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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2026 · 21 January · Morning Shift · Physics · Electrostatics

A point charge of 10^ -8 C is placed at origin. The work done in...

A point charge of 10−8C10^{-8} \mathrm{C} is placed at origin. The work done in moving a point charge 2μC2 \mu \mathrm{C} from point A(4,4,2) m\mathrm{m} to B(2,2,1) m\mathrm{m} is ____\_\_\_\_ J. \left(\frac{1}{4 \pi \epsilon_{o}}=9 \times 10^{9}\right.$ in SI units )

Options

  1. A45×10−645 \times 10^{-6}
  2. B15×10−615 \times 10^{-6}
  3. C0
  4. D30×10−630 \times 10^{-6}

Correct answer

D. 30×10−630 \times 10^{-6}

Step-by-step solution

Distance from origin to A: rA=42+42+22=6r_A = \sqrt{4^2 + 4^2 + 2^2} = 6 m
Distance from origin to B: rB=22+22+12=3r_B = \sqrt{2^2 + 2^2 + 1^2} = 3 m
Work done: W=q2(VB−VA)=q2×q14πε0(1rB−1rA)W = q_2(V_B - V_A) = q_2 \times \frac{q_1}{4\pi\varepsilon_0}\left(\frac{1}{r_B} - \frac{1}{r_A}\right)
=2×10−6×9×109×10−8×(13−16)= 2 \times 10^{-6} \times 9 \times 10^9 \times 10^{-8} \times \left(\frac{1}{3} - \frac{1}{6}\right)
=18×10−5×16=30×10−6= 18 \times 10^{-5} \times \frac{1}{6} = 30 \times 10^{-6} J
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