A capacitor is two plates that hold charge apart. How much it holds per volt you apply is its capacitance: $C = \dfrac{Q}{V}$, read aloud C equals Q over V. $C$ is the capacitance in farads (F), $Q$ is the charge stored on one plate in coulombs, and $V$ is the voltage across the plates in volts. The farad is enormous — a one-farad capacitor at one volt holds a whole coulomb — so real parts are labelled in microfarads ($\mu\mathrm{F} = 10^{-6}\ \mathrm{F}$) and nanofarads ($\mathrm{nF} = 10^{-9}\ \mathrm{F}$). Convert BEFORE the sockets, every time.

Charging a capacitor costs energy, and the bill is $E = \tfrac{1}{2} C V^2$ — E equals one half C V squared, with $E$ the stored energy in joules. Both oddities earn their place. The half: the first coulomb steps onto an empty plate for free, the last one has to shove past everything already there, so you pay the AVERAGE voltage, which is half the final one. The square: double the voltage and you quadruple the energy — which is why a camera flash charges to hundreds of volts rather than fattening the capacitor.

Push the units through and the check is one-way, as always. Farads times volts gives coulombs; farads times volts squared gives coulomb-volts, and a coulomb-volt is a joule. If your rearrangement will not cancel to the answer's units, it is wrong — no appeal. Units that DO work out never prove you right.