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The electric energy density for a parallel plate capacitor is $\frac{1}{2}\varepsilon E^2$ where $\varepsilon$ is the dielectric constant and E is the magnitude of electric field. In my textbook it says that this is also true for any capacitor in vacuum. How can we prove this statement?

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The energy stored in a capacitor is just: $$W = \frac{1}{2}CV^2=\frac{1}{2}\frac{\varepsilon A}{d}(Ed)^2=\frac{1}{2}\varepsilon A d E^2=\frac{1}{2}V\varepsilon E^2$$ So de energy density per unit of volume is $\frac{1}{2}\varepsilon E^2$. The answer is extracted from https://en.wikipedia.org/wiki/Capacitor

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    $\begingroup$ I believe this doesn't explain why the geometry of the capacitor doesn't matter, you have skipper the integration of $V$ for which it was assumed a parallel plate case. Check (en.wikipedia.org/wiki/Capacitance#Capacitors). It would be nice to add a general proof of $W=CV^2/2$. $\endgroup$ – ohneVal May 29 '18 at 9:03
  • $\begingroup$ Can it be proved for a capacitor with any geometric shape? $\endgroup$ – D. Drake May 29 '18 at 15:09

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