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JEE Main 2025 (28 Jan, Shift 2) · Energy Stored & Force between Plates

Energy Stored & Force between Plates question from JEE Main 2025 (28 Jan, Shift 2)

Q9JEE Main 2025 · 28 JAN · SHIFT 2Medium

Energy Stored & Force between Plates

A parallel plate capacitor of capacitance 1 µF is charged to a potential difference of 20 V. The distance between plates is 1 µm. The energy density between plates of capacitor is:

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Answer: (a)

  1. E=Vd=2010−6=2×107 V/mE = \frac{V}{d} = \frac{20}{10^{-6}} = 2 \times 10^{7}\text{ V/m}
  2. u=12ε0E2=12×8.85×10−12×4×1014u = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^{14}
  3. u=1770≈1.8×103 J/m3u = 1770 \approx 1.8 \times 10^{3}\text{ J/m}^3
  4. So the answer is (a) 1.8×1031.8 \times 10^{3} J/m3^3.
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