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

This is my attempt at solving it:

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

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 $C_1 = \frac{Q_1}{84V}$ and $C_1 + 12 \mu F = \frac{Q_{total}}{28V}$.

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

Unlike series connection where the +ve and -ve plates are connected, so the excess charge is wiped out.

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

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