I want to determine the charge in a capacitor. I didn't remember the formula for it, \$Q = C \cdot V\$, so I tried to derive it. This is how I went forwards. I've made the assumption that the capacitor has no initial charge. \$v = \frac{dw}{dq} \Leftrightarrow dw = v \space dq \Rightarrow \int dw = \int v \space dq \Rightarrow w = v \space q\$.

Since the energy stored in a capacitor is \$w = \frac{1}{2} C v_c^2\$, I tried to plug this in which results in the equation for charge: \$q = \frac{1}{2} C v_c\$.

This is half of the expected value, so where it the faulty logic here?

If I start by using the definition of current, I get the correct equation:

\$i = \frac{dq}{dt} \Leftrightarrow dq = i \space dt \$

\$i_c = C \frac{dv_c}{dt} \Leftrightarrow i_c \space dt = C \space dv_c\$

\$dq = C \space dv_c \Rightarrow q = C \space v_c\$

  • 2
    \$\begingroup\$ Oh no... this "half of energy" thing again.. \$\endgroup\$
    – Eugene Sh.
    Nov 30, 2017 at 20:32
  • 1
    \$\begingroup\$ \$v\$ is not a constant, so your integral of \$ v\:dq\$ is wrong. \$\endgroup\$
    – Chu
    Nov 30, 2017 at 20:40
  • \$\begingroup\$ @Chu Of course! Silly me, there had to be something wrong with the math. Thanks! \$\endgroup\$
    – eirik-ff
    Nov 30, 2017 at 21:21

1 Answer 1


May as well document what Chu pointed out:

$$\begin{align*} \textrm{d} w &= v\:\textrm{d} q& Q&=C\: V\\ &\therefore\\ \int \textrm{d} w &= \int_0^Q v\:\textrm{d} q\\\\ W &= \int_0^Q v\:\textrm{d} q & &=\int_0^Q \frac{q}{C}\:\textrm{d} q\\\\ &= \frac{1}{C} \int_0^Q q\:\textrm{d} q & &= \frac{1}{C} \left[\frac{1}{2}q^2\right]\bigg|_0^Q\\\\ &= \frac{1}{C} \left[\frac{1}{2}Q^2\right] & &= \frac{1}{C} \left[\frac{1}{2}C^2V^2\right]\\\\ &= \frac{1}{2} C \: V^2 \end{align*}$$

That's the work/energy stored on a capacitor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.