# Gauss’s Law Problem – Spherical Shell with Non-uniform Charge

**1. The problem statement, all variables and given/known data**

Consider a spherical shell with inner radius r1=0.30 m and outer radius r2=1.00 m. The hollow inside the shell contains no charge; and charge is distributed on the inside surface of the shell and within the shell itself, such that the electrical field inside the shell itself is everywhere outward pointing and of uniform constant magnitude 28 N/C.

a) What is the charge per unit area on the inner surface at r=r1?

b) What is the charge per unit volume at radius r=0.65 m (within the material of the shell)?

**2. Relevant equations**

[itex]\phi[/itex]ₑ = [itex]\oint[/itex] E · dA = Q/[itex]\epsilon[/itex]ₒ

area of sphere = 4πr^2

**3. The attempt at a solution**

I managed to get the a) part of the question using Gauss’s Law. I made the Gaussian surface directly on the inner surface of the sphere (an infinitesimally small thickness, you could say). Since the electric field is constant everywhere on the surface, I could rewrite the equation as:

E4πr^2[itex]\epsilon[/itex]ₒ = Q

Since I’m looking for the charge per unit area, I rewrote it like this, and got the right answer:

[itex]\mu[/itex] = Q/A = q/(4πr^2)

[itex]\mu[/itex] = E[itex]\epsilon[/itex]ₒ

For the b) part of the question, I’m really struggling.

What I tried to do this: Since the volumetric charge density is ρ = Q/V, we could say that for an infinitesimally small volume, ρ = dQ/dV.

From this logic, I rewrote my equation for charge in terms of volume instead of radius. Then, I wanted to take the derivative of this function in terms of volume. Then, I reformulated the function in terms of radius. Here’s my steps:

Q = E4πr^2[itex]\epsilon[/itex]ₒ

V = (4/3)πr^3

r = [itex]\sqrt[3]{3V/4π}[/itex]

Q = E4π(3V/4π)^(2/3)[itex]\epsilon[/itex]ₒ

dQ/dV = 2πr^2E[itex]\epsilon[/itex]ₒ/r

It’s possible that my calculus got rusty, but this function is not giving me the right answer.

Can somebody lead me in the right direction please?

http://ift.tt/1jktyUe

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