Q60P - Introduction to Electrodynamics

Question

A point charge q is at the center of an uncharged spherical conducting

shell, of inner radius a and 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)?

Step-by-Step Solution

Answer

The work done to move a point charge q at the center of an uncharged spherical conducting shell of inner radius a and outer radius b to infinity is  $\frac{q^2}{8\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)$.

Step 1: Given data

A point charge q is at the center of an uncharged spherical conducting

shell, of inner radius a and outer radius b.

Step 2: Work done to spread charge -q on a sphere of radius r with another charge q at the center

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

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

Step 3: Work done to remove center charge

A point charge q induces -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 -q and radius a is

$W_2=-\frac{1}{8\pi {\epsilon }_0}\frac{q^2}{a}$

From equation (1), work done to bring the inner spherical shell with -q and radius b  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  $=\frac{q^2}{8\pi {\epsilon }_0}\left(\frac{1}{a}-\frac{1}{b}\right)$.