## Thursday, January 23, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 3

Notes to a video lecture on http://www.unizor.com

Problems 3

Problem A

An infinitesimally thin sphere α of radius R is electrically charged with uniform density of electric charge σ (coulombs per square meter).
A point P is inside this sphere at a distance h (meters) from its center, so h is less than R.
What is the intensity of the electric field produced by this sphere at point P?

Solution

The magnitude of the field intensity at point P produced by any infinitesimal piece of sphere α is inversely proportional to a square of its distance to point P and directly proportional to its charge, and the charge, in turn, is proportional to its area with density σ being a coefficient of proportionality.

Let's define a system of Cartesian coordinates in our three-dimensional space with the origin at the center O of the center of the charged sphere α with Z-axis along segment OP.

Then the coordinates of point P, where we have to calculate the vector of electric field intensity, are (0,0,h).

From the considerations of symmetry, the vector of intensity of the field, produced by an entire electrically charged sphere α, at point P inside it should be directed along a line OP from point P to a center of a sphere, that is along Z-axis. Indeed, for any infinitesimal area of sphere α near point (x,y,z) there is an area symmetrical to it relatively to Z-axis near point (−x,−y,z), which produces the intensity vector of the same magnitude, the same vertical (parallel to OP) component of it and opposite horizontal component. So, all horizontal components will cancel each other, while vertical ones can be summarized by magnitude.
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.

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 z=r and z=r+dr and calculate the vertical component of the intensity vector produced by it. Then we will integrate the result from r=−R to r=R.
This choice is based on a simple fact that for every small piece of this spherical ring its distance to point P 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.

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 σ. The area of a spherical ring equals to a difference between areas of two spherical caps.
The area of a spherical cap equals to 2π·R·H, where H is the height of a cap.
The spherical cap formed by a plane cutting a sphere at z=r has height R−r. The spherical cap formed by a plane cutting a sphere at z=r+dr has height R−r−dr.
Therefore, the area of a spherical ring between these two cutting planes is
area(r,dr) =
= 2π·R·
[(R−r)−(R−r−dr)] =
= 2π·R·
dr

The charge concentrated in this spherical ring is
dQ(r) = 2π·σ·R·dr

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:
L² = (R² − r²) + (h−r)² =
= R² + h² −2h·r

The magnitude of the intensity vector produced by this spherical ring is, therefore,
dE(h,r) = k·dQ(r)/L² =
= 2π·k·σ·R·
d/ L² =
= 2π·k·σ·R·
d/ (R²+h²−2h·r)

We are interested only in vertical component of this vector, which is equal to
dEz(h,r)= dE(h,r)·sin(PAB) =
dE(h,r)·(h−r)/L =
= 2π·k·σ·R·
dr·(h−r) / L³ =
= 2π·k·σ·R·
dr·(h−r) / (R²+h²−2h·r)3/2

Integrating this by r from −R to R can be done as follows.
First of all, let's substitute
x = R²+h²−2h·r
Then
r = (R²+h²−x)/2h
dr = −dx/2h
The limits of integration for x are from R²+h²+2h·R=(R+h)² to R²+h²−2h·R=(R−h)².
Now the expression to integrate looks like
dEz(h,x) = C·(h²−R²+x)·x−3/2·dx
where constant C equals to
C = −π·k·σ·R/(2h²)

Let's integrate the above expression in the limits specified.
First, find the indefinite integral.
C·(h²−R²+x)·x−3/2·dx =
= −2C·(h²−R²)·x−1/2 + 2C·x1/2 =
= 2C·[(R²−h²)·x−1/2 + x1/2]

This expression for indefinite integral should be used to calculate the definite integral in limits for x from (R+h)² to (R−h)².

Assuming that point P is inside a sphere, that is h is less than radius R,
((R+h)²)1/2 = R+h
((R−h)²)1/2 = R−h

Therefore, substituting the upper limit into an expression for an indefinite integral, we get
2C·[(R²−h²)/(R−h) + (R−h)] = 4C·R
Substituting the lower limit, we get
2C·[(R²−h²)/(R+h) + (R+h)] = 4C·R
The difference between these two expressions is zero, which means that the intensity of the electric field inside a uniformly charged sphere is zero.

Problem B

An infinitesimally thin sphere α of radius R is electrically charged with uniform density of electric charge σ (coulombs per square meter).
A point P is outside this sphere at a distance h (meters) from its center, so h is greater than R.
What is the intensity of the electric field produced by this sphere at point P?

Solution

Start as in the previous problem up to indefinite integral
2C·[(R²−h²)·x−1/2 + x1/2]

Assuming that point P is outside a sphere, that is h is greater than radius R,
((R+h)²)1/2 = R+h
((R−h)²)1/2 = h−R

Therefore, substituting the upper limit into an expression for an indefinite integral, we get
2C·[(R²−h²)/(h−R) + (h−R)] = −4C·R
Substituting the lower limit, we get
2C·[(R²−h²)/(R+h) + (R+h)] = 4C·R
The difference between them is the intensity of the electric field produced by a sphere at a point outside it:
E(h) = −8C·R =
= 8π·k·σ·R²/(2h²) =
= 4π·k·σ·R²/h²

Notice that
area(Sphere) = 4π·R²
Therefore, Q = 4π·σ·R²
where σ is the density of electric charge on a sphere, represents a total charge of a sphere.
Hence,
4π·k·σ·R²/h² = k·Q/h²
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.

## Tuesday, January 21, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 2

Notes to a video lecture on http://www.unizor.com

Problems 2

Problem A

An infinite infinitesimally thin plane α is electrically charged with uniform density of electric charge σ (coulombs per square meter).
What is the intensity of the electric field produced by this plane at point P positioned at a distance h (meters) from its surface?

Solution

The magnitude of the field intensity at point P produced by any infinitesimal piece of plane α is inversely proportional to a square of its distance to point P and directly proportional to its charge, and the charge, in turn, is proportional to its area with density σ being a coefficient of proportionality.

Let's define a system of cylindrical coordinates in our three-dimensional space with the origin at the projection O of the point P onto our electrically charged plane α with Z-axis along segment OP and polar coordinates on plane α with r for radial distance OA between any point A on this plane and the origin of coordinates O and φ for a counterclockwise angle from the positive direction of some arbitrarily chosen base ray OX within plane α, originated an point O, to radius OA.

Then the coordinates of point P, where we have to calculate the vector of electric field intensity, are (0,0,h). The coordinates of any point A on plane α will be (r,φ,0).

From the considerations of symmetry, the vector of intensity of the field, produced by an entire infinite electrically charged plane α, at point P outside of it should be directed along a perpendicular OP from point P to a plane. Indeed, for any infinitesimal area of plane α there is an area symmetrical to it relatively to point O, which produces the intensity vector of the same magnitude, the same vertical (parallel to OP) component of it and opposite horizontal (within a plane α) component. So, all horizontal components will cancel each other, while vertical ones can be summarized by magnitude.
Therefore, we should take into account only projections of all individual intensity vectors from all areas of a plane onto a perpendicular OP from point P onto plane α, as all other components will cancel each other.

The approach we will choose is to take an infinitesimal area on a plane in a form of a ring centered at point O of infinitesimal width dr with inner radius r and outer radius r+dr and calculate the vertical component of the intensity vector produced by it. Then we will integrate the result from r=0 to infinity.
This choice is based on a simple fact that for every small piece of this ring its distance to point P 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.

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 σ. The area of a ring equals to
area(r,dr) = π[(r+dr)²−r²] =
= 2π·r·
dr + π·(dr)²

We can drop the infinitesimal of the second order π·(dr)² and leave only the first component - infinitesimal of the first order, that we plan to integrate by r from 0 to infinity.

The charge concentrated in this ring is
dQ(r) = 2π·σ·r·dr

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:
L² = h² + r²
The magnitude of the intensity vector produced by this ring is, therefore,
dE(h,r) = k·dQ(r)/L² =
= 2π·k·σ·r·
d/ L² =
= 2π·k·σ·r·
d/ (h² + r²)

We are interested only in vertical component of this vector, which is equal to
dEz(h,r)= dE(h,r)·sin(PAO) =
dE(h,r)·h/L =
= 2π·k·σ·r·
dr·h / L³ =
= 2π·k·σ·r·
dr·h / (h² + r²)3/2

Integrating this by r from 0 to ∞ can be done as follows.
First of all, let's substitute
x = r/h
Then
r = h·x
dr = h·dx
The limits of integration for x are the same, from 0 to ∞.
Now the expression to integrate looks like
dEz(h,x) =
= 2π·k·σ·h³·x·
d/ h³(1 + x²)3/2 =
= 2π·k·σ·x·
d/ (1 + x²)3/2

Before going into details of integration, note that this expression does not depend on distance h from point P to an electrically charged plane α. This is quite remarkable!
No matter how far point P is from plane α, the intensity of electric field at this point is the same.

To integrate the last expression for a projection onto Z-axis of the intensity of electric field produced by an infinitesimal area of plane α, introduce another substitution:
y = x² + 1
Then
dx = dy/2
The limits of integration for y are from 1 to ∞.
The expression to integrate becomes
dEz(y) = π·k·σ·d/ y3/2 =
= π·k·σ·y−3/2·
dy

This is easy to integrate. The indefinite integral of yn is yn+1/(n+1). Using this for n=−3/2, we get an indefinite integral of our function
−2·π·k·σ·y−1/2 + C
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 α:
E = [1,∞]π·k·σ·y−3/2·dy = 2π·k·σ

Let's note again that this value is independent of the distance h of point P, where we measure the intensity of the electric field, from an electrically charged plane α. It only depends on the density σ of electric charge on this plane.

As for direction of the intensity vector, as we suggested above, it's always perpendicular to the plane α.
Hence, we can say that the electric field produced by a uniformly charged plane is uniform, at each point in space it is directed along a perpendicular to a plane and has a magnitude E=2π·k·σ, where σ represents the density of electric charge on a plane and k is a Coulomb's constant.

Problem B

An infinite infinitesimally thin plane α is electrically charged with uniform density of electric charge σ (coulombs per square meter).
What is the work needed to move a charge q (coulombs) from point M positioned at a distance m (meters) from its surface to point N positioned at a distance n (meters) from its surface?

Solution

Notice that positions of points M and N are given only in terms of their distance to a charged plane α, 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.

So, our work only depends on the distance along the vertical and can be calculated as
WMN = E·(n−m)·q =
= 2π·k·σ·q·(n−m)

## Friday, January 17, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Field Poten...

Notes to a video lecture on http://www.unizor.com

Electric Field Potential

Coulomb's Force
The general form of the Coulomb's Law, when two electrically charged point-objects, A and B, are involved, is
F = k·qA·qB / R²
where
F 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 newtons(N)
qA is electric charge of point-object A in coulombs(C)
qB is electric charge of point-object B in coulombs(C)
R is the distance between charged objects in meters(m)
k is a coefficient of proportionality, the Coulomb's constant, equals to 9.0·109 in N·m²/C²

We have introduced a concept of electric field intensity as a force acting on a probe point-object B, charged with +1C of electricity, from a field produced by the main object A. This force is a characteristic of a field at a point where a probe object is located and is equal to
E = k·qA / R²

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.

Work of Coulomb's Force
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.

Recall from the Mechanics part of this course that the work of the force F, acting at an angle φ to a trajectory on the distance S, is
W = F·S·cos(φ)

For a non-uniform motion and variable force all components of this formula are dependent on some parameter x, like time or distance:
dW(x) = F(x)·dS(x)·cos(φ(x))
where we have to use infinitesimal increments of work dW(x) done by force F(x) on infinitesimal distance dS(x).
The angle φ(x) is the angle between a vector of force F(x) and a tangential to a trajectory at point S(x).

Using a concept of scalar product of vectors and considering force and interval of trajectory as vectors, the same definition can be written as
dW(x) = (F(x)·dS(x))

The latter represents the most rigorous definition of work.
Integration by parameter x from x=xstart to x=xend can be used to calculate the total work

W[xstart , xend ] performed by a variable force F(x), acting on an object in a non-uniform motion, on certain distance S(x) along its trajectory, as the parameter x changes from xstart to xend.

In case of a motion of an electrically charged object in an electrical field the force is the Coulomb's force.
Let's analyze the work needed to move such an object in the field of electrically charged point-object from one position to another.

Let the charge of the main point-object in the center of the electrical field be Q. We move a probe object of charge q along a radius from it to the center of a field from distance r1 to r2.
The Coulomb's force on a distance x from the center equals to
F(x) = k·Q·q / x²
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 +).

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 dx on a segment from r1 to r2.
W[r1,r2] = [r1,r2]k·Q·q·d/ x²
Since the indefinite integral (anti-derivative) of 1/ is −1/x, the amount of work is
W[r1,r2] = k·Q·q·(1/r1−1/r2)
This work is additive. If we move from a distance of r1 to a distance r2 and then from a distance r2 to a distance r3, 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 r3 without stopping at r2
W[r1,r2] + W[r2,r3] = W[r1,r3]

Case 2. Circular Motion
Consider now that we move a probe object circularly, not changing the distance from the center of the electric field.
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.
Obviously, this work, as we move the probe object circularly, is also additive.

Case 3. General
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.

Conservative Forces
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.

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 conservative forces, and electrostatic forces are conservativeGravitational forces are also of the same type.
As an example of non-conservative 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.

Electric Field Potential
The electric field potential is a quantitative characteristic of an electric field, defined for each position in this field, 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.

In the radial field produced by the point-object charged with Q amount of electricity the electric field potential at any point in the field depends only on a distance of this point to a center of the field.
Using the formula above for r1=∞ and r2=r for a probe object charged with q=+1C of electricity, we obtain the formula for an electric potential at distance r from a center of the field, where a point-object charged with Q amount of electricity is located
V(r) = −k·Q / r

Notice that the derivative of potential V(r) by distance r from a center gives the field intensity:
V'(r) = k·Q / r²
So, knowing the potential at each point of the radial field, we can determine the intensity at each point.

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 r1 to distance r2 by, first, moving it to infinity, which results in amount of work W1 = −V(r1), then from infinity to distance r2, which results in amount of work W2 = V(r2). 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 r1 to r2 as calculated above:
W = W1 + W2 = −V(r1)+V(r2) =
= k·Q·q·
(1/r1−1/r2) = W[r1,r2]

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.

To find the amount of work needed to move a charge q from a point in the electric field with a potential V1 to a point with potential V2 we use the formula W = q·(V2−V1)

## Monday, January 13, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Intensity

Notes to a video lecture on http://www.unizor.com

Electric Field Intensity

Electric field intensity is the force (based on the Coulomb's Law) of an electric field of some electrically charged object A, exhorted on a probe point-object B, charged with one coulomb of positive electricity (+1C), positioned at some point in the electric field around a main object A.

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 A.

This is a measure of the intensity of the electric field 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 A and a position in space relatively to this object.

In many cases the word "intensity" is replaced with a word "strength" or just dropped from the conversation. So, terms electric field intensityelectric field strength or in some cases simply electric field are synonymous.

First of all, electric field intensity is a force and, therefore, a vector. 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.

The magnitude of the vector of electric field intensity can be calculated based on the Coulomb's Law.
The general form of the Coulomb's Law, when two electrically charged point objects, A and B, are involved, is
E = k·qA·qB / R²
where
E 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 N - newtons
qA is electric charge of point-object A in C - coulombs
qB is electric charge of point-object B in C - coulombs
R is the distance between charged objects in m - meters
k is a coefficient of proportionality, equals to 9.0·109 in N·m²/C²

Since the electric charge of a probe point-object B, that we use to measure the intensity of an electric field of some charged point-object A, is qB=+1C, the magnitude of the electric field intensity of point-object A with electric charge qA at a distance R from it is
E = k·qA / R²

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.

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 intensity of the combined electric field of all main point-objects.

Example 1
Consider a pair of point-objects at points A and B, charged with positive charge +q each, at distance 2d from each other. What would be an intensity E(x) of their combined electric field at a point P on a perpendicular bisector of a segment AB at distance x from a midpoint M of this segment?

Magnitude of the intensity of electric field from each object is
EA = EB = k·q/(d²+x²)
One of them is directed from A to P, another - from B to P. They are at angle to each other, so we need the rule of parallelogram to find a resulting force. Let ∠APM be φ.
tan(φ) = d/x
Representing each field intensity vector as the sum of two vectors - one along the line MP and another along the line AB, and taking into consideration only the components along MP (the other two nullify each other), we can calculate the resulting intensity
E = EA·cos(φ) + EB·cos(φ) =
= 2k·q·cos(φ) / (d²+x²) =
= 2k·q·x /
[(d²+x²)3/2]

Example 2
Consider an infinitely thin rod of a length 2d charged with electricity such that the density of electrical charge (amount of charge per unit of length) equals to λ.
Our task is to determine the electrical field intensity at any point outside this rod.

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 x=−d to x=+d.

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. Point P(a,b) is the one, where the field intensity is to be established. Y-coordinate y=b represents the distance from point P 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.

The probe charge of +1C is at point P(a,b) and the field intensity at this point is the sum of all forces exhorted by the pieces of rod onto this probe charge.

Consider infinitesimal piece of the rod of the length dx positioned at X-coordinate x. The plan is to determine the force it exhort onto the probe charge of +1C at point P(a,b) and integrate the result from x=−d to x=d.

The electric charge of the piece of the rod of the length dx is λ·dx, where λ is the given density of electric charge in the rod.
The square of the distance r from this piece of the rod to point P(a,b) is
r² = (a−x)² + b²

Now we can apply the Coulomb's Law to determine the infinitesimal electric force dF between this piece of the rod of the length dx and a probe charge of +1C at point P(a,b)
dE = k·λ·d/ [(a−x)²+b²]
The above is the magnitude of the electrical force. Its direction is along the line connecting a piece of the rod with point P(a,b).

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 Fx along the X-axis and vertical force Fy along the Y-axis. Then we can separately integrate each component to get two components of the final force.

Simple math gives us the following expressions for component of the force F
dEx = dE·(a−x) / r =
= k·λ·
dx·(a−x) /
[(a−x)²+b²]3/2
dEy = dE·b / r =
= k·λ·
dx·b /
[(a−x)²+b²]3/2

Now you see how important is Mathematics to succeed in Physics!

Let's integrate each force, horizontal and vertical, on x∈[−d,d] interval.

First, let's find indefinite integral for Ex
k·λ·dx·(a−x)/[(a−x)²+b²]3/2 =
...substitute y=(a−x)²+b²
−0.5·k·λ·y−3/2dy =
= k·λ·y−1/2

Definite integral for y should be taken in limits from (a+d)²+b² to (a−d)²+b², which results in the following expression for ExEx = k·λ·{[(a−d)²+b²]−1/2[(a+d)²+b²]−1/2}

Incidentally,
rright = [(a−d)²+b²]1/2
is the distance from point P(a,b) to the right end of the rod and
rleft = [(a+d)²+b²]1/2
is the distance to the left end.
So, the formula for horizontal component of the resulting field force is
Ex = k·λ·[1/rright − 1/rleft]
Interestingly, as the length of the rod increases to infinity, the horizontal component of the field strength diminishes to zero, as both 1/rright and 1/rleft 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.

Now let's address the vertical component of the electric field intensity Ey.
First, let's calculate the indefinite integral
k·λ·dx·b / [(a−x)²+b²]3/2 =
...substitute (a−x)/b=tan(y)
...tan(y)=(a−x)/b
...(a−x)²+b² = b²·[tan²(y)+1] =
...= b²/cos²(y)

...[(a−x)²+b²]−3/2 = b−3·cos3(y)
...dx=−b/cos²(y)·dy
= −(k·λ/b)·cos(y)·dy =
= (k·λ/b)·sin(y) + C

As far as limits of integration, if x∈[−d,d] then y∈[y1,y2], where
y1 = arctan((a+d)/b) and
y2 = arctan((a−d)/b)
To find the definite integral in the limits above, we will use a trigonometric identity
sin(arctan(z)) = z / √(1+z²)

Now we can express the vertical component of the field intensity as
Ey = (k·λ/b)·[sin(y2)−sin(y1)]
where y1 and y2 are defined above.
Ey = (k·λ/b)·{(a−d)·[(a−d)²+b²]−1/2 − (a+d)·[(a+d)²+b²]−1/2}

## Friday, January 10, 2020

### Unizor - Physics4Teens - Electromagnetism - Electric Field - Problems 1

Notes to a video lecture on http://www.unizor.com

Problems 1

Problem A
At three vertices PQ and R of an equilateral triangle with the length of a side d there are three electrical charges qPqQ and qR. All charges are equal in magnitude to q, but the first two are positive, while the third is negative:
qP = qQ = −qR = q
Determine the magnitude and direction of the combined electrical attraction force acting on charge qR from qP and qQ.

Solution
FP = −k·q²/ (from qP on qR)
FQ = −k·q²/ (from qQ on qR)
These forces are at angle π/3 to each other. Adding them as vectors. The direction of a combined force is from point R to a midpoint between P and Q.
Its magnitude is
FP+Q = −k·q²·√3/

Problem B
Two point-objects of mass m each are hanging at the same level above the ground on two parallel vertical threads in the gravitational field with a free fall acceleration g.
When they are charged with the same amount of electricity q, they move away from each other to a distance d from each other, and each thread will make some angle with a vertical.
Determine this angle as a function of known parameters.

Solution
Fe is an electrical repelling force between objects.
φ is an angle that each thread makes with a vertical.
Fe = k·q²/d² (Coulomb's Law)
T·cos(φ) = m·g (= weight)
T·sin(φ) = Fe (= repelling)
tan(φ) = Fe /(m·g)
tan(φ) = k·q² /(m·g·d²)

Problem C
We assume the Bohr's classical model of an atom.
The atom of hydrogen has 1 proton and 1 electron rotating around it on a distance R.
The electric charge of a proton is positive +q, for an electron it is −q.
The mass of an electron is m.
What is the angular speed and frequency of rotation of electron?

Solution
Fe is an electrical attracting force of proton to electron that keeps electron on its orbit.
Fe = k·q²/R² (Coulomb's Law)
ω is angular speed of an electron.
m·R·ω² = Fe (Newton's Law)
ω = √k·q²/(R³·m)
ν = ω/(2π)
Let's substitute real data (all numbers are in SI units):
k = 9.0·109 N·m²/C²
q = 1.602·10−19 C
m = 9.109·10−31 kg
R = 25·10−12 m (empirical)
ν ≅ 2.028·1016 rev/sec

Problem D
A proton of mass m=1.673·10−27kg and charge of q=1.602·10−19C is launched from a distance a=0.5m towards a nucleus of Uranium that has N=92 protons.
What should be the minimal initial speed v of bombarding proton to overcome the repelling force of all protons inside the nucleus of Uranium and get closer to it on a distance b=0.001m from a nucleus?

Solution
Initial kinetic energy of a bombarding proton E=m·v²/2 should be equal to a work A 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.
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.
Let x be a variable distance of a proton to a nucleus.
F(x) = k·q·N·q/x²=k·N·q²/x²
Then the work done by this force on a segment from x to x+dx is
dA = k·N·q²·dx/x²
The total amount of work done by electrical repelling force is
[a,b]k·N·q²·dx/x² =
= k·N·q²·
[(1/b)−(1/a)]
This is supposed to be equal to kinetic energy of a bombarding proton at the beginning of its motion
m·v²/2 = k·N·q²·[(1/b)−(1/a)]
This gives the value of speed v
v = √2·k·N·q²·[(1/b)−(1/a)]/m
Substituting the values listed above:
v ≅ 68.64 m/sec

## Tuesday, January 7, 2020

### Unizor - Physics4Teens - Electromagnetism - Electrical Field - Coulomb's...

Notes to a video lecture on http://www.unizor.com

Coulomb's Law

As we know, excess of electrons above the number of protons in an object is what we call negative charge. Deficiency of electrons is called positive charge.
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.

It means that there are some forces around electrically charged objects that act in space around them. That is precisely what the term force field means. So, there is an electrical force field that surrounds each electrically charged object.

The next task is to determine the strength of the electrical forces in the electrical field.
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.

Just as a thought experiment, imagine two point objects A and B with one excess electron in each e1A and e1B. The object A will repel object B with some strength F. If the number of excess electrons in object A is increased to two (e1A and e2A), the forces of repelling B must be added: one part from e1A to e1B and another from e2A to e1B. So, the repelling force acting on B will be doubled.

Similar arguments used M times for M excess electrons in object A will lead to multiplication of the initial force by M. If we increase the number of excess electrons in object B to N, we will have to multiply our force by N.
As a result, the total force will be proportional to a product M·N.

We know that the unit of electric charge coulomb is proportional to a charge of one electron. More precisely, the charge of 1 electron is 1.602176634·10−19C.
Therefore, if objects A and B have electric charge qA and qB in coulombs, the force of attraction or repelling between them is proportional to qA·qB.

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.

Consider a set of tiny springs attached to each electron of the charged object A. Each such spring represents a force developed by one electron.
Obviously, the greater the distance between a probe object B and object A - the smaller is the density of springs in the space. It is intuitively obvious that the force of attraction or repelling acting on object B is proportional to a density of springs in the area where B is located, the less springs are observed where B is - the smaller the force will be and vice versa.

In turn, the density of springs is inversely proportional to a square of a distance from object A because the area of a sphere equals to 4πR2, where R is a radius.

Therefore, the force of an electrical field at a distance R from its source (a charged object) is inversely proportional to a square of R.

Summarizing all the above, we suspect that the force of attraction or repelling between two charged objects with charges qA and qB at a distance R from each other should be proportional to
qA·qB / R²

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.
In the SI system, where the force is expressed in newtons, electric charge in coulombs and the distance in meters the force of attraction or repelling is
F = k·qA·qB / R²
where
F 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 N - newtons
qA is electric charge of object A in C - coulombs
qB is electric charge of object B in C - coulombs
R is the distance between charged objects in m - meters
k is a coefficient of proportionality, equals to 9.0·109 in N·m²/C²

The above is the Coulomb's Law, discovered by French physicist Charles-Augustin de Coulomb in 1785.

The Coulomb's Law 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 positive, it's a repulsive force. If the charges are of opposite "sign", one positive with deficiency of electrons and another negative with excess of electrons, the force is negative, it's attractive force.
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.

As we see, the Coulomb's Law for electric field looks very similar to the Newton's Law for gravitational field.
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.

Another purely quantitative difference between these two fields is the magnitude of the force.
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.
Let's compare these two forces using the Coulomb's Law and the Newton's Law, using standard SI units.
Electric charge of an electron is 1.602176634·10−19 C.
Mass of an electron is 9.1093837015·10−31 kg.

Electrical force
(repelling)

Fe = k·qA·qB / R², where
k=9.0·109
qA=qB=1.602176634·10−19
R=0.001
Resulting electrical repelling force is
Fe ≅ 2.31·10−22 N

Gravitational force
(attracting)

Fg = G·mA·mB / R², where
G=6.674·10−11
mA=mB=9.1093837015·10−31
R=0.001
Resulting gravitational attracting force is
Fg ≅ 5.53·10−65 N

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.