Question

The capacitance of a variable capacitor joined with a battery of 100 V is changed from 2 μF to 10 μF. What is the change in the energy stored in it ?(A) 2 X 10⁻² J
(B) 2.5 X 10⁻² J
(C) 6.5 X 10⁻² J
(D) 4 X 10⁻² J

Answer 1

Hey Students,

◆ Answer - (D)

∆E = 4×10^-2 J

◆ Explaination -

# Given -

V = 100 V

C1 = 2 μF = 2×10^-6 F

C2 = 10 μF = 10×10^-6 F

# Solution -

Change in energy stored in capacitor is given by -

∆E = 1/2 (C2-C1)V^2

∆E = 1/2 × (10×10^-6 - 2×10^-6) × 100^2

∆E = 4×10^-2 J

Therefore, the change in the energy stored in capacitor is 4×10^-2 J.

Thanks dear...