Three identical parallel conducting plates A, B and C are placed as shown.+Q charge is given to B. Plates A and C can be connected to earth via switch $S_{1}$ and $S_{2}$ respectively as shown,+Q charge is given to B.

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If $S_{1} a n d S_{2}$ are closed together, a charge of amount $\frac{Q}{3}$ will pass through $S_{1}$ and a charge of amount $\frac{2 Q}{3}$ will pass through $S_{2}$

If $S_{1} a n d S_{2}$ are closed together, a charge of amount $\frac{Q}{3}$ will pass through $S_{2}$ and a charge of amount $\frac{2 Q}{3}$ will pass through $S_{1}$

If $S_{1}$ is closed with $S_{2}$ open, a charge of amount Q will pass through $S_{1}$

If $S_{2}$ is closed with $S_{1}$ open, a charge of amount Q will pass through $S_{2}$

answer is A, B, C.

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Detailed Solution

$(Q - \frac{Q}{3}) i . e ., \frac{2 Q}{3}$ $\Rightarrow q = \frac{- Q}{3}$ $C_{1} = \frac{A ε_{0}}{2 d}, C_{2} = \frac{A ε_{0}}{d}$
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Charge on plate $B_{1}$ + charge on plate $B_{1} = Q$
1. Now if only is closed then charge on plate C will still be zero
charge on plate $B_{1} = 0$
total charge + Q is on plate $B_{1}$
-Q amount of charge will flow through $S_{1}$
So as to form a capacitor $C_{1}$
II. Similarly, if only $S_{2}$ is closed -Q amount of charges will flow through $S_{2}$
in both of above cases
Charge of amount Q (sign is not taken) is flown
Question Image
Now, if $S_{1} a n d S_{2}$ both are closed let the distribution charges on the capacitors are
Now $\frac{- q}{C_{1}} - \frac{(Q + q)}{C_{2}} = 0$

$\Rightarrow \frac{- Q}{3}$ amount of charge is flown through $S_{1}$ and amount of charge is flown through $S_{2}$