The maximum energy stored by a capacitor is when it’s (theoretically) fully charged, or just before it reaches dielectric breakdown, if you will.

It seems, then, that the maximum energy ($E$) $=QV$, where $Q$ is the maximum charge attained and $V$ is the maximum ‘voltage’.

However, the correct equation for the maximum energy is $E=\frac{1}{2}QV$. Why?

I’m looking for a simple, intuitive explanation because my reasoning that $E=QV$ seems sound to me.

I know about Why is there a $\frac 1 2$ in $\frac 1 2 mv^2$?, but I’m looking for a reason to disprove my specific argument that $E=QV$ and not the “Oh, it comes from integration.” response.


3 Answers 3


You can calculate the energy by assuming you charge the capacitor with a constant current.

Then the energy input to the capacitor is


Since I is constant, you know that (for a linear capacitor) V(t) is a ramp function. So from geometry (the area of a triangle formula) the integral is


Now, $It$ is the total charge that's been given to the capacitor and $V(t)$ is the final voltage after charging, so this is the same as the $\frac{1}{2}QV$ formula you are familiar with.

Physical explanation

Imagine you start with an uncharged capacitor. The first tiny element of charge you drive onto the plates takes practically no energy, because the voltage starts at 0. As you add more charge to the plates the voltage increases and it takes more energy to add each additional element of charge. Finally, the last element takes $V \cdot dQ.$ energy to push onto the plates.

If all the charge took equal energy to push onto the plates, you would end up with your formula, $E=QV\;.$ But since the first bits of charge you put on took less energy, you end up with an overall average of just $E=\frac{1}{2}QV\;.$ To figure out why the pre-factor is exactly 1/2, you could either go to the integration formula, or use the geometric argument which is how I mentally calculated the integral anyway.

  • 2
    $\begingroup$ OP states that he wants a physical answer and not the 'Oh, it comes from integration' response. $\endgroup$
    – Kyle Kanos
    May 12, 2014 at 19:17
  • $\begingroup$ @KyleKanos, OK, I added an attempt to explain it physically. $\endgroup$
    – The Photon
    May 12, 2014 at 19:39
  • $\begingroup$ @ThePhoton Yes your physical explanation is perfectly what I'm looking for. Thanks $\endgroup$
    – hb20007
    May 12, 2014 at 19:41

This'll make you mad, but the more typical formulation is $$ E = \frac12 CV^2 $$ which makes it a little more obvious that the half is what you get when after integrating $\int CV dV$.


The half comes from exactly the same source. It's the same idea if you put Q = CV, and then you have C~m, and V~v in $E=\frac 12 mv^2$.


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