Question

In a system of two concentric spherical conducting shells charge $q$ is given to both inner and outer shells as shown in figure. Inner shell is connected to earth by a switch. Find the amount of heat produced when switch is closed.

• A. $\frac{k{q}^2}{4a}$
• B. $\frac{k{q}^2}{8a}$
• C. $\frac{9k{q}^2}{8a}$
• D. $\frac{6k{q}^2}{7a}$

Answer 1

The correct option is C $\frac{9k{q}^2}{8a}$

Initial potential energy of the system

$U_{initial}=\sum U_{self}+U_{interaction}$

Substituting the values,

$\Rightarrow U_{initial}=\frac{k{q}^2}{2a}+\frac{k{q}^2}{2\times 2a}+\left(\frac{k{q}^2}{2a}\right)$

$\therefore U_{initial}=\frac{5k{q}^2}{4a}$

When the inner sphere is grounded let's assume $q^{\prime }$ be the final charge on inner sphere such that electric potential on it, $V_{inner}=0$,

$\Rightarrow \frac{k{q}^{\prime }}{a}+\frac{kq}{2a}=0$

$\Rightarrow q^{\prime }=-\frac{q}{2}$

Final energy $U_{final}$ of the system:

$U_{final}=\sum U_{self}+U_{interaction}$

$\Rightarrow U_{final}=\frac{k{q}^{\prime 2}}{2a}+\frac{k{q}^2}{2\left(2a\right)}+\frac{k{q}^{\prime }}{2a}q$

$\Rightarrow U_{final}=\frac{k{q}^2}{8a}+\frac{k{q}^2}{4a}-\frac{k{q}^2}{4a}$

$\Rightarrow U_{final}=\frac{k{q}^2}{8a}$

Amount of Heat produced
$=(U_i-U_f)$

$=\frac{5k{q}^2}{4a}-\frac{k{q}^2}{8a}$

$=\frac{9k{q}^2}{8a}$

Hence, option (c) is the correct answer.