Question

Two identical capacitors having plate separation $d_0$ are connected parallel to each other across the 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 the battery is removed. Now the first plate of the first capacitor and the second plate of the second capacitor start moving with constant velocity $u_o$ toward left. The magnitude of the current flowing in the loop during this process is given as $\frac{Q{u}_0}{x{d}_0}$. Find $x$ :

Answer 1

The each plate of a capacitor has equal and opposite charge .
If charge $q$ flow through the loop , the charge distribution should be as shown in the figure. Let each plate move a distance $x$ from its initial position.
Potential across top capacitor is $V_1=E_1(d_0+x)=\frac{(Q/2-q)(d_0+x)}{A{ϵ}_0}$
and potential across bottom capacitor is $V_2=E_2(d_0-x)=\frac{(Q/2+q)(d_0-x)}{A{ϵ}_0}$
As the battery is removed, so $V_1-V_2=0$
or $\frac{(Q/2-q)(d_0+x)}{A{ϵ}_0}=\frac{(Q/2+q)(d_0-x)}{A{ϵ}_0}$
or $\frac{Q{d}_0}{2}-q{d}_0+\frac{Qx}{2}-qx=\frac{Q{d}_0}{2}+q{d}_0-\frac{Qx}{2}-qx$
or $2q{d}_0=Qx\Rightarrow q=\frac{Qx}{2d_0}$
Current $i=\frac{dq}{dt}=\frac{Q}{2d_0}\frac{dx}{dt}=\frac{Q{u}_0}{2d_0}$
thus, $x=2$