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JEE Main 2019 (9 Apr, Shift 1) · Energy Stored & Force between Plates

Energy Stored & Force between Plates question from JEE Main 2019 (9 Apr, Shift 1)

Q28JEE Main 2019 · 9 APR · SHIFT 1Medium

Energy Stored & Force between Plates

A capacitor with capacitance 5μF is charged to 5μC. If the plates are pulled apart to reduce the capacitance to 2μF, how much work is done?

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Answer: (c)

  1. Isolated capacitor: QQ constant, W=Q22(1C2−1C1)W = \frac{Q^2}{2}\left(\frac{1}{C_2} - \frac{1}{C_1}\right).
  2. W=(5×10−6)22(12×10−6−15×10−6)W = \frac{(5 \times 10^{-6})^2}{2}\left(\frac{1}{2 \times 10^{-6}} - \frac{1}{5 \times 10^{-6}}\right)
  3. W=12.5×10−12×0.3×106=3.75×10−6 JW = 12.5 \times 10^{-12} \times 0.3 \times 10^{6} = 3.75 \times 10^{-6}\text{ J}
  4. So the answer is (c) 3.75×10−63.75 \times 10^{-6} J.
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