Question

A parallel plate air capacitor has a initial capacitance $C$. If plate separation is slowly increased from $d_1$ to $d_2$, then mark the correct statement(s). (Take potential of the capacitor to be constant, i.e., throughout the process it remains connected to battery.)

• A. Work done by electric force= negative of work done by the external agent.
• B. Work done by external force $-\int \overrightarrow{F}.\overrightarrow{d}x$, where $\overrightarrow{F}$ is the electric force of attraction between the plates at plate separation $x$.
• C. Work done by electric force $\ne$ negative of work done by external agent.
• D. Work done by battery $=$ two times the change in electric potential energy stored in capacitor.

Answer 1

Answer: (A), (B), (D)

The correct options are
A Work done by electric force= negative of work done by the external agent.
B Work done by external force $-\int \overrightarrow{F}.\overrightarrow{d}x$, where $\overrightarrow{F}$ is the electric force of attraction between the plates at plate separation $x$.
D Work done by battery $=$ two times the change in electric potential energy stored in capacitor.

$C=\frac{{\epsilon }_{0}A}{d_1},C^{{}^{\prime }}=\frac{{\epsilon }_{0}A}{d_2}$
Work done by battery is
$W_b=V\times chargeflown$
charge flow Q=$CV$, $C=\frac{{ϵ}_{0}A}{d}$
charge flow Q=$\frac{{\epsilon }_{0}A}{d}V$
$W_b={\epsilon }_{0}A{V}^2[\frac{1}{d_2}-\frac{1}{d_1}]$
Change in potential energy of capacitor is
$\Delta U=\frac{1}{2}(C^{\prime }-C{)V}^2$
$=\frac{1}{2}{\epsilon }_{0}A{V}^2[\frac{1}{d_2}-\frac{1}{d_1}]$