Hey... I have a quick question for you guys about electric potential.(adsbygoogle = window.adsbygoogle || []).push({});

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=[tex]\oint[/tex][tex]\frac{kq}{r}[/tex]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?

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# Potentials from continuous distributions.

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