What is the average electric energy density b | vedclass

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Question

(A) The electric field $E$ produced by a single charged plate with surface charge density $\sigma = q/A$ is given by $E = \frac{\sigma}{2 \epsilon_0}

Vedclass · Accepted Answer

$\frac{q^2}{2 \epsilon_0 A^2}$

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(A) The electric field $E$ produced by a single charged plate with surface charge density $\sigma = q/A$ is given by $E = \frac{\sigma}{2 \epsilon_0} = \frac{q}{2 A \epsilon_0}$.In a parallel plate capacitor,the electric field between the plates is the sum of the fields due to both plates: $E_{total} = \frac{q}{2 A \epsilon_0} + \frac{q}{2 A \epsilon_0} = \frac{q}{A \epsilon_0}$.The electric energy density $u$ is defined as $u = \frac{1}{2} \epsilon_0 E^2$.Substituting the value of $E_{total}$ into the energy density formula:$u = \frac{1}{2} \epsilon_0 \left( \frac{q}{A \epsilon_0} \right)^2 = \frac{1}{2} \epsilon_0 \left( \frac{q^2}{A^2 \epsilon_0^2} \right) = \frac{q^2}{2 \epsilon_0 A^2}$.

What is the average electric energy density between the plates of a charged capacitor? (Here,$q$ = charge on the capacitor and $A$ = area of the capacitor plates.)

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