Energy in capacitors: A charge of 2.00 μC flows onto the plates of a capacitor when it is connected to a 12.0-V potential source. What is the minimum amount of work that must be done in charging this capacitor?

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by $E=\frac{1}{2}CV^2=\frac{Q^2}{2C}$, where $Q$ is the charge on the capacitor and $V$ is the potential difference across it. In this case, $Q=2.00\mu C$ and $V=12.0$V, so the energy stored in the capacitor is: 
$$E=\frac{(2.00\mu C)^2}{2\cdot\frac{\epsilon_0A}{d}}=\frac{(2.00	imes10^{-6}C)^2}{2\cdot\frac{8.85	imes10^{-12}m^{-1}}{0.001m}}\approx 12.0\mu J$$
Therefore, the minimum amount of work that must be done to charge the capacitor is also 12.0 μJ, since all of the energy stored in the capacitor comes from the work done by the potential source. Thus, the answer is C.

Energy in Capacitors: a Charge of 2