Two identical capacitors are connected as shown and having initial charge $Q_{0}$ . Separation between plates of each capacitor is $d_{0}$ . Suddenly, the left plate of upper capacitor and right plate of lower capacitor starts moving with speed $v$ towards left while other plate of each capacitor remains fixed.
(given $\frac{Q_{0} V}{2 d_{0}} = 10 A$ ). The value of current (in $A$ ) in the circuit is

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Solution

From conservation of charge,
$q_{1} + q_{2} = 2 Q_{0} . . . . (1)$
At time $t$
$C_{1} = \frac{ϵ_{0} A}{d_{0} + V t}, C_{2} = \frac{ϵ_{0} A}{d_{0} - V t}$

$\frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{2}}$

$\frac{q_{1}}{q_{2}} = \frac{d_{0} - V t}{d_{0} + V t}$
Substituting in $(1)$
$\Rightarrow q_{2} + q_{2} (\frac{d_{0} - V t}{d_{0} + V t}) = 2 Q_{0}$
$\Rightarrow q_{2} (\frac{2 d_{0}}{d_{0} + V t}) = 2 Q_{0}$
$\Rightarrow q_{2} = \frac{2 Q_{0}}{2 d_{0}} (d_{0} + V t) I = \frac{d q_{2}}{d t} = \frac{Q_{0} V}{d_{0}} = 20 A$

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From conservation of charge, q1+q2=2Q0....(1) At time t C1=ϵ0Ad0+Vt, C2=ϵ0Ad0−Vt q1C1=q2C2 q1q2=d0−Vtd0+Vt Substituting in (1) ⇒q2+q2(d0−Vtd0+Vt)=2Q0 ⇒q2(2d0d0+Vt)=2Q0 ⇒q2=2Q02d0(d0+Vt)I=dq2dt=Q0Vd0=20 A