Energy per unit volume a capacitor having area A and separation d kept potential difference V is given by -dfrac-1-2varepsilon -0-dfrac-V-2-d-2-dfrac-1-2-CV-2dfrac-1-2-varepsilon -0dfrac-V-2-d-2-dfrac-Q-2-2C-

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Toppr Admin · Accepted Answer

Energy stored in capacitor of capacitance $C = \frac{1}{2} C V^{2}$

Capacitance of a parallel plate capacitor = $C = \frac{A ϵ_{0}}{d}$

Volume of parallel plate capacitor= $A d$

Thus energy stored per unit volume = $\frac{\frac{1}{2} C V^{2}}{A d} = \frac{1}{2} ϵ_{0} \frac{V^{2}}{d^{2}}$

Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by :12ε0V2d212CV212ε0V2d2Q22C

Energy stored in capacitor of capacitance C=12CV2 Capacitance of a parallel plate capacitor = C=Aϵ0dVolume of parallel plate capacitor=AdThus energy stored per unit volume = 12CV2Ad=12ϵ0V2d2

Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by :
$\frac{1}{2 ε_{0}} \frac{V^{2}}{d^{2}}$
$\frac{1}{2} C V^{2}$
$\frac{1}{2} ε_{0} \frac{V^{2}}{d^{2}}$
$\frac{Q^{2}}{2 C}$

Energy stored in capacitor of capacitance $C = \frac{1}{2} C V^{2}$

Capacitance of a parallel plate capacitor = $C = \frac{A ϵ_{0}}{d}$

Volume of parallel plate capacitor= $A d$

Thus energy stored per unit volume = $\frac{\frac{1}{2} C V^{2}}{A d} = \frac{1}{2} ϵ_{0} \frac{V^{2}}{d^{2}}$