Question

Initially the potential difference of a 8 μF capacitor is 30 V. Then it is changed to 40 V. What is the increase in energy?

(a) 28 × 10^-4 J

(b) 18 × 10^-4 J

(c) 8 × 10^-4 J

(d) 24 × 10^-4 J

The question was posed to me during an internship interview.

This intriguing question comes from Energy Stored in a Capacitor in division Electrostatic Potential and Capacitance of Physics – Class 12

Answer 1

The correct option is (a) 28 × 10^-4 J

The explanation: The required equation for the increase in energy is given by:

U2 – U1 = \(\frac {1}{2}\) × C (V2^2 – V1^2)

              = \(\frac {1}{2}\) × 8 × 10^-6 × (40^2 – 30^2)

              = \(\frac {1}{2}\) × 8 × 10^-6 × 700

              = 4 ×  700 × 10^-6

              = 2800 ×  10^-6

              = 28 × 10^-4 J

Therefore, the increase in energy is 28 × 10^-4 J.