For a capacitor with area A and potential dif | vedclass

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(A) The energy density (energy per unit volume) $u$ of an electric field $E$ is given by the formula: $u = \frac{1}{2}{\varepsilon _0}{E^2}$.For a par

Vedclass · Accepted Answer

$\frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}}$

Vedclass · Answer

(A) The energy density (energy per unit volume) $u$ of an electric field $E$ is given by the formula: $u = \frac{1}{2}{\varepsilon _0}{E^2}$.For a parallel plate capacitor,the electric field $E$ between the plates is related to the potential difference $V$ and the distance $d$ by the relation: $E = \frac{V}{d}$.Substituting this value of $E$ into the energy density formula,we get: $u = \frac{1}{2}{\varepsilon _0}{\left( \frac{V}{d} \right)^2} = \frac{1}{2}{\varepsilon _0}\frac{{{V^2}}}{{{d^2}}}$.

For a capacitor with area $A$ and potential difference $V$ separated by a distance $d$,the energy per unit volume is:

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