Ncert Exercises & Solutions

Q 2.10) A 12pF capacitor is connected to a 50V battery. How much electrostatic energy is stored in the capacitor?

Given,

Capacitance of the capacitor, C = 12pF = 12 x 10^-12F

Potential difference, V = 50 V

Electrostatic energy stored in the capacitor is given by the relation,

$E = \frac{1}{2} C V^{2} = \frac{1}{2} \times 12 \times 10^{- 12} \times (50)^{2} J = 1.5 \times 10^{- 8} J$

Therefore, the electrostatic energy stored in the capacitor is 1.5 x 10^-8J. was disconnected.