A metal sphere of radius R ,carrying charge q ,is surrounded by a

thick concentric metal shell (inner radius a,outer radius b,as in Fig. 2.48). The

shell carries no net charge.

(a) Find the surface charge density $\mathit{\sigma }$at R ,at a ,and at b .

(b) Find the potential at the center, using infinity as the reference point.

(c) Now the outer surface is touched to a grounding wire, which drains off charge

and lowers its potential to zero (same as at infinity). How do your answers to (a) and (b) change?

(a)

Consider a metal sphere of radius R,carrying charge q,is surrounded by a

thick concentric metal shell (inner radius a ,outer radius b ,).

At , Rthe charge density is,

$\sigma =\frac{q}{4{\mathrm{\pi R}}^2}$

At a , write the expression for surface charge density is,

$\frac{-q}{4\pi a^2}$

Because the charge is $-q$at that distance.

At b , write the expression for surface charge density is,

$\frac{+q}{4\pi b^2}$

Because the charge is $+q$at that distance.

(b)

$$\begin{gathered} E\left(r\right)=\left\{\begin{array}{ll}0,& r<R\\ \frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}& R<r<a\\ 0,& a<r<b\\ 0,& b<r\end{array}\right. \\ ={\int }_{\infty }^b\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}dr-{\int }_b^{R}0dr-{\int }_a^R\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{q}{r^2}dr+0 \\ =\frac{q}{4{\mathrm{\pi \epsilon }}_0\left(\begin{array}{c}{\displaystyle \frac{1}{b}+\frac{1}{R}}\end{array}-{\displaystyle \frac{1}{a}}\right)} \end{gathered}$$

Therefore, the potential at the center is$E\left(r\right)=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{b}+\frac{1}{R}-\frac{1}{a}\end{array}\right).$

Now consider the outer surface is touched to a grounding wire, which drains off charge

and lowers its potential to zero (same as at infinity).

Thus ${\sigma }_R=\frac{q}{4{\mathrm{\pi R}}^2}$

Here ${\sigma }_R$is the charge density at distance R.

$\sigma =\frac{q}{4{\mathrm{\pi a}}^2}$

Here, ${\sigma }_a$is the charge density at distance a

Here, ${\sigma }_b$is the charge density at distance b

Now,

role="math" localid="1654687503947" $E\left(r\right)=\left\{\begin{array}{ll}0,& r<R\\ \frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}& R<r<a\\ 0,& a<r<b\\ 0,& r<b\end{array}\right.\begin{array}{}\end{array}$

Determine the potential at the center.

$$\begin{gathered} V_{\left(center\right)}=-{\int }_{\infty }^{center}E\left(r\right)dr \\ ={\int }_a^R\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2} \end{gathered}$$