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Physics · Electrostatics

JEE Main 2026 — 21 January, Morning Shift — Question 42

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{A}(4,4,2) \mathrm{m} to B(2,2,1)m\mathrm{B}(2,2,1) \mathrm{m} is ____\_\_\_\_ J. ( 14πϵ0=9×109\frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} in SI units)

  1. Option A:

    45×10−645 \times 10^{-6}

  2. Option B:

    0

  3. Option C:

    30×10−630 \times 10^{-6}

    Correct
  4. Option D:

    15×10−615 \times 10^{-6}

Answer: C

Step-by-step solution

Work done by external agent : Wext =ΔU;\mathrm{W}_{\text {ext }}=\Delta \mathrm{U} ; ΔU→\Delta \mathrm{U} \rightarrow Change in potential energy in taking the charge from initial to final configuration ⇒Wext =14πϵ0q1q2rf−14πϵ0q1q2ri\Rightarrow W_{\text {ext }}=\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{f}}-\frac{1}{4 \pi \epsilon_{0}} \frac{q_{1} q_{2}}{r_{i}}

Now, rf=(2−0)2+(2−0)2+(1−0)2=3 mr_{f}=\sqrt{(2-0)^{2}+(2-0)^{2}+(1-0)^{2}}=3 \mathrm{~m}

ri=(4−0)2+(4−0)2+(2−0)2=6 m∴ Wext=(9×109)×(10−8×2×10−6)[13−16]=3×10−5=30×10−6 J\begin{aligned} \mathrm{r}_{\mathrm{i}} & =\sqrt{(4-0)^{2}+(4-0)^{2}+(2-0)^{2}}=6 \mathrm{~m} \therefore \quad \mathrm{~W}_{\mathrm{ext}} & =\left(9 \times 10^{9}\right) \times\left(10^{-8} \times 2 \times 10^{-6}\right)\left[\frac{1}{3}-\frac{1}{6}\right] & =3 \times 10^{-5} & =30 \times 10^{-6} \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Electrostatics
Topic
Electrostatic Potential Energy and Electric Potential