A point charge qis at the center of an uncharged spherical conducting

shell, of inner radius aand outer radius b. Question:How much work would it take to move the charge out to infinity (through a tiny hole drilled in the shell)?

A point charge qis at the center of an uncharged spherical conducting

shell, of inner radius aand outer radius b.

$W=-\frac{1}{8\tau \tau {\epsilon }_0}\frac{q^2}{r}.......\left(1\right)$

Here, ${\epsilon }_0$is the permittivity of free space.

A point charge qinduces -q charge at the inner surface and +q charge on the outer surface of the conducting spherical shell.

Work done to bring the point charge with nothing else nearby is

$W_1=0$

From equation (1), work done to bring the inner spherical shell with -qand radius a is

role="math" localid="1657520476370" $W_2=-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{a}$

From equation (1), work done to bring the inner spherical shell with -qand radiusb is
$W_3=-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{b}$

Thus, the net initial charge is

$W_i=W_1+W_2+W_3$

Substitute the values in the above equation and get

$$\begin{gathered} W_i=0-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{a}+\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{b} \\ =\frac{q^2}{8\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right) \end{gathered}$$

This is the initial energy of the configuration. After the point charge is removed, the spherical shell becomes neutral and thus the final energy of the configuration is zero. Hence, the work done to remove the charge is

$W=0-W_i$

$=\frac{q^2}{8\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)$

Thus, the work done to remove the point charge is $=\frac{q^2}{8\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)$.