A capacitor of unspecified capacitance $C_1$ so that the potential difference across it is $84.0\ \mathrm V$. It is then disconnected from the voltage source and connected in parallel with a $12\ \mathrm{\mu F}$ capacitor. The potential difference across this combination is $28.0\ \mathrm V$. What is $C_1$?

This is my attempt at solving it:

We have that $C_1=\frac Q{84\ \mathrm V}$ and that $C_1+12\ \mathrm{\mu F}= \frac Q{28\ \mathrm V}$. Then $Q=28\ \mathrm V(C_1+12\ \mathrm{\mu F})$ and by substituting into the first equation, we find that $C_1=6\ \mathrm{\mu F}$.

I don't understand why the charge $Q$ is the same before and after we connect the first capacitor to the second capacitor. I thought that should only happen if they were connected in series. I feel like the set up should be $C_1=\frac{Q_1}{84\ \mathrm V}$ and $C_1+12\ \mathrm{\mu F}=\frac{Q_\text{total}}{28\ \mathrm V}$.


1 Answer 1


Connecting capacitors in parallel means that the positive plates are connected together and the negative plates are connected together. The charge on each capacitor probably changes, but the total amount of positive and negative charge is the same as before.

Unlike series connection where the positive and negative plates are connected, so the excess charge is wiped out.

Your second method should give the same answer, if you finish the calculation.


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.