(adsbygoogle = window.adsbygoogle || []).push({}); [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([tex]\vec{x}[/tex]) = [tex]\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho(\vec{x'})d^{3}x'}{|\vec{x}-\vec{x'}|}[/tex]

[tex]\rho(\vec{x'}) = \frac{Q}{4\pi R^{2}}[/tex]

3. The attempt at a solution

if thexvector is along the z axis (can exploit spherical symmetry), then |x-x'| = [tex]\sqrt{x^{2} + R^{2} - 2xRcos\theta}[/tex]. (using cosine law)

Then

V([tex]\vec{x}[/tex]) = [tex]\frac{1}{4\pi\epsilon_{0}}\frac{Q}{2}\int_{0}^{\pi}\frac{sin\theta d\theta}{\sqrt{x^{2} + R^{2} - 2xRcos\theta}}[/tex]

V(x) = [tex]\frac{Q}{4\pi\epsilon_{0}x}[/tex]

or

V(r) = [tex]\frac{Q}{4\pi\epsilon_{0}r}[/tex]

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) = [tex]\frac{Q}{4\pi\epsilon_{0}R}[/tex]

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

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# Homework Help: Potential inside (and outside) a charged spherical shell

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