Question

A parallel plate capacitor has its two plates connected to an ideal spring of force constant K. Relaxed length of spring is L and it is made of non-conducting material. The area of each plate is A. The capacitor has a charge $q_0$ on it. To discharge the capacitor through the resistance R, switch S is closed. Assume that there is no friction and the plates always remains parallel to each other.

If the time constant of the circuit is very large and discharge process is very slow, heat dissipated in the resistance is

• A. $\frac{q_0^2L}{2\epsilon_0A}-\frac{q_0^4}{4\epsilon_0^2A^2K}$
• B. $\frac{q_0^2L}{2\epsilon_0A}-\frac{q_0^4}{8\epsilon_0^2A^2K}$
• C. $\frac{q_0^2L}{2\epsilon_0A}-\frac{q_0^4}{4\epsilon_0^2A^2K}$
• D. $\frac{q_0^2L}{2\epsilon_0A}-\frac{q_0^4}{8\epsilon_0^2A^2K}$

Answer 1

Answer: (B)

$\frac{q_0^2L}{2\epsilon_0A}-\frac{q_0^4}{8\epsilon_0^2A^2K}$

Answer 2

The system consists of a parallel plate capacitor and a spring. The capacitor has an initial charge $q_0$. The plates are connected to a spring with relaxed length L and force constant K. The electrostatic force between the plates is attractive, $F_e = \frac{q^2}{2\epsilon_0 A}$. We assume the spring is connected such that it provides a restoring force towards the relaxed length L. Based on the options, we infer that the spring force is $K(L-x)$ inwards, where x is the distance between the plates. The initial state is mechanical equilibrium with charge $q_0$, so $\frac{q_0^2}{2\epsilon_0 A} = K(L-x_0)$, which gives $x_0 = L - \frac{q_0^2}{2\epsilon_0 A K}$. This requires $L > \frac{q_0^2}{2\epsilon_0 A K}$.

The initial total energy of the system is the sum of the initial electrical energy and the initial spring potential energy: $E_{initial} = \frac{1}{2} \frac{q_0^2}{C_0} + \frac{1}{2} K(L-x_0)^2 = \frac{1}{2} q_0^2 \frac{x_0}{\epsilon_0 A} + \frac{1}{2} K(L-x_0)^2$. Substituting $x_0$ and $L-x_0$, we get $E_{initial} = \frac{q_0^2 L}{2\epsilon_0 A} - \frac{q_0^4}{4\epsilon_0^2 A^2 K} + \frac{q_0^4}{8\epsilon_0^2 A^2 K} = \frac{q_0^2 L}{2\epsilon_0 A} - \frac{q_0^4}{8\epsilon_0^2 A^2 K}$.

When the capacitor is fully discharged, $q=0$, and the plates reach the final equilibrium position where the spring force is zero, i.e., $x_f = L$. The final total energy is $E_{final} = 0$.

In the case of very slow discharge, the system remains in mechanical equilibrium throughout the process. By conservation of energy, the heat dissipated in the resistor is equal to the decrease in the total energy of the system: $H = E_{initial} - E_{final} = \frac{q_0^2 L}{2\epsilon_0 A} - \frac{q_0^4}{8\epsilon_0^2 A^2 K}$.