A 12 pF capacitor is connected to a 50 V battery, the electrostatic energy stored in the capacitor in nJ is
- 15
- 7.5
- 0.3
- 150
The Correct Option is A
Solution and Explanation
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 = $\frac{1}{2}(12 \times 10^{-12} \text{ F})(50 \text{ V})^2 = 15 \times 10^{-9} \text{ J} = 15 \text{ nJ}$
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