A parallel plate capacitor of capacitance ‘C’ has charges on its plates initially as shown in the figure, now at t=0, the switch ‘S’ is closed. Select the CORRECT alternative(s) for this circuit diagram.

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In steady state the charges on the outer surfaces of plates ‘A’ and ‘B’ will be the same in magnitude and sign

In steady state the charges on the inner surfaces of plates ‘A’ and ‘B’ will be the same in magnitude and opposite in sign

In steady state the charges on the outer surfaces of plates ‘A’ and ‘B’ will be the same in magnitude and opposite in sign

The work done by the battery till time steady state is reached is $\frac{5 ε^{2} C}{2}$

answer is A, C, D.

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Detailed Solution

Suppose charge flown through the battery is Q, then charge distribution will be as:
The electric field in the region between A and B is = $\frac{Q - 2 ε C}{2 A ε_{0}} - \frac{ε Q - Q}{2 A ε_{0}} = \frac{2 Q - 3 ε C}{2 A ε_{0}}$
P.D. between the plates, $\frac{2 Q - 3 ε C}{2 A ε_{0}} . d = ε$
$\Rightarrow \frac{2 Q - 3 ε C}{2} \frac{1}{C} = ε$
$\Rightarrow 2 Q = 5 ε C \Rightarrow Q \frac{5 ε C}{2}$
w.d. by battery $= ε Q = \frac{5 ε^{2} C}{2}$