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Question

(A) The electric energy density $u$ is defined as the energy per unit volume,given by $u = \frac{1}{2} \varepsilon_0 E^2$.For a parallel plate capacit

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The electric energy density $u$ is defined as the energy per unit volume,given by $u = \frac{1}{2} \varepsilon_0 E^2$.For a parallel plate capacitor,the electric field $E$ between the plates is given by $E = \frac{\sigma}{\varepsilon_0}$,where $\sigma = \frac{q}{A}$ is the surface charge density.Substituting $E$ into the energy density formula:$u = \frac{1}{2} \varepsilon_0 \left( \frac{\sigma}{\varepsilon_0} \right)^2 = \frac{\sigma^2}{2\varepsilon_0}$.Substituting $\sigma = \frac{q}{A}$:$u = \frac{(q/A)^2}{2\varepsilon_0} = \frac{q^2}{2\varepsilon_0 A^2}$.Thus,the correct option is $A$.

The mean electric energy density between the plates of a charged capacitor is (here $q$ = charge on the capacitor and $A$ = area of the capacitor plate).

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