Question

What capacitance is needed to store 3.00μC of charge at a voltage of 120 V?

Answer 1

The capacitance needed to store $3\mu C$ of charge when $120$ volts is applied can be found by taking the charge divided by the voltage.

That is

$\displaystyle C=\frac{3.00 \mu C\times \left[\left[\frac{10^{-6}C}{1\mu C}\right]\right]}{120V}$

$\displaystyle C=2.50\times10^{-8}F\left\{\frac{1nF}{10^{-9}F}\right\}$

$C=25.0nF$

which is the capacitance needed.