# Why does total charge stay the same when capacitors are connected in parallel?

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 $$C_1+12\ \mathrm{\mu F}=\frac{Q_\text{total}}{28\ \mathrm V}$$.