b>O<sub>2</sub></b></i> is 16+16=32. Atomic weight of water <i><b>H<sub>2</sub>O</b></i> is 2+16=18.<br /><br /><br /><br />Now we can take components of any chemical reaction proportional to the <br />atomic weight of corresponding molecules, which will result in <br />proportional number of molecules. For example, not being able to <br />experiment with one molecule of methane <i><b>CH<sub>4</sub></b></i> and two molecules of oxygen <i><b>O<sub>2</sub></b></i>, we can experiment with <i><b>16 gram</b></i> of methane and <i><b>64 gram</b></i><br /> of oxygen, and the proportionality of the number of molecules will be <br />preserved - for each molecule of methane there will be two molecules of <br />oxygen.<br /><br /><br /><br />As you see, taking amount of any mono-molecular substance in grams <br />equaled to the atomic weight of the molecules of this substance (called a<br /> <i>mole</i>) assures taking the same number of molecules, regardless of the substance. This number is the Avogadro Number and is equal to <i><b>N=6.02214076·10<sup>23</sup></b></i>.<br /><br />Thus, one <i>mole</i> of methane <i><b>CH<sub>4</sub></b></i> (atomic weight of <i><b>C</b></i> is 12, atomic weight of <i><b>H</b></i> is 1) weighs 16g, one <i>mole</i> of silicon <i><b>Si<sub>2</sub></b></i> (atomic weight of <i><b>Si</b></i> is 14) weighs 28g, one <i>mole</i> of copper oxide <i><b>CuO</b></i> (atomic weight of <i><b>Cu</b></i> is 64, atomic weight of <i><b>O</b></i><br /> is 16) weighs 80g etc. And all those amounts of different substances <br />have the same number of molecules - the Avogadro number (approximately, <br />of course).<br /><br /><br /><br />The theory behind the atomic bonds inside a molecule is quite complex <br />and is beyond the scope of this course. Based on this theory and <br />experimental data, for many kinds of atomic bonds there had been <br />obtained an amount of energy needed to break these bonds, that is its <br />inner chemical energy.<br /><br />Thus, chemical energy of atomic bonds inside a mole of methane <i><b>CH<sub>4</sub></b></i><br /> is 1640 kilo-joules (because a molecule of methane has 4 bonds between <br />carbon and each atom of hydrogen, each bond at 410KJ), inside a molecule<br /> of oxygen <i><b>O<sub>2</sub></b></i> - 494 kilo-joules (1 bond between 2 atoms oxygen at 494KJ), inside a molecule of carbon dioxide <i><b>CO<sub>2</sub></b></i> is 1598 kilo-joules (2 bonds between carbon and each atom of oxygen, each 799KJ), inside a molecule of water <i><b>H<sub>2</sub>O</b></i> is 920 kilo-joules (2 bonds between oxygen and each atom of hydrogen, each 460KJ).<br /><br /><br /><br />Let's go back to methane burning:<br /><br /><i><b>CH<sub>4</sub> + 2O<sub>2</sub> = 2H<sub>2</sub>O + CO<sub>2</sub></b></i><br /><br />This chemical reaction converts 1 mole of methane (16g) and 2 moles of <br />oxygen (64g) into 1 mole of carbon dioxide (44g) and 2 moles of water <br />(36g).<br /><br />The energy we have to spend to break the atomic bonds of 1 mole of methane and 2 moles of oxygen, according to above data, is<br /><br /><i><b>E<sub>in</sub> = 1640 + 2·494 = 2628 KJ</b></i><br /><br />The energy we have to spend to break atomic bonds of 2 moles of water and 1 mole of carbon dioxide, according to above data, is<br /><br /><i><b>E<sub>out</sub> = 2·920 + 1598 = 3438 KJ</b></i><br /><br />The net energy is<br /><br /><i><b>E<sub>net</sub> = 2628 − 3438 = −810 KJ</b></i><br /><br />This net energy is the amount of thermal energy released by burning 16g <br />of methane, using 64g of oxygen, obtaining as a result 44g of carbon <br />dioxide and 36g of water.<br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0tag:blogger.com,1999:blog-3741410418096716827.post-52607433522978610102019-07-18T14:41:00.001-07:002019-07-18T14:41:54.070-07:00Unizor - Physics4Teens - Energy - Atoms<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><i>Notes to a video lecture on http://www.unizor.com</i><br /><br /><br /><br /><u>Atoms and Chemical Reaction</u><br /><br /><br /><br />Discussing <i>mechanical energy</i>, we analyzed the movement of objects.<br /><br />When talking about <i>thermal energy (heat)</i>, we had to go deeper inside the objects and analyzed the movement of <i>molecules</i>, the smallest parts of objects that retained the properties of objects themselves.<br /><br />Now we go even deeper, inside the molecules, in search of new kinds of energy.<br /><br /><br /><br />The components of <i>molecules</i> are called <i>atoms</i>. Currently there are more than 100 kinds of <i>atoms</i>, classified in the Mendeleev's Periodic Table.<br /><br /><br /><br />Different combinations of these <i>atoms</i> in different quantities make up all kinds of <i>molecules</i>, each with its own properties.<br /><br /><br /><br />In some cases a single <i>atom</i> makes up a <i>molecule</i>. For example, a single atom of iron (denoted by symbol <b>Fe</b>) makes up a molecule of iron.<br /><br />In some other cases a pair of <i>atoms</i> of the same type makes up a <i>molecule</i>. For example, two atoms of oxygen (denoted by symbol <b>O</b>) make up a molecule of oxygen (denoted by symbol <b>O<sub>2</sub></b>).<br /><br />In more complicated cases a few <i>atoms</i> of different types make up a <i>molecule</i>. For example, two atoms of hydrogen (denoted by symbol <b>H</b>) and one atom of oxygen (<b>O</b>) make up a molecule of water (<b>H<sub>2</sub>O</b>).<br /><br /><br /><br />One of the most complicated molecules that contains many elements in <br />different quantities is a molecule of protein that has about half a <br />million of atoms.<br /><br /><br /><br /><b>Chemical energy</b> is a <i>potential energy</i> of bonds between <i>atoms</i> that hold them together in a <i>molecule</i>.<br /><br /><b>Chemical reaction</b> is a process of re-arranging of <i>atoms</i> in a group of <i>molecules</i>, getting, as a result, a group of other <i>molecules</i>.<br /><br />During <b>chemical reactions</b> some bonds between atoms are broken and some are created. Therefore, the <u>energy might be either released or consumed</u> in the process of <b>chemical reaction</b>.<br /> This energy, stored in the molecules as potential energy of atomic <br />bonds and released or consumed during chemical reaction, is classified <br />as <i>chemical energy</i>.<br /><br /><br /><br />Let's consider a few examples of chemical energy.<br /><br /><br /><br />1. <i>Coal burning</i><br /><br />One molecule of carbon, that consists of one carbon atom <i><b>C</b></i>, and one molecule of oxygen, that consists of two oxygen atoms <i><b>O<sub>2</sub></b></i>, when brought together and lit up, will join into one molecule of carbon dioxide <i><b>CO<sub>2</sub></b></i><br /> in the process of burning. After the chemical reaction of burning is <br />initiated, it will maintain itself, as the process of burning produces a<br /> flame that lights up new molecules of carbon, joining them with oxygen.<br /><br />The chemical reaction<br /><br /><i><b>C + O<sub>2</sub> = CO<sub>2</sub></b></i><br /><br />is endothermic (consumes heat energy) in the very beginning, when we <br />have to light up the carbon, but, as soon as the reaction started, it <br />becomes exothermic, that is it produces heat energy, because the <br />potential energy of atoms inside molecules of carbon and oxygen together<br /> is greater than potential energy of atoms inside a molecule of carbon <br />dioxide.<br /><br /><br /><br />2. <i>Making water from hydrogen and oxygen</i><br /><br />Two molecules of hydrogen, each consisting of two hydrogen atom <i><b>H<sub>2</sub></b></i>, and one molecule of oxygen, that consists of two oxygen atoms <i><b>O<sub>2</sub></b></i>, when brought together and lit up, will join into two molecules of water <i><b>H<sub>2</sub>O</b></i><br /> in the process of hydrogen burning. After the chemical reaction of <br />burning is initiated, it will maintain itself, as the process of burning<br /> produces a flame that lights up new molecules of hydrogen, joining them<br /> with oxygen.<br /><br />The chemical reaction<br /><br /><i><b>2H<sub>2</sub> + O<sub>2</sub> = 2H<sub>2</sub>O</b></i><br /><br />is endothermic (consumes heat energy) in the very beginning, when we <br />have to light up the hydrogen, but, as soon as the reaction started, it <br />becomes exothermic, that is it produces heat energy, because the <br />potential energy of atoms inside two molecules of hydrogen and one <br />molecule of oxygen together is greater than potential energy of atoms <br />inside two molecules of water.<br /><br /><br /><br />3. <i>Photosynthesis</i><br /><br />This is a complicated process, during which the light from sun, air <br />components (such as carbon dioxide, nitrogen and oxygen), water and <br />whatever is in the soil are converted by the plants into chemical energy<br /> that maintains their life. This is an endothermic process, and, as its <br />result, plants grow. In most cases they consume carbon dioxide from the <br />air, break it into carbon and oxygen, consume the water from the soil, <br />break it into hydrogen and oxygen (they need sun's radiation energy to <br />break the molecules of <i><b>CO<sub>2</sub></b></i> and <i><b>H<sub>2</sub>O</b></i>),<br /> use the carbon, hydrogen and part of oxygen to produce new organic <br />molecules they consist of and release the unused oxygen back into <br />atmosphere.<br /><br /><br /><br />4. <i>Battery</i><br /><br />Battery consists of three major components: <b>anode</b>, <b>cathode</b> and <b>electrolyte</b><br /> in-between anode and cathode. As a result of a chemical and <br />electro-magnetic reaction between the molecules of anode and <br />electrolyte, some electrons are transferred from anode to electrolyte. <br />Then, as a result of a chemical and electro-magnetic reaction between <br />the molecules of cathode and electrolyte, some electrons are transferred<br /> from electrolyte to cathode. As the result, there are extra electrons <br />on the cathode, which were taken from the anode, thus creating <br />electrical potential.<br /><br /><br /><br />These simple examples explain the general mechanism of chemical energy, <br />released or consumed in the course of chemical reaction, that transforms<br /> molecules by rearranging their atoms' composition. As a result of a <br />chemical reaction and change in the atomic composition of molecules, <br />potential energy of bonds between atoms in molecules is changing. If the<br /> total potential energy of the resulting molecules is greater than the <br />potential energy of the bonds inside original molecules, the process is <br />endothermic, it consumes energy. In an opposite case the process is <br />exothermic, it produces energy.<br /><br /><br /><br />The exothermic process of extracting chemical energy using chemical <br />reaction is the key to getting energy from gasoline in the car engine, <br />producing heat and light in the fireplace by burning wood, it's the <br />source of energy in all living organisms, including humans. We exist <br />because our body knows how to extract chemical energy from the food.<br /><br /><br /><br />Unizorhttp://www.blogger.com/profile/06592791874048701921noreply@blogger.com0