A capacitor is two plates that do not touch. Push charge onto one and an equal charge leaves the other, and a voltage appears between them. How much charge it takes to raise that voltage by one volt is the capacitor's capacitance.

$C = \dfrac{Q}{V}$ — read aloud C equals Q over V. $C$ is the capacitance in farads; $Q$ is the charge stored, in coulombs; $V$ is the voltage across the plates, in volts. One farad is one coulomb per volt, which is an enormous capacitance — real components live in microfarads (µF, a millionth) and nanofarads (nF, a thousand-millionth), and that is where the arithmetic gets dangerous. Rearranged, $Q = CV$ going one way and $V = \dfrac{Q}{C}$ coming back.

Charging a capacitor takes work, and the work does not vanish — it sits in the electric field between the plates. $E = \tfrac{1}{2} C V^{2}$, read aloud E equals half C V squared, with $E$ the stored energy in joules, and $C$ and $V$ as before. Two characters in that line do all the damage. The ½ is there because the voltage climbed from zero to V while the charge went in, so the average voltage doing the work was half the final one — leave it out and you report twice the energy. The square means doubling the voltage banks four times the energy, which is why capacitor banks are specified by their voltage rating first and their capacitance second.

Units guide, they do not confess. Push them through: coulombs over volts is a farad, so $C = Q/V$ survives the check; farads times volts squared is a joule, so $\tfrac{1}{2}CV^{2}$ survives too. If the units refuse to cancel into the answer's units, the rearrangement is wrong, no appeal — but units that do work out never prove you right. The check runs one way only, and it will never catch a missing ½.