Three identical parallel conducting plates A, B and C are placed as shown. Switches `S_1 and S_2` are open, and can connect A and C to earth when closed. +Q charges is given to B.

Options

A. If `S_1`is closed with `S_2` open, a charge of amount Q will pass through `S_1`.
B. If instead `S_2` were closed with `S_1` open, a charge of amount Q will pass through `S_2`.
C. If `S_1 and S_2` are closed together, a charge of amount `2Q//3` will pass through `S_2`.
D. If `S_1 and S_2` are closed together, a charge of amount `2Q//3` will pass through `S_1`, and a charge of amount `Q//3` will pass through `S_2`.

Answer

A, B, C

Explanation

a.,b.,c.
When either `A` or `C` is earthed (but not both together), a parallel plate capcitor is formed with`B`, with `+-Q` charges on the inner surfaces. (The other plate, which is not earthed, plays no role.) Hence, charge of amount `+Q` flows to the earth. When both are earthedtogether, `A` and `C` effectively become connected. The plates now form two capacitors in paralle, with capacitances in the ration `1:2`, and hence share charge `Q` in the same ratio.
`C_(1)=(epsilon_(0)A)/(2d),C_(2)=(epsilon_(0)A)/(d)(C_(1))/(C_(2))=(1)/(2)`
Now we can write
`(q_(1))/(q_(2))=(C_(1))/(C_(2))=(1)/(2)`
because potential difference of both the capacitors is same and `q_(1)+q_(2)=Q`.
Solving these, we get `q_(1)=Q//3,q_(2)=2Q//3`.