# Finding Electric Fields and Electrostatic Potential

1. Oct 4, 2013

### John.Martinez

Problem #1

A slab of thickness 2a (extending infinitely in the y and z directions) with non-uniform charge density is parallel to the yz-plane. The charge density is given by

ρ(x) = γx2, -a < x < a where γ is a constant.
0, a < |x|.

(a) Use Gauss's law plus symmetry to find the electric field E(r) everywhere in space.

(b) Give the electrostatic potential V(r) associated with this electric field when the potential vanishes at the origin, V(0)=0.

(c) For your electric field, evaluate the divergence of E.

(d) For your scalar potential, evaluate the gradient of V.

Relevant equations for Problem #1

(1.1) $\int$E(dot)dA = 1/ε0*Qencl.

(1.2) V(r) = -$\int$E(dot)dl, from some point to r.

(1.3) E = -∇V

Attempt at Problem #1

(a) I started by using (1.1), and by symmetry, finding that the left side of (1.1) is 4∏*s*x*|E|. The right side is then 1/ε0*∫ρ dτ over some volume.

I figure that the field is pointing uniformly in the x directions, and thus I integrate over x, with s, ∅(phi) held constant, and x taking the place of z, (where the typical definition of dτ over a cylinder is dτ=s*ds*d∅*dz)

Right side = 1/ε0*γ*s*2∏*∫x2dx from 0 to x

=1/ε0*γ*s*2∏/3*x3 i_hat.

Left side = Right side
|E| = 1/(6*ε0)*γ*x2 i_hat.

(b) I choose the origin to be the point at which I begin evaluating the right side of (1.2).

V(r) = -∫1/(6*ε0)*γ*x'2 i_hat. from 0 to x

= -1/(18*ε0)*γ*x3 i_hat.

(c) Divergence of E is 1/(12*ε0)*γ*x.

(d) Gradient of V = -E = -1/(6*ε0)*γ*x2 i_hat.

Problem #2

An infinitely long cylindrical volume of radius R contains a charge density ρ(s)=ks4 where k is a constant and s is the distance from the axis of the cylinder. Note that this is NOT a constant density.

(a) Find the electric field everywhere in space.

(b) From your result in part (a) find the electrostatic potential inside the cylinder, assuming that the potential vanishes along the axis of symmetry, V(s) = 0 when s=0.[/b]

Relevant equations for Problem #2

(2.1) $\int$E(dot)dA = 1/ε0*Qencl.

(2.2) V(r) = -$\int$E(dot)dl, from some point to r.

Attempt at Problem #2

Looking at the question, I believe I need to use a Gaussian cylinder of radius r, where r > R.

(a) Using (2.1), the right side of (2.1) is 1/ε0 * ∫ρ dτ. Holding ∅ as a constant, and z as a constant (from 0 to length L), and pulling the constant k out from ρ,

Right side = 1/ε0 * 2∏*k*L*∫s'4 ds' from 0 to s.

Right side = 1/(5ε0) * 2∏*k*L*s5.

Left side, by symmetry = |E|*2∏*s*L.

Left side = Right side: |E|*2∏*s*L = 1/(5ε0) * 2∏*k*L*s5,

|E| = 1/(5ε0)*k*s4 s_hat.

(b) Using (2.2),

V(s) = -∫1/(5ε0)*k*s'4 s_hat, from 0 to s.

V(s) = -k/(25*ε0)*s5 s_hat.

Any advice on these questions would be greatly appreciated. I am fairly certain I have them right, however I have no way of checking, and certain aspects of the topic evade me.

Thanks in advance for any help you may offer. I love physics, but I am finding this semester to be particularly rough on me. I'm studying every day, and making sure to keep trusting in my ability to figure out these complex problems.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2013

### BruceW

Hi John, welcome to physicsforums, dude!
for problem 1(a) you have the right method, but I think there is a mistake in there. The left-hand-side of 1.1 should not be 4∏*s*x*|E| In this equation, you seem to be integrating over an area which is not perpendicular to the x direction? But as you say later, the electric field is only in the x direction (due to symmetry).

Also, I don't understand how you got this equation for the right-hand side: 1/ε0*γ*s*2∏*∫x2dx You are supposed to be integrating the charge density over some volume, right? And you say you are using dτ=s*ds*d∅*dx (this is good), so the integral you are doing is ρ(x)/ε0 *s*ds*d∅*dx right? But this does not integrate to the answer which you wrote for the right-hand side. (remember that 's' is a variable too).

3. Oct 6, 2013

### John.Martinez

Regarding your first issue, I am under the impression that I am supposed to integrate over the area of the surface which the field is passing through. In this case, the field is passing through the circular portion of a Gaussian cylinder, as the side lengths of the cylinder do not contribute to the overall field. As such, the field should just be 4*pi*s*E, which is my new result.

For the RHS, I saw the error in integration you mentioned, and my new result is

4*pi*s*E = gamma*s/(12*epsilon_0) x^3 i_hat

4. Oct 7, 2013

### BruceW

but for the left side, 2*pi*s is not the area of a circle. And for the right side, you are integrating γx2/ε0 *s*ds*d∅*dx so there should be a separate integration over x and s.

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