# Derivation of energy in capacitor

While searching for something totally unrelated to this, I came a cross a website that derived it in this fashion: the instanteous power in a capacitor is given by $$p_c= v_c(t)\cdot i_c$$

since $$i_c(t) = C\frac{dv_c}{dt}$$, this becomes $$p_c = v_c(t)\cdot C\frac{dv_c}{dt}$$

No issues so far....but, he then proceeds to write: $$\frac{dw_c(t)}{dt}=\frac{d}{dt}[\frac{1}{2}Cv_c^2(t)]$$.

power is the derivative of energy, so I get the left hand side of the equation. However, how does $$C\frac{dv_c}{dt}\cdot v_c(t)=\frac{d}{dt}[\frac{1}{2}Cv_c^2(t)]$$ on the right hand side of the equation?

• Your question is Energy and you start off with power . Recheck.. Integrate (t) to get E and then you get E=1/2CV² Jul 24, 2020 at 21:49

$$\\frac{d}{dt}v_c^2(t) = 2v_c(t)\frac{dv_c(t)}{dt}\$$
therefore $$\v_c(t)\frac{dv_c(t)}{dt} = \frac{1}{2}\frac{d[v_c^2(t)]}{dt}\$$
• @Tony Not sure what you are getting at. $v_c(t)$ is the instantaneous voltage across the cap at time t, while charging. What do you mean by 'charged up with DC?' Jul 24, 2020 at 22:16