Question

A thick conductor having inner radius $a$ and outer radius $b$ has a small passage as shown in the figure.


A charge $+Q$ is placed at the centre. The work done by external agent in taking the charge from centre to infinity is

• A. $-\frac{Q^2}{4\pi {\epsilon }_{0}a}-\frac{Q^2}{4\pi {\epsilon }_{0}b}+\frac{Q^2}{4\pi {\epsilon }_0(a-b)}$
• B. $-\frac{Q^2}{8\pi {\epsilon }_{0}b}$
• C. $\frac{Q^2}{8\pi {\epsilon }_{0}a}$
• D. $\frac{Q^2}{8\pi {\epsilon }_{0}a}-\frac{Q^2}{8\pi {\epsilon }_{0}b}$

Answer 1

The correct option is D $\frac{Q^2}{8\pi {\epsilon }_{0}a}-\frac{Q^2}{8\pi {\epsilon }_{0}b}$

The induced charge will be as shown in the figure,


Initial electrostatic energy of the given system at the center of the shell is

$U_i=(U_{\text{self}}{)}_1+(U_{\text{self}}{)}_2+\left[\text{Ineraction energy}\right].....\left(1\right)$

Where,
$(U_{\text{self}}{)}_1=$Electrostatic potential energy at the centre of the shell due to charge induced at inner surface$=\frac{Q^2}{8\pi {\epsilon }_{0}a}$


$(U_{\text{self}}{)}_2=$Electrostatic potential energy at the centre of the shell due to charge induced at outer surface$=\frac{Q^2}{8\pi {\epsilon }_{0}b}$

Interaction energy $E=$ Electrostatic potential energy due to induced charges and the centre charge$=-\frac{Q^2}{4\pi {\epsilon }_{0}a}-\frac{Q^2}{4\pi {\epsilon }_{0}b}+\frac{Q^2}{4\pi {\epsilon }_{0}b}$

Substituting the values in equation $1$ we get,
$U_i=\frac{Q^2}{8\pi {\epsilon }_{0}a}+\frac{Q^2}{8\pi {\epsilon }_{0}b}-\frac{Q^2}{4\pi {\epsilon }_{0}a}-\frac{Q^2}{4\pi {\epsilon }_{0}b}+\frac{Q^2}{4\pi {\epsilon }_{0}b}$

$\Rightarrow U_i=\frac{Q^2}{8\pi {\epsilon }_{0}b}-\frac{Q^2}{8\pi {\epsilon }_{0}a}$

When the charge is taken to infinite the electrostatic potential energy of the system will be zero.i.e.,
$U_f=0$

Work done by external agent in taking the charge from centre to infinity is
$$W_{ext}=U_f-U_i$$Substituting the values, we have

$W_{ext}=0-(\frac{Q^2}{8\pi {\epsilon }_{0}b}-\frac{Q^2}{8\pi {\epsilon }_{0}a})$

$\therefore W_{ext}=-\frac{Q^2}{8\pi {\epsilon }_{0}b}+\frac{Q^2}{8\pi {\epsilon }_{0}a}$

Hence, option (a) is correct option.