Electrostatics

JEE Main level, single correct, medium

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Question 225 of 311

Two points PP and QQ are maintained at the potentials of 10 V10 \mathrm{~V} and −4 V-4 \mathrm{~V} respectively. The work done in moving 100 electrons from PP to QQ is

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Answer: (D) 2.24×10−16 J2.24 \times 10^{-16} \mathrm{~J}

W=QdV=Q(Vq−VP)=−100×(1.6×10−19)×(−4−10)=+100×1.6×10−19×14=+2.2410−16 J\begin{aligned} & W=Q d V=Q\left(V_q-V_P\right)=-100 \times\left(1.6 \times 10^{-19}\right) \times(-4-10) \\ & =+100 \times 1.6 \times 10^{-19} \times 14=+2.2410^{-16} \mathrm{~J} \end{aligned}

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