Question

Separation between plates of a parallel plate capacitor of capacitance $\frac{{ϵ}_{o}A}{d}$ connected to a battery of EMF E is doubled very slowly. Choose the correct option(s).

• A. Heat generated in circuit is $\frac{E^2{ϵ}_{o}A}{2d}$
• B. Work done by battery when the separation is increasing is $-\frac{E^2{ϵ}_{o}A}{2d}$
• C. Work done by electrostatic force is positive during process.
• D. Heat generated in circuit is zero.

Answer 1

Answer: (B), (D)

The correct options are
B Work done by battery when the separation is increasing is $-\frac{E^2{ϵ}_{o}A}{2d}$
D Heat generated in circuit is zero.

Work done by battery is $W=E△q$
It is clear that $△q$ is negative here.

$△q=\frac{CE}{2}-CE=-\frac{CE}{2}=-\frac{{ϵ}_{o}EA}{2d}$
$W=-\frac{{ϵ}_o{E}^{2}A}{2d}.$


As the process is done very slowly, heat generated in process is zero.