What is the energy density of a parallel plat | vedclass

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Question

(A) The energy density $u$ of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.For a parallel plate capacitor,the electric field $E$

Vedclass · Accepted Answer

$\frac{Q^2}{2\varepsilon_0 A^2}$

Vedclass · Answer

(A) The energy density $u$ of an electric field is given by $u = \frac{1}{2} \varepsilon_0 E^2$.For a parallel plate capacitor,the electric field $E$ between the plates is $E = \frac{\sigma}{\varepsilon_0} = \frac{Q}{A \varepsilon_0}$.Substituting this value of $E$ into the energy density formula:$u = \frac{1}{2} \varepsilon_0 \left( \frac{Q}{A \varepsilon_0} \right)^2$$u = \frac{1}{2} \varepsilon_0 \left( \frac{Q^2}{A^2 \varepsilon_0^2} \right)$$u = \frac{Q^2}{2 \varepsilon_0 A^2}$.

What is the energy density of a parallel plate capacitor with area $A$ and charge $Q$?

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