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JEE Main 2009 (AIEEE) · Electrostatic Potential Energy

Electrostatic Potential Energy question from JEE Main 2009 (AIEEE)

Q57JEE Main 2009 · AIEEEEasy

Electrostatic Potential Energy

Two points PP and QQ are maintained at the potentials of 10 V10\text{ V} and −4 V-4\text{ V}, respectively. The work done in moving 100100 electrons from PP to QQ is:

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Answer: (c)

  1. W=q(VQ−VP)W = q(V_Q - V_P) with q=−100e=−1.6×10−17q = -100e = -1.6\times10^{-17} C.
  2. VQ−VP=−4−10=−14V_Q - V_P = -4 - 10 = -14 V
  3. W=(−1.6×10−17)(−14)=2.24×10−16W = (-1.6\times10^{-17})(-14) = 2.24\times10^{-16} J
  4. So the answer is (c).
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