# Why isn't energy stored in a capacitor equal to just $QV$ when it is fully charged? [duplicate]

I just learned that the energy stored in a capacitor is electrostatic potential energy. Also potential energy is a state function and is independent of the path. So, when the capacitor is fully charged to $Q$ and a potential difference of $V$, the potential energy of the system would be $$Q(V_+)-Q(V_-) = Q(\Delta V) = U$$ where, $V_+$ is the potential of the positive plate and $V_-$ is the potential of the negatively charged plate.

This is how we calculate the potential energy of system of charges. Why does this reasoning fail in case of capacitors?

## 1 Answer

Let's say we have a capacitor with capacitance $C$ with stored charge $Q_0$ at an initial potential difference of $V_0$.

Now we are going to discharge it say over a simple resistor and determine how much energy was stored. During discharge in a small time interval $\Delta t$ the energy $\Delta E$ dissipated from the capacitor will be:

$\Delta E = V I \Delta t = V \Delta Q$

with $V$ the instantaneous voltage over the capacitor and $\Delta Q$ the discharge. If $V$ were constant in time you would arrive at your result but it isn't. In fact $V$ decreases over time as the capacitor gets discharged. Less and less energy is retrieved from the capacitor per unit time as it discharges.

To find the energy dissipated you need to sum (integrate) all the small unit energy losses from fully charged to fully discharged. Using $\Delta Q = -C \Delta V$ you get:

$\int \Delta E = -C\int_{V_0}^0V \Delta V=\frac{1}{2}CV_0^2=\frac{1}{2}Q_0V_0$.