# Solve Spherical Shell Gauss Problem with Differential Form Only

1. Mar 17, 2013

### gavman

1. The problem statement, all variables and given/known data
A hollow spherical shell carries charge density $\rho=\frac{k}{r^2}$ in the region $a<=r<=b$, where a is the inner radius and b is the outer radius. Find the electric field in the region $a<r<b$.
I'm not allowed to use integral form of Gauss's law, must use differential form.

2. Relevant equations
Relating charge density to electric field divergence $\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}$

3. The attempt at a solution
Using the integral method, I believe the electric field is $\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}}\frac{r-a}{r^2}\hat{r}$
I then went ahead and determined that spherical co-ordinates are the way to go, and that the E-field has only an $\hat{r}$ component and put together
$\vec{∇}\cdot\vec{E}=\frac{1}{r^2}\frac{∂(r^2\vec{E}(\vec{r}))}{∂r}=\frac{k}{r^2\epsilon_{0}}$

By inspection I decided that I should have $\vec{E}(\vec{r})=\frac{k}{\epsilon_{0}r}\hat{r}$. While this does satisfy $\vec{∇}\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}$, it is not in agreement with the integral form.

So my question is, am I using these relationships incorrectly? I'm very confused, since when I integrate ∇E from a to r, I get the same answer as the Gauss integral form, but not sure why this could be, since I understand that the integral of ∇E is equal to the charge enclosed by a Gaussian surface. Since I'm just interested in the E-field at a point in the shell, why would I need to integrate anyway?

2. Mar 17, 2013

### TSny

Hello, gavman.

By inspection you found a particular solution to the differential equation. You can add to it any solution to the associated homogeneous differential equation. Then you can satisfy the appropriate boundary condition at r = a.

3. Mar 18, 2013

### gavman

Hi TSny
Ok, I have $\frac{k}{\epsilon_{0}r^2}$ as a solution to the associated DE. To get this I used ∇E=0 then substituted my particular solution for E. Do I then just state that at the boundary, E=0 and then build an equation from my two E(r) equations which satisfies this condition? I'm just not sure if I have made an unreasonable assumption.
thanks.

4. Mar 18, 2013

### TSny

Yes, that sounds right. Of course, you are invoking the boundary condition that E = 0 at r = a and you might ask where that comes from. It's easy to use Gauss' law to get E = 0 everywhere r<a. But since you are not to use Gauss' law, I'm not sure how you are supposed to get the boundary condition at r= a.

5. Mar 18, 2013

### ehild

In the inner range (r<a), ρ=0. The solution of ∇E=0 is E=C/r2, C is a constant. But there is no singularity at r=0, so C must be zero.

ehild

6. Mar 18, 2013

### TSny

Ah. Good. That nails the boundary condition.

7. Mar 18, 2013

### gavman

Great, thanks for your help guys, appreciate it.