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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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answer is A, B, C, D.

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Detailed Solution

Capacitance of capacitor is $S = C_{0} =$ $\frac{k \in_{0} a . L}{d}$

$C = \frac{\in_{0} a x}{d} + \frac{k \in_{0} a (L - x)}{d}$

C = $\frac{a \in_{0}}{d} [x + k (L - x)]$ = $\frac{a \in_{0}}{d} [k L - (k - 1) x]$ = $\frac{a \in_{0}}{d} [k L - (k - 1) v t]$

So, C decreases linearly with time
Charge on capacitor $Q = C_{0} V_{0} =$ $\frac{k \in_{0} a L}{d}$ $V_{0} = c o n s t a n t .$
Potential difference across plate is

V = $\frac{Q}{C} = \frac{C_{0} V_{0}}{C}$ $\Rightarrow$ V $\infty$ $\frac{1}{C}$

$V = \frac{V_{0}}{\frac{a \in_{0}}{d} [k L - (k - 1) v t]}$

Potential energy $U = \frac{1}{2}$ $Q V = \frac{1}{2}$ $C_{0} V_{0} V \Rightarrow U \infty V$

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