Two identical capacitors connected as shown and having initial charge `Q_(0)` separation between plates of capacitor is `d_(0)`. Suddenly the left plate of upper capacitor start moving with velocity V towards left and right plate of capacitor remains fixed, (given `(Q_(0)v)/(2d)=1A`). Find the value of current (in amp) in the circuit

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.
.
`((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))`.