The capacitance of a variable capacitor connected to a 100 V battery is changed from 2 μF to 5 μF. What is the change in the energy stored in it?
Answer & explanation
Correct answer: option 3
U = CV2 / 2
Here V is constant hence change in energy will be $ U_2 - U_1 = \frac{1}{2} (C_2 - C_1)V^2 = \frac{1}{2} \times 3\times 10^{-6} \times 100^2 = 1.5\times 10^{-2}J$