Q38P - Introduction to Electrodynamics

Question

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?

Step-by-Step Solution

Answer

(a)The surface charge density At R ,$\sigma =\frac{q}{4{\mathrm{\pi R}}^2},$At a , $\sigma =\frac{-q}{4{\mathrm{\pi a}}^2}$, At b , $\sigma =\frac{+q}{4{\mathrm{\pi b}}^2}$

(b)The potential at the center, using infinity as the reference point $E\left(r\right)=\frac{q}{4{\mathrm{\pi \epsilon }}_0}\left(\begin{array}{c}\frac{1}{b}+\frac{1}{R}-\frac{1}{a}\end{array}\right)$

(c)The answer to (a) and (b) change is $V_{\left(center\right)}=-{\int }_a^R\frac{q}{4{\mathrm{\pi \epsilon }}_0{\mathrm{r}}^2}dr.$

Step 1: Determine the expressions for the charge density.

(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 , R the 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.

Step 2: Determine potential at center

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

Step 3: Solution for subpart (c)

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

${\sigma }_b=0$

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

Now, 

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

Determine the potential at the center.

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