Two identical capacitors haveng plate separation `d_(0)` are connected parallel to each other across the points A and B as shown in. A charge Q is imparted to the system by removed Now the first plate of the first capacitor and the second plate of the second capacitor start moving with constant velocity `u_(0)` toward left. Find the magnitude of the current flouing in the loop during this process.
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Answer

`(Qu_(0))/(2d_(0))`

Explanation

Let each plate move a distance `x` from its initial position.
Let `q` charge flow in the loop. Using `KVL`, we have.
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`((Q/2-q)(d+x))/(epsilon_(0)A)-((Q/2+q)(d-x))/(epsilon_(0)A)=0`
or `q=(Qx)/(2d_(0)),`
`I=(dq)/(dt)=Q/(2d_(0))((dx)/(dt))=(Qu_(0))/(2d_(0))`.