Chapter 23 Problem 19

Problem

A nonconducting sphere of radius $r_{0}$ carries a total charge $Q$ distributed uniformly throughout its volume. Determine the electric potential as a function of the distance $r$ from the center of the sphere for

(a) $r>r_{0}$

(b)$r<r_{0}$

Take $r>V=0$ at $r=\infty$.

(c)Plot $V$ versus $r$ and $r>E$ versus $r$.

Solution

(a)

Spherically symmetric :

$E(r>r_{0})={\frac {Q}{4\pi \epsilon _{0}r^{2}}}$

$V(r\geq r_{0})=-\int _{\infty }^{r}{\frac {Q}{4\pi \epsilon _{0}r^{2}}}dr$ $=\left.{\frac {Q}{4\pi \epsilon _{0}r}}\right|_{\infty }^{r}$ $={\frac {Q}{4\pi \epsilon _{0}r}}$

(b)

use Gauss's Law (with a spherical Gaussian surface)

$4\pi r^{2}E={\frac {Q}{\epsilon _{0}}}{\frac {{\frac {4}{3}}\pi r^{3}}{{\frac {4}{3}}\pi r_{0}^{3}}}$ $\rightarrow E(r<r_{0})={\frac {Qr}{4\pi \epsilon _{0}r_{0}^{3}}}$

use the electric field to calculate the potential

$V(r<r_{0})=V(r_{0})-\int _{r_{0}}^{r}{\frac {Qr}{4\pi \epsilon _{0}r_{0}^{3}}}dr$ $={\frac {Qr}{4\pi \epsilon _{0}r_{0}}}-\left.{\frac {Qr^{2}}{8\pi \epsilon _{0}r_{0}^{3}}}\right|_{r_{0}}^{r}$ $={\frac {Qr}{8\pi \epsilon _{0}r_{0}}}\left(3-{\frac {r^{2}}{r_{0}^{2}}}\right)$

(c)

$V_{0}=V(r=r_{0})={\frac {Q}{4\pi \epsilon _{0}r_{0}}}$

$E_{0}=E(r=r_{0})={\frac {Q}{4\pi \epsilon _{0}r_{0}^{2}}}$

for $r<r_{0}$:

$V/V_{0}={\frac {{\frac {Q}{8\pi \epsilon _{0}r_{0}}}\left(3-{\frac {r^{2}}{r_{0}^{2}}}\right)}{\frac {Q}{4\pi \epsilon _{0}r_{0}}}}={\frac {1}{2}}\left(3-{\frac {r^{2}}{r_{0}^{2}}}\right)$

$E/E_{0}={\frac {\frac {Qr}{4\pi \epsilon _{0}r_{0}^{3}}}{\frac {Q}{4\pi \epsilon _{0}r_{0}^{2}}}}={\frac {r}{r_{0}}}$

for $r>r_{0}$:

$V/V_{0}={\frac {\frac {Q}{4\pi \epsilon _{0}r}}{\frac {Q}{4\pi \epsilon _{0}r_{0}}}}={\frac {r_{0}}{r}}=(r/r_{0})^{-1}$

$E/E_{0}={\frac {\frac {Q}{4\pi \epsilon _{0}r^{2}}}{\frac {Q}{4\pi \epsilon _{0}r_{0}^{2}}}}={\frac {r_{0}^{2}}{r^{2}}}=(r/r_{0})^{-2}$

(c)