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Q.

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

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a

charges on the two capacitors as a function of time are Qd0−vt2d0,Qd0+vt2d0

b

charges on the two capacitors as a function of time are Qd02d0−vt,Qd02d0+vt

c

current in the circuit will increase as time passes on

d

current in the circuit will be constant

answer is A.

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Detailed Solution

Let q1 and q2 be the instantaneous charges on capacitors. Since they are in parallel, thenq1C1=q2C2 and q1+q2=Q C1=ε0Ad0+vt,C2=ε0Ad0−vt So q1q2=C1C2=d0−vtd0+vt or q2d0−vtd0+vt+q2=Q So q2=Qd0+vt2d0 and q1=Qd0−vt2d0Hence, option (a) is correct and option (b) is incorrect.i=−dq1dt or dq2dt or i=Qv2d0 Which does not depend on time. So option (d) is correct and option (c) is incorrect.
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