Between the plates of a parallel plate capacitor of plate area $\mathrm{A}$ and capacity $0.025 \mu…
Between the plates of a parallel plate capacitor of plate area $\mathrm{A}$ and capacity $0.025 \mu \mathrm{F}$, a metal plate of area, $\mathrm{A}$ and thickness equal to $\frac{1}{3}$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to $100 \mathrm{~V}$, then the amount of work done to remove the metal plate from the capacitor is
$62.5 \mu \mathrm{J}$
$30.2 \mu \mathrm{J}$
$52.6 \mu \mathrm{J}$
$35.4 \mu \mathrm{J}$
Solution
Initially
$c^{\prime}=\frac{\varepsilon A}{d-\frac{d}{3}}=\frac{3}{2} \frac{\varepsilon_0 A}{d}=\frac{3}{2} c$ [C=capacitor of plate without conduction]
$\begin{aligned} & v_i=v_f=100 v \\ & u_i=\frac{1}{2} c v^2 \\ & u_f=\frac{1}{2} c v^2\end{aligned}$
$v_i-v_f=\frac{1}{2}\left(\frac{3}{2} c-c\right) v^2=\frac{1}{2} \times \frac{c}{2} \times v^2$
$\begin{aligned} & =\frac{0.025 \times 10^4}{4} \mu \mathrm{J} \\ & =\frac{250}{4}=62.5 \mu \mathrm{J}\end{aligned}$