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Physics Question on Capacitors and Capacitance

The work done in placing a charge of 8×10−18 C8 \times 10^{-18} \, C on a capacitor of capacity 100 μF100\, \mu F is

A

32×10−32J32 \times 10^{-32} J

B

16×10−32J16 \times 10^{-32} J

C

3.1×10−26J3.1 \times 10^{-26} J

D

4×10−10J4\times 10^{-10} J

Answer

32×10−32J32 \times 10^{-32} J

Explanation

Solution

The work done is stored as the potential energy. The potential energy stored in a capacitor is given by W=12Q2c=12(8×10−18)2100×10−6W=\frac{1}{2} \frac{Q^{2}}{c}=\frac{1}{2} \frac{\left(8 \times 10^{-18}\right)^{2}}{100 \times 10^{-6}} =32×10−32J=32 \times 10^{-32} J