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

Question

Energy per unit volume for a capacitor having area A and separation d kept at potential difference V is given by (A)  (1/2) ε0 (V2/d2)  (B)  (1/2ε 0) (V2/d2)  (C)  (1/2) CV2  (D)  (Q2/2C)

Tardigrade Admin · Accepted Answer

Energy density =(1/2) ε0 E2 Since E=(V/d) Therefore energy density =(1/2) ε0((V2/d2))

Q. Energy per unit volume for a capacitor having area $A$ and separation d kept at potential difference $V$ is given by

$\frac{1}{2} ε_{0} \frac{V ^{2}}{d ^{2}}$

$\frac{1}{2 ε _{0}} \frac{V ^{2}}{d ^{2}}$

$\frac{1}{2} C V^{2}$

$\frac{Q ^{2}}{2 C}$

Energy density $= \frac{1}{2} ε_{0} E^{2}$
Since $E = \frac{V}{d}$
Therefore energy density $= \frac{1}{2} ε_{0} (\frac{V ^{2}}{d ^{2}})$

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

21​ε0​d2V2​

2ε0​1​d2V2​

21​CV2

2CQ2​