Question

A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is

• A. 15
• B. 7.5
• C. 0.3
• D. 150

Answer 1

The Correct Option is A

Energy stored in a capacitor (E) = $\frac{1}{2}CV^2$, where C is capacitance and V is voltage.

C = 12 pF = 12 × 10^-12 F V = 50 V

E = $\frac{1}{2}(12 \times 10^{-12} \text{ F})(50 \text{ V})^2 = 15 \times 10^{-9} \text{ J} = 15 \text{ nJ}$