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: (a). U = 1/2CV² U = 1/2 × 12 × 10⁻¹² × 50² = 6 × 10⁻¹² × 2500 U = 1.5 × 10⁻⁸ J = 15 nJ So the answer is (a) 15.

A 12 pF capacitor is connected to a 50 V battery, the electrostatic… | NEET 2024

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Energy Stored & Force between Plates

A $12\text{ pF}$ capacitor is connected to a $50\text{ V}$ battery, the electrostatic energy stored in the capacitor in $\text{nJ}$ is

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Answer: (a)

$U = \frac{1}{2}CV^2$
$U = \frac{1}{2} \times 12 \times 10^{-12} \times 50^2 = 6 \times 10^{-12} \times 2500$
$U = 1.5 \times 10^{-8}\text{ J} = 15\text{ nJ}$
So the answer is (a) 15.

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