Question

The two identical plates are given charges as shown in figure. If the plate area of either face of each plate is $A$ and separation between the plates is $d$, then find the amount of heat liberated after closing the switch.

• A. $\frac{q^{2}d}{{ϵ}_{0}A}$
• B. $\frac{q^{2}d}{4{ϵ}_{0}A}$
• C. $\frac{2q^{2}d}{{ϵ}_{0}A}$
• D. $\frac{q^{2}d}{2{ϵ}_{0}A}$

Answer 1

The correct option is D $\frac{q^{2}d}{2{ϵ}_{0}A}$

Before closing the switch:

Charge distribution on outer plate will be
$q_{outer}=\frac{3q+q}{2}=2q$


This means charge on the inner plates will be respectively $3q-2q=q$ and $q-2q=-q$.
This forms a capacitor system.

So, initial energy stored in capacitor will be

$E_i=\frac{q^2}{2C}$

Where, $C=\frac{{ϵ}_{0}A}{d}$

When the switch is closed:

Potential difference across the plates will become zero. As a result, no energy will remain stored in the capacitor and the energy initially stored will be dissipated as heat.

So, heat dissipated will be

$H=\frac{q^2}{2C}$$=\frac{q^{2}d}{2{ϵ}_{0}A}$

Therefore, option $\left(d\right)$ is correct.