Solveeit Logo
Login

Question

Physics Question on electrostatic potential and capacitance

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

A

12ε0V2d2 \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}

B

12ε0V2d2 \frac{1}{2 \varepsilon_0} \frac{V^2}{d^2}

C

12CV2 \frac{1}{2} CV^2

D

Q22C \frac{Q^2}{ 2C}

Answer

12ε0V2d2 \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}

Explanation

Solution

Energy density = 12ε0V2d2 \frac{1}{2} \varepsilon_0 \frac{V^2}{d^2}