Two identical parallel plate capacitors $A$ and $B$ are connected in series through a battery of potential…

Two identical parallel plate capacitors $A$ and $B$ are connected in series through a battery of potential difference $V$ (see figure). Area of each plate is $a$ and initially plates of capacitors are separated by a distance $d .$ Now, separation between plates of capacitor $B$ starts increasing at constant rate $v$, find the rate by which work is done on the battery when separation between plates of capacitor $B$ is $2 d$.
  1. $\frac{a \varepsilon_{0} v V^{2}}{d^{2}}$
  2. $\frac{a \varepsilon_{0} v V^{2}}{2 d^{2}}$
  3. $\frac{a \varepsilon_{0} v V^{2}}{9 d^{2}}$
  4. $\frac{2 a \varepsilon_{0} v V^{2}}{d^{2}}$

Solution

Let at any instant separation between plates of capacitor $B$ is $x$, then
$C_{e q}=\frac{\frac{a \varepsilon_{0}}{d} \cdot \frac{a \varepsilon_{0}}{x}}{\frac{a \varepsilon_{0}}{d}+\frac{a \varepsilon_{0}}{x}}=\frac{a \varepsilon_{0}(d x)}{d x(d+x)}=\frac{a \varepsilon_{0}}{d+x}$
$Q=C_{e q} V=\frac{a \varepsilon_{0}}{d+x} V$
$\frac{d Q}{d t}=-\frac{a \varepsilon_{0}}{(d+x)^{2}} V \frac{d x}{d t}=-\frac{a \varepsilon_{0}}{(d+x)^{2}} V v$
Rate of work done on the battery $=-\left(\frac{d Q}{d t}\right) V=\frac{a \varepsilon_{0} v V^{2}}{9 d^{2}} \quad($ when $x=2 d)$ *

Asked in: JEE Mains - Capacitance - Test 2

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