# Potentials from continuous distributions.

1. Mar 31, 2008

### scorpion990

Hey... I have a quick question for you guys about electric potential.
I have a spherical shell with a constant charge distribution. The total charge(Q), along with the shell's radius is given. Also, V(infinity) is defined to be 0 in this case.

I'm told to find:
a. The potential at r = the radius of the shell
b. The potential at r =0.
Now, I'm aware that V = kq/r, and I understand where the formula comes from. I'm just having trouble applying it to various situations..

At r = the radius of the shell, I can treat the shell as a point charge, correct? And therefore, V = kQ/r.

Part b confuses me a bit. First of all, there is no charge at the center. Second of all, if I were to "plug and chug" without really thinking too much, I'd be tempted to plug r=0 into: V = kQ/r. Of course, this would give that V = infinity.

IDEA: I just had a thought. My E&M class is definitely calculus-based, but we usually aren't expected to do much more than one-dimensional integrals on exams. Is it possible to calculate potential as:
V=$$\oint$$$$\frac{kq}{r}$$ds?

kq/r is a constant, and the surface integral with respect to s would just equal the surface area of the original sphere, correct? Anybody mind helping me out?

2. Mar 31, 2008

### awvvu

V = KQ/r is only applicable for point charges or spherically symmetrical charge distributions (when outside them).

You can use the definition of electric potential as:
$$V = -\int_{\infty}^{0} E dr$$

If you can find the electric fields inside and outside the sphere (they're different, so you'll have to split the integral up from infinity to R and R to 0), then this will give you the potential. And apparently it's KQ/R (R = radius of shell).

I think doing the surface integral that way works also... but you need to be careful, because you're not integrating KQ/r through the surface, but some little contribution dQ from each piece of the surface. You need to find a way to relate dQ to the total charge Q.

Last edited: Mar 31, 2008
3. Mar 31, 2008

### zarbanx

u r right for the first part bt for the second part(according to shell theorem by newton which applies in this case)the electric field inside the shell is zero and thus the potential at the surface is the potential at the centre

4. Apr 3, 2008

### scorpion990

Oh... sorry... I wrote that integral incorrectly.I actually integrated with respect to the dq across the sphere, got some nice cancellations, and got KQ/R. =)

Thank you!