Potential inside (and outside) a charged spherical shell

[russdot]

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

Homework Statement

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.

Homework 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}}

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...

 

[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'?

 

[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).

 

[gabbagabbahey]

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

V(\vec{x})=\int \frac{\sigma (\vec{r&#039;}) d^2 x&#039;}{|\vec{x}-\vec{x&#039;}|}

where \sigma (\vec{r&#039;})= Q/4 \pi R^2 is the surface charge density.

 

[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&#039; d\theta&#039; d\phi as the surface area element.
I have solved the problem though, thanks for the assistance!

 

[gabbagabbahey]

No problem, but I wasn't much assistance