# How are charges distributed on a capacitor in the following circuit?

Four metallic plates are used to form two identical capacitors as shown in the figure.

Initially the plates were uncharged. Two batteries of different emf are connected as shown.
We have to find the charge on each capacitor after steady state is achieved.
So first I took charges on lower plates of X and Y to be $$q$$ and $$q'$$.Then applied Kirchhoff's law.
$$5-\frac qC-2+\frac{q'}{C}=0$$ where C is the capacitance of the given capacitors.
Now I have no idea how to proceed further. I thought a lot about it, but still I am having difficulty in analyzing how charges are given out and distributed by each battery. (Ignore the dots present in each capacitor).
Given answer is $$q=-q'=\frac{3C}{2}$$

• If capacitors in series have same C the charges q and q' should be identical and with same orientation.... May 12, 2021 at 11:53
• @Krešimir Bradvica ,If that is the case then the kirchhoffs equation doesn't make sense as it becomes 3=0 May 12, 2021 at 12:06
• I just said that capacitor terms in the equation should have same sign..... May 12, 2021 at 15:22