Difference between voltage and potential energy in a capacitor I'm confused about the difference between voltage and potential energy in a capacitor. Suppose you have a capacitor with a voltage V and capacitance C, and you release a particle with charge $+Q$ from the positive end of the capacitor. Is the potential energy of the particle that gets converted to kinetic energy $(1/2)CV^2$ or $QV$? What is the difference between these two quantities?
 A: Some definitions might be useful:


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*Potential: the potential energy per unit charge, $V = \frac{U}{q}$. Potential depends only on the environment and the location, not on what is placed at that location.

*Voltage: a difference in potential, $\Delta V$, between two points in the same environment. You can think of this as the change in potential energy per unit charge for a test charge moving between the two points.


So the first answer I would give you is that potential energy depends on the test charge ($+Q$ in your example), but voltage does not, because it's per unit charge.
But I think what you really mean to ask is, why isn't the potential energy of a capacitor, $\frac{1}{2}CV^2$, the same as the potential energy of a charge moving across the capacitor, $QV$? That's because the potential energy of a capacitor represents the energy that had to be put in to move all the charges that are already in the capacitor.


*

*The first charge $Q$ to be moved from one plate to the other didn't need any energy to do it, because at the beginning, the plates were uncharged, and thus at the same potential.

*After one charge had been moved, there was a potential difference $V_1 = Q/C$. In order to overcome that potential difference, the next charge needed energy $QV_1 = Q^2/C$.

*After the second charge had been moved, there was a potential difference $V_2 = 2Q/C$. So the third charge needed energy $QV_2 = 2Q^2/C$.


...and so on. Adding all these up gives
$$\frac{Q^2}{C} + \frac{2Q^2}{C} + \frac{3Q^2}{C} + \cdots + \frac{NQ^2}{C} = \frac{(N^2 + N)Q^2}{2C} \approx \frac{(NQ)^2}{2C}$$
where $NQ$ is the total charge on the capacitor. ($N$ particles, each of charge $Q$.) Of course in practice, we consider infinitesimal elements of charge, and do an integral instead:
$$\int_0^{Q_\text{total}} \frac{q}{C}\mathrm{d}q = \frac{Q_\text{total}^2}{2C}$$
