# Potential inside (and outside) a charged spherical shell

1. Sep 28, 2008

### russdot

[solved] Potential inside (and outside) a charged spherical shell

1. The problem statement, all variables and given/known data
Use the integral (i) to determine the potential V(x) both inside and outside a uniformly charge spherical surface, with total charge Q and radius R.

2. Relevant equations
(i) V($$\vec{x}$$) = $$\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho(\vec{x'})d^{3}x'}{|\vec{x}-\vec{x'}|}$$
$$\rho(\vec{x'}) = \frac{Q}{4\pi R^{2}}$$

3. The attempt at a solution
if the x vector is along the z axis (can exploit spherical symmetry), then |x-x'| = $$\sqrt{x^{2} + R^{2} - 2xRcos\theta}$$. (using cosine law)
Then
V($$\vec{x}$$) = $$\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2}\int_{0}^{\pi}\frac{sin\theta d\theta}{\sqrt{x^{2} + R^{2} - 2xRcos\theta}}$$

V(x) = $$\frac{Q}{4\pi\epsilon_{0}x}$$
or
V(r) = $$\frac{Q}{4\pi\epsilon_{0}r}$$
Which is correct for outside the spherical shell, but I can't figure out how |x - x'| would be any different (or what else should be different) for inside the spherical shell to get
V(r) = $$\frac{Q}{4\pi\epsilon_{0}R}$$

I realize there are better/easier ways of doing this, but the question says to use that integral specifically...

2. Sep 28, 2008

### gabbagabbahey

Shouldn't you have:

$$V(x)=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2} \int_0^{x} \int_{0}^{\pi }\frac{sin \theta ' dx' d \theta '}{\sqrt{(x')^{2} + R^{2} - 2(x')Rcos \theta '}}$$

I understand that the integration over $\phi '$ gave you $2 \pi$ , but what happened to the integration over $x'$?

3. Sep 28, 2008

### russdot

Well $$d^{3}x'$$ turns into $$r^{2}sin\theta dr d\theta d\phi$$ and since r is constant at R (spherical shell) then an R^2 comes out of the integral and cancels the R^2 in the denominator from the charge density rho = Q / (4 pi R^2).

I also figured out the problem, after integration:
$$V(x) = \frac{Q}{4\pi \epsilon_{0}} \frac{1}{2xR} [\sqrt{(R + x)^{2}} - \sqrt{(R - x)^{2}}]$$
and I forgot to consider the different cases for $$\sqrt{(R - x)^{2}}$$ when x > R (outside spherical shell) and x<R (inside).

Last edited: Sep 28, 2008
4. Sep 28, 2008

### gabbagabbahey

Since your dealing with a surface charge, and not a volume charge, you should have:

$$V(\vec{x})=\int \frac{\sigma (\vec{r'}) d^2 x'}{|\vec{x}-\vec{x'}|}$$

where $\sigma (\vec{r'})= Q/4 \pi R^2$ is the surface charge density.

5. Sep 28, 2008

### russdot

Yes, you're right. The form I used was ambiguous, sorry about that!
I guess I just jumped right to using $$R^{2}sin\theta' d\theta' d\phi$$ as the surface area element.
I have solved the problem though, thanks for the assistance!

6. Sep 28, 2008

### gabbagabbahey

No problem, but I wasn't much assistance

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