Question

Separation between the plates of a parallel plate capacitor connected to a battery (zero resistance) of constant e.m.f. is increased with constant (very slow) speed by external forces During the process W is the work done by external forces $\Delta U$ is the change in potential energy of the capacitor, $W_b$ is the work done by the battery and H is the heat loss in the circuit Then

• A. $W+W_b=\Delta U$
• B. $H\ne 0$
• C. $H=\Delta U$
• D. W = 0

Answer 1

The correct option is D W = 0

Since there will be some loss in form of electromagnetic radiations, hence energy conservation equation must take into account that loss. Option A is incorrect.

Heat loss will be 0 since no resistor is present.

Change in potential energy will be non zero since capacitance changes and hence energy by $\frac{1}{2}C{V}^2$

Option D is correct.