Two identical capacitors having plate separation $d_{0}$ are connected parallel to each other across points A and B as shown in figure. A charge Q is imparted to the system by connecting a battery across A and B and battery is removed. Now first plate of first capacitor and second plate of second capacitor starts moving with constant velocity $u_{0}$ towards left. Find the magnitude of current flowing in the loop during the process.

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$\frac{Q}{2 d_{0}} u_{0}$

$\frac{Q}{d_{0}} u_{0}$

$\frac{2 Q}{d_{0}} u_{0}$

$\frac{Q}{3 d_{0}} u_{0}$

answer is A.

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Detailed Solution

Let each plate moves a distance x from its initial position. Let q charge flows in the loop. Using Kirchhoff's voltage law

$\begin{array}{l} \frac{(\frac{Q}{2} - q) (d_{0} + x)}{ϵ_{0} A} - \frac{(\frac{Q}{2} + q) (d_{0} - x)}{ϵ_{0} A} = 0 \\ \Rightarrow q = \frac{Qx}{2 d_{0}} \\ ∴ I = \frac{dq}{dt} = \frac{Q}{2 d_{0}} (\frac{dx}{dt}) = \frac{Q}{2 d_{0}} u_{0} \end{array}$