Electrostatics

JEE Main level, single correct, medium

184 of 311

Question 184 of 311

A point charge 2×10–2C2\times {10}^{–2}C is moved from PP to SS in a uniform electric field of 30NC–130N{C}^{–1} directed along positive xx-axis. If coordinates of PP and SS are (1,2,0)mand(0,0,0)m(1,2,0)m\text{and}(0,0,0)m respectively, the work done by electric field will be

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Answer: (C) −600mJ-600mJ


As electric field is uniform, so force acting on the charged particle will be uniform.

Therefore,

WE=F⃗⋅s⃗=qE⃗⋅s⃗{W}_{E}=\vec{F}\cdot \vec{s}=q\vec{E}\cdot \vec{s}
=(2×10−2)[30i^⋅[(0i^+0j^+0k^)−(i^+2j^+0k^)]]=\left(2\times {10}^{-2}\right)\left[30\hat{i}\cdot \left[\left(0\hat{i}+0\hat{j}+0\hat{k}\right)-\left(\hat{i}+2\hat{j}+0\hat{k}\right)\right]\right]
=−60×10−2J=-60\times {10}^{-2}J
=−600mJ=-600mJ

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