91Ó°ÊÓ

(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,

$\mathrm{σ}=\frac{q}{4{\mathrm{Ï€¸é}}^2}$

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

$\frac{-q}{4\mathrm{Ï€}{a}^2}$

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

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

$\frac{+q}{4\mathrm{Ï€}{b}^2}$

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

(b)

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

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

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 ${\mathrm{σ}}_R=\frac{q}{4{\mathrm{Ï€¸é}}^2}$

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

$\mathrm{σ}=\frac{q}{4{\mathrm{Ï€²¹}}^2}$

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

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

Now,

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

Determine the potential at the center.

$$\begin{gathered} V_{\left(center\right)}=-{∫}_{∞}^{center}E\left(r\right)dr \\ ={∫}_a^R\frac{q}{4{\mathrm{Ï€Î\mu }}_0{\mathrm{r}}^2} \end{gathered}$$