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

enter image description here

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 $q=-q'=\frac{3C}{2}$

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

1 Answer 1


I agree with Mr. Bradvica that the charges should be equal, but the left side capacitor is positive on the bottom, and the other is positive on top. The 3 volt potential difference is divided between them.

  • $\begingroup$ Can you please explain why and how.As I am unable to proceed with equations $\endgroup$ May 12, 2021 at 14:14

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