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MHT CET 2026 (12 May, Shift 2) · Energy Stored & Force between Plates

Energy Stored & Force between Plates question from MHT CET 2026 (12 May, Shift 2)

Q5MHT CET 2026 · 12 MAY · SHIFT 2Medium

Energy Stored & Force between Plates

A capacitor of capacity 5 μF5\ \mu\text{F} is charged by a battery of e.m.f. 1010 V. At an instant, the potential difference across the capacitor is 44 V and the time rate of change of potential difference across the capacitor is 0.60.6 V/s. Then the time rate at which energy is stored in the capacitor at that instant is

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Answer: (a)

  1. U=12CV2⇒dUdt=CVdVdtU = \frac{1}{2}CV^2 \Rightarrow \frac{dU}{dt} = CV\frac{dV}{dt}
  2. dUdt=5×10−6×4×0.6\frac{dU}{dt} = 5 \times 10^{-6} \times 4 \times 0.6
  3. dUdt=12×10−6 J/s\frac{dU}{dt} = 12 \times 10^{-6}\text{ J/s}
  4. So the answer is (a) 12×10−612 \times 10^{-6} J/s.
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