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

Asked in Re-NEET 2024 · Energy stored and energy density

(1) 15
(2) 7.5
(3) 0.3
(4) 150

Answer: (1) 15

Step-by-step solution

U=1/2CV²=1/2×12×10⁻¹²×(50)².

U=6×2500×10⁻¹²=1.5×10⁻⁸ J =15 nJ.

Why the other options are wrong

(2) Uses 1/4CV², half the stored energy.
(3) No single slip gives 0.3 nJ; it is fifty times too small, as if V were not squared: 1/2×12×10⁻¹²×50=0.3 nJ.
(4) A factor of 10 high; 1/2×12×2500=15000 pJ, which is 15 nJ, not 150.

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