A circuit is connected as shown in the figure with the switch S open. When the switch is closed the total amount of charge that flows from $Y$ to $X$ is?
(C) 27 $\mu$C
(D) 81 $\mu$C
1) Figure out initial charges on both capacitors. (18 $\mu$C on both)
2) Figure out final charges on both capacitors. (9 and 36 $\mu$C on left & right capacitors respectively)
How does the charge move around? What directions do they take?
Attempt at this problem:
When a capacitor is charged, conventional current flows into the positive terminal of the capacitor, and when the capacitor is discharged, current flows out of the positive terminal.
So, if positive charge enters the positive terminal, charges the capacitor, then exit of positive charge from negative terminal should charge the capacitor as well? Then, if positive charge exits from positive terminal, discharging the capacitor, then entrance of positive charge from negative terminal should discharge the capacitor?
If I utilize this logic here, +9$\mu$C (micro-coloumbs) flows from $Y$ to $X$ to the left capacitor (9$\mu$C is lost by this capacitor). Then if +18$\mu$C flows from $Y$ to $X$ to the right capacitor, the charge on it goes to 36$\mu$C (final) from the 18$\mu$C (initial).
Hence total charge flown from $Y \rightarrow X$ is $9 + 18 = 27\mu$C (which is the correct answer to this question).
Is this logic correct? If not, is this an OK analogy to the correct idea?
Questions taken from: JEE(A) 2007 (Part - 1)