chapter22

===Chapter 22 Electric Field II: Continuous Charge Distributions ===

As proven in the previous chapter, Coulomb's Law is used to determine the force exerted by charge q1 on another charge q2. Using this law, we can determine the electric field produced by various continuous charge distributions. Continuous charge distributions can be defined as charges that are so close together, that they can be thought of as being a single volume, surface or line of charge. When dealing with the amount of charge per unit, we can take the quantity charge and divide it by the volume, area or length. This is outlined below:

Volume charge density: Surface charge density: Linear charge density:



The symbols described above are, respectively, row, omega and lambda.

22-1 Calculating E From Coulomb's Law
Now, to find the electric field using Coulomb's Law, generically, is quite easy: This basically means that the electric field can be determined by the amount of charge in the object emanating the field, divided by the square of the distance away from the charge. However, this formula only applies when we are dealing with point charges, when dealing with things that are bigger; we first need to take an element of the field produced by an element of the charge. This makes the final equation: 

E on the Axis of a Finite Line of Charge: A line is the simplest shape to deal with, so that is what we will be looking at first. To make the math easier, we will also be assuming that the center of the line is at x = 0 and that the two ends of the line are at x = +/- L/2.

In order to find the electric field produced from this line of charge, we need a point P on the X-axis, where Xp >L/2. After selecting our point, we use Coulomb’s law to get the electric field next to the line at point P. First, we must recognize that dx is a distance Xp - X away from point P, we can replace r with Xp - X. Then, we can recognize that lambda * dx can replace dq, allowing everything to be in terms of the same variable creating the following equation. Then, we take the integral from each side, allowing us to find the charge over the entire length of the line, rather than at a specific element. If we substitute u = Xp - X (meaning du = -dx), our math can become much simpler. In the end, we get the final sequence of equations. This provides the ground for the final equation, the one that optimal to find E on the axis of the axis of a finite line of charge. Problem: A uniform line charge of linear charge density lambda = 3.5nC/m extends from x = 0 to x = 5 m. A) What is the total charge? B) Find the electric field on the x-axis at x = 6m, x = 9m and x = 250m. C) Find the field at x = 250m, using the approximation that the charge is a point charge at the origin and compare your result with that for the exact calculation in part B.

E off the Axis of a Finite Line of Charge: 

Dealing with E off the axis of a Finite line of Charge is slightly more complicated, mainly because we are dealing with a 'y' component to the equation now and we need to consider that r^2 = (x^2 + y^2). As before, we substitute dq for lambda dx for. However, this time we need to find the x and y components before we find what the total electric field is. Solving for the ‘y’ component, we can multiply dE by cos (θ). By that recognizing cos (θ) = y/r, the math can become simpler when integrating, because we recognize that y is constant, so it can be factored out to produce the equation the first equation that solves for Ey below. Then, by using trigonometric substitution, we can see that x = y tan(θ) and thus dx = y sec^2(θ) and that y = r cos(θ) and thus 1/r = cos(θ)/y. By replacing each of the values, we get the final formula in the sequence below.

Solving for Ex is a very similar procedure, so to save time we have skipped all of the extra steps and decided to provide the final equation below and you are free to derive the formula on your own to improve your understanding of it. Problem: A uniform line charge extends from x = -2.5cm to x = 2.5cm and has a linear charge density of lambda = 6.0 nC/m. A) Find the total charge. <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">B) Find the electric field on the y-axis at y = 4cm, y = 12cm, and y = 4.5m <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">C) Find the electric field at y = 4.5m using the approximation that the charge is a point charge at the origin and compare your result with that for the exact calculation in part B.

<span style="background-color: transparent; color: black; display: block; font-family: Arial; font-size: 13px; text-align: left; text-decoration: none;">E off the Axis of an Infinite Line of Charge: <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">To find the electric field next to an infinite line of charge, we use almost the same method as a finite line of charge. All we do is change the limits of the equation from x = +/- L/2 to +/- infinity. We see that this becomes equal to <span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">θ = +/- π /2 because the range has to be a 45 degree angle. When solving based in these new limits, we get Ex = 0 and Ey = 2k <span style="font-family: Arial,sans-serif;">λ /y.

<span style="background-color: transparent; color: black; display: block; font-family: Arial; font-size: 13px; text-align: left; text-decoration: none;">E on the axis of a Ring Charge: <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">When dealing with finding E on the axis ring charge, it is actually simpler than finding E off the axis of a finite line of charge. This is because all the y components cancel out as shown below.

<span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none;">

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">Using a similar technique what we have done before, only realizing that in this case cos(θ ) = x/r and that r^2 = x^2 + a^2, we get the following formulas. <span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">Then, by recognizing that both x and a are constants, we can take everything but the dq can be taken out of the formula producing the formula: <span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">Using the equation, we see that in the center of the ring of charge, E = 0 because the entirety of the ring cancels each other out (and the fact that x=0), and up until x = a, E gets larger then goes back down because at that point the denominator becomes larger than the numerator. Using a similar method as an infinite line of charge, when x >> a, it reduces down to a point charge because a is effectively 0.

Problem: <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">A 2.75 uC charge is uniformly distributed on a ring of radius 8.5cm. <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">A) Find the electric field on the x-axis at x = 1.2cm, x = 3.6cm, and x = 4m. <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">B) Find the electric field at x = 4m using the approximation that the charge is a point charge at the origin and compare your result with that for the exact calculation in part A.

<span style="background-color: transparent; color: black; display: block; font-family: Arial; font-size: 13px; text-align: left; text-decoration: none;">E on the Axis of a Uniformly Charged Disk: <span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none; vertical-align: auto;">On a disk with a uniform charge, we use the same type of method as a ring of charge. Once again, the y components will cancel out, so we only need to deal with the x.

<span style="background-color: transparent; color: black; font-family: Arial; font-size: 13px; text-decoration: none;">

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">In the figure shown above, we need to deal with rings of different radii, a, with a width of da. The area of this ring is dA = 2π a(da), and the charge is dq = σ(dA) = 2πσ(a)da. Substituting this into Coulomb's Law, we get:

dEcos(θ) = 2kπσa(da)/r2 dEx = 2xkπσa(da)/r3

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">When substituting r = √(x^2 + a^2) we get dEx = 2xkπσa(da)/(x2+a2)3/2

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">Then when we integrate, with the bounds equal to 0 to R, we get:



Problem: A uniform disc charge of area charge density omega = 6.2nC/m^2 with a radius of 5m. A) What is the total charge? B) Find the electric field on the x-axis at x = 10m, x = 15m and x = 300m. C) Find the field at x = 300m, using the approximation that the charge is a point charge at the origin and compare your result with that for the exact calculation in part B.

<span style="background-color: transparent; color: black; display: block; font-family: Arial; font-size: 13px; text-align: left; text-decoration: none;">E Due to an Infinite Plane of Charge:

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">For finding the electric field due to an infinite plane of charge, we use the same formula that we found above, only that we allow and allow the ratio R/x to go to infinity. When we do this, we get the equation:

Ex =2kπσ

<span style="color: black; font-family: Arial,sans-serif; font-size: 10pt;">This is true because the electric field is independent of x. When looking at negative values of x, on the other side of the infinite plane, we get the same equation, just negative. When moving along the x-axis, when x = 0, there is a discontinuity, where the change is:

4kπσ

22-2 Gauss's Law
While in Chapter 21, electric fields have only been described visually, we now can describe them mathematically using Gauss's Law.

Electric Flux ϕ is used to describe the number of field lines through a surface. The equation for electric flux is shown below
 * __Electric Flux:__**

ϕ = EA

Where E is the electric field and A is the area of the surface. If the surface area is not perpendicular to the electric field, the dot takes care of this by adding a cosθ.

When the area is not constant, we need to find electric flux in terms of an element of the area and field. This allows us to express electric flux in terms of a surface integral.

Problem: In a particular region of the earths atmosphere, the electric field above the earths surface has been measured to be 150 N/C downward at an altitude of 250 m and 170 N/C downwards at an altitude of 400 m. Calculate the volume charge density, the <span style="font-family: 'Calibri','sans-serif'; font-size: 14px;">ρ assuming it to be uniform between 250 and 400 m ( you may neglect the curvature of earth. Why?).

Figure 22-17 shows an example of a simple Gaussian surface that we are already used to, a sphere around a point charge. We can now prove using Gauss's Law that.
 * __Quantitative Statement of Gauss's Law__**

22-3 Calculating E From Gauss's Law
__**Plane Symmetry:**__ If a surface has plane symmetry, it means that means that all points on an infinite plane surface are the same.

In the figure above, there is an infinite plane of charge, with uniform surface charge density<span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> σ. E is always perpendicular to the plane, and when two points are equally distanced away but on opposite sides of the plane, they have the same magnitude but opposite direction. On all of the curved surfaces, E is perpendicular to N, therefore there is no flux through it. The Flux on the other two surfaces are EA. The total flux through the closed surface is 2EA. The net charge inside is<span style="font-family: 'Times New Roman','serif'; font-size: 16px;"> σ A, therefore:



and if you solve for E, you get

This is an easier way of showing what we did in section 22-1, by using Gauss's Law instead of Coulombs Law.

__**Spherical Symmetry:**__ When dealing with Spherical Symmetry, we assume that all charges are distributed on the sphere evenly. When applying Gauss's Law to this, we pick a spherical Gaussian surface, making the area. This makes it the familiar situation of a sphere around a point charge. When we apply Gauss's Law, we get

Problem: Consider two concentric conducting spheres: The outer sphere is hollow and has a charge -7Q on it. The inner sphere is solid, it has a charge +2Q on it. a) How is the charge distributed on the outer sphere. That is, how much charge on the outer surface and how much charge is on the inner surface? b) Suppose a wire is connected between the inner and outer spheres. After electrostatic equilibrium is established, how much total charge is on the outside sphere? How much charge is on the outer surface of the outer sphere? How much charge is on the inner sphere? Does the electric field at the surface of the inside sphere change when the wire is connected? c) Suppose you return to the original conditions in part a. We now connect the outer sphere to ground and disconnect it. How much total charge will be on the outer sphere? How much charge will be on the outer surface of the outer sphere and how much will be on the inner surface?

When looking at a uniform spherical shell with radius R and charge Q, we can do a similar set up to Spherical Symmetry. However, an exception is caused when we look at the electric field in the shell. Because there is no overall charge inside the Gaussian surface (inside the shell) the electric field is zero. This can also be thought of as all of the charge on the shell canceling itself out.
 * __E Due to a Thin Spherical Shell of Charge:__**

As before, the charge is when outside the sphere. however, a new exception is caused when inside the sphere. This is because there is still a charge in the Gaussian surface, however, because the row is constant, there is less charge. We can find the new charge by formula below. In this new formula, 'R' is the original radius of the sphere and 'r' is the radius that we are looking at. This also tells us that when going towards the center of the sphere, we have a linear decrease as shown below.
 * __E Due to a Uniformly Charged Sphere:__**

When looking at a coaxial surface, or a thin infinite line of charge, we can find the electric field by using a cylindrical Gaussian surface. This makes the area 2πRL. Because, we can reduce this down to , the same formula that we found for an infinite line of charge using Coulomb's Law. However, this only works when you know that you are far from the ends of the line of charge. If you know that you are close to one of the ends, then you can't use this formula as you can't assume that the electric field is perpendicular to the surface or that it is constant everywhere on it. However, this case isn't used in the AP exam, so we will not be looking at this in detail.
 * __Cylindrical Symmetry:__**

Problem: Figure 22-39 shows a portion of an infinitely long cable in cross section. The inner conductor carries a charge of 6 nC/m; the outer conductor is uncharged. a) Find the electric field for all values R where R is the distance from the axis of the cylindrical system. b) What are the surface charge densities on the inside and outside surface of the outer conductor?

Laboratory
The purpose of our lab was to graph the electric field at different positions on a piece of conductive paper, generated by a 1.5 battery. The set up and procedure can be found at the following link: []

Data: While gathering data, we found that the data points were semetrical over the horizontal axis, so we were able to cut the number of data points in half. Graph

The graph shows a huge electric field on the positive terminal and an electric field of 0 on the negative terminal. There is a huge negative slope between the points. The ends of the terminals, the one node has higher influence on teh field due to it being closer and closer reflects the positive and negative at the ends. []


 * CATEGORY ||  SCORE (1-4)  ||  POINTS (0-20)  ||
 * Content || 3.5 || 17.5 ||
 * Organization || 4 || 20 ||
 * Accuracy || 4 || 20 ||
 * Appearance || 4 || 20 ||
 * Participation || 4 || 20 ||
 * TOTAL || 97.5 ||