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Q.

A parallel plate capacitor has a dielectric slab in it. The slab just fills the space inside the capacitor. The capacitor is charged by a battery and the battery is disconnected. Now the slab is started to pull out uniformly at t = 0. If at time t, capacitance of the capacitor is C, potential difference across plate is V, and energy stored in it is U, then which of the following graphs are correct ? 

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a

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b

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c

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d

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answer is A, B, C, D.

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Detailed Solution

Capacitance of capacitor is S = C0=k0 a.Ld

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C=0axd+k0 a(Lx)d

C = a0d[x+k(Lx)]   = a0d[kL(k1)x]   =  a0d[kL(k1)vt]

So, C decreases linearly with time 
Charge on capacitor  Q = C0V0=k0 aLd V0= constant.
Potential difference across plate is

V = QC=C0V0C  V 1C

V=V0a0d[kL(k1)vt]

Potential energy  U=12 QV=12   C0V0 VU V

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