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A charge Q is imparted to two identical ...

A charge Q is imparted to two identical capacitors in paralle. Separation of the plates in each capacitor is `d_0`. Suddenly, the first plate of the first capacitor and the second plate of the second capacitor start moving to the left with speed v, then

A

charges on the two capacitors as a function of time are `(Q(d_0-vt))/(2d_0), (Q(d_0+vt))/(2d_0)`.

B

charges on the two capacitors as a function of time are `(Qd_0)/(2(d_0-vt)), (Qd_0)/(2(d_0+vt))`.

C

current in the circuit will increase as time passes on

D

current in the circuit will be constant.

Text Solution

Verified by Experts

The correct Answer is:
A, D
a.,d.
Leq `q_(1)` and `q_(2)` be the instantaneous charges on capacitors. Since they are in parallel, then `(q_(1))/(C_(1))=(q_(2))/(C_(2))` and `q_(1)+q_(2)=Q`
`C_(1)=(epsilon_(0)A)/(d_(0)+vt),C_(2)=(epsilon_(0)A)/(d_(0)-vt)`
So `(q_(1))/(q_(2))=(C_(1))/(C_(2))=(d_(0)-vt)/(d_(0)+vt)` or `q_(2)((d_(0)-vt)/(d_(0)+vt))+q_(2)=0`
So `q_(2)=(Q(d_(0)+vt))/(2d_(0))` and `q_(1)=(Q(d_(0)-vt))/(2d_(0))`
Hence, option (a) is correct and option (b) is incorrect.
`i=(-dq_(1))/(dt)` or `(dq_(2))/(dt)` or `i=(Q_(v))/(2d_(0))`
Which does not depend on time. So option (d) is correct and option (c) is incorrect.
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