If you take two capacitors, one with capacitance C and the other with 2C, and charge them to voltages V and 2V volts, assuming CV = Q, they'd have charges Q and 4Q.

Now when these two are connected in parallel with oppositely charged plates connected, charge flow takes place till terminal potentials are equal. I assumed it'd be something like this:

enter image description here

It seems like the electron density on the negatively charged plate(which I've shaded according to charge intensity) decreases as charge flows to the positive plate(This takes place from both of the negative plates; I've only shown one.). Eventually, it'll be enough to flip the polarity of the capacitor with the smaller charge, and bring down the charge on the other one so that the voltages end up equal.

It seems as though this doesn't check out, though. What am I missing here?

Edit: I've got to find the final energy of the configuration; it's actually $\frac{3CV^2}{2}$, but I'm always getting $\frac{5CV^2}{6}$.

  • 1
    $\begingroup$ Can you elaborate more on why your answer doesn't "check out"? It's unclear what your question really is. $\endgroup$ Jul 8, 2020 at 12:46
  • $\begingroup$ Info added, but I just wanted to know if that was the way it worked. $\endgroup$
    – harry
    Jul 8, 2020 at 13:44

2 Answers 2


Your general technique is correct. The equilibrium configuration will be one in which:

  • The voltage across each of the capacitors is the same, and
  • The amount of charge on each half of the circuit is the same as it was initially. (Note that the left "half" of the circuit is disconnected from the right "half", so there is no way for charge to move from one side to the other.)

From this information, it is possible to deduce the amount of charge on each of the capacitors in the equilibrium, and then to find the total energy stored in the capacitors.


If you connect a plate with +4Q to one with -Q the result will be 3Q divided between the two plates with Q on capacitor C and 2Q on capacitor 2C, giving a voltage of Q/C = 1 volt (on each capacitor in this case).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.