1.

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 JThe 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»

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.



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