Q.

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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a

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

b

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

c

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

d

The work done by the battery till time steady state is reached is  5ε2C2

answer is A, C, D.

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Detailed Solution

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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 = Q2εC2Aε0εQQ2Aε0=2Q3εC2Aε0
P.D. between the plates,  2Q3εC2Aε0.d=ε
2Q3εC21C=ε
2Q=5εCQ5εC2
w.d. by battery =εQ=5ε2C2
 

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