CapacitanceHard
Question
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 ?
Options
A.
B.
C.
D.
Solution
Capacitance of capacitor is = C0 = 
C =
C =
[x + l(L − x)] = [kL − (k − 1)x] = 
[kL − (k − 1) vt]
So, C decreases linearly with time
Charge on capacitor Q = C0V0 =
V0 = constant.
Potential difference across plate is V =
⇒ V ∞ 
V =
Potential energy U =
QV = 
C0V0.V
C =
C =
[x + l(L − x)] = [kL − (k − 1)x] = [kL − (k − 1) vt]So, C decreases linearly with time
Charge on capacitor Q = C0V0 =
V0 = constant.Potential difference across plate is V =
⇒ V ∞ V =
Potential energy U =
QV = C0V0.VCreate a free account to view solution
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