Question

Capacity of a capacitor is 48 µF. When it is charged from 0.1 C to 0.5 C, change in energy stored is

• A. 2500 J
• B. 2.5 x 10-3 J
• C. 2.5 × 106 J
• D. 2.42 x 10-2 J

Answer 1

Answer: (A)

2500 J

Answer 2

The energy stored in a capacitor is given by:

$U = \frac{q^2}{2C}$

The change in energy when the charge changes from $q_1$ to $q_2$ is:

$\Delta U = \frac{q_2^2 - q_1^2}{2C}$

Substitute the given values: $C = 48\,\mu\text{F} = 48\times10^{-6}\,\text{F}$, $q_1 = 0.1\,\text{C}$, $q_2 = 0.5\,\text{C}$.

$\Delta U = \frac{(0.5)^2 - (0.1)^2}{2 \times 48\times10^{-6}} = \frac{0.25 - 0.01}{96\times10^{-6}} = \frac{0.24}{96\times10^{-6}}$

$\Delta U = \frac{0.24}{96\times10^{-6}} = 0.0025 \times 10^{6} = 2500\,\text{J}$