Consider a parallel plate capacitor of area A | vedclass

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(A) The energy density $u$ (energy per unit volume) stored in an electric field $E$ is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$. The

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$\frac{1}{2} \varepsilon_0 E^2 Ad$

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(A) The energy density $u$ (energy per unit volume) stored in an electric field $E$ is given by the formula: $u = \frac{1}{2} \varepsilon_0 E^2$. The total volume $V$ of the space between the plates of a parallel plate capacitor is the product of the area $A$ and the separation $d$,so $V = Ad$. The total potential energy $U$ stored in the capacitor is the product of the energy density and the volume: $U = u 	imes V$. Substituting the expressions,we get: $U = (\frac{1}{2} \varepsilon_0 E^2) 	imes (Ad)$. Therefore,the potential energy stored is $U = \frac{1}{2} \varepsilon_0 E^2 Ad$.

Consider a parallel plate capacitor of area $A$ (of each plate) and separation $d$ between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates,then the potential energy stored in the capacitor is $:-$

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