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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\$

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    \$\begingroup\$ Oh no... this "half of energy" thing again.. \$\endgroup\$
    – Eugene Sh.
    Nov 30, 2017 at 20:32
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    \$\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

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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.

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