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JEE Main 2026 (4 Apr, Shift 1) · Energy Stored & Force between Plates

Energy Stored & Force between Plates question from JEE Main 2026 (4 Apr, Shift 1)

Q6JEE Main 2026 · 4 APR · SHIFT 1Difficult

Energy Stored & Force between Plates

A parallel plate air capacitor is connected to a battery. The plates are pulled apart at uniform speed vv. If xx is the separation between the plates at any instant, then the time rate of change of electrostatic energy of the capacitor is proportional to xαx^{\alpha}, where α\alpha is ______.

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Answer: (a)

  1. Battery connected: VV constant, U=12CV2=ε0AV22xU = \frac{1}{2}CV^2 = \frac{\varepsilon_0 AV^2}{2x}.
  2. dUdt=−ε0AV22x2dxdt=−ε0AV2v2x2\frac{dU}{dt} = -\frac{\varepsilon_0 AV^2}{2x^2}\frac{dx}{dt} = -\frac{\varepsilon_0 AV^2 v}{2x^2}
  3. With vv constant, dUdt∝x−2\frac{dU}{dt} \propto x^{-2}.
  4. So the answer is (a) α=−2\alpha = -2.
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