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Physics · Electrostatics

JEE Main 2026 — 8 April, Evening Shift — Question 13

Two point charges q1=3μCq_1 = 3\mu\mathrm{C} and q2=−4μCq_2 = -4\mu\mathrm{C} are placed at points (2i^+3j^+3k^)(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+3\hat{\mathbf{k}}) and (i^+j^+k^)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) respectively. Force on charge q2q_2 is N. (14πϵ0=9×109SI Units)\left(\frac{1}{4\pi\epsilon_0}=9\times10^9\mathrm{SI\,Units}\right)

  1. Option A:

    (12i^+24j^+24k^)×10−3(12\hat{\mathbf{i}}+24\hat{\mathbf{j}}+24\hat{\mathbf{k}})\times10^{-3}

  2. Option B:

    (4i^+8j^+8k^)×10−3(4\hat{\mathbf{i}}+8\hat{\mathbf{j}}+8\hat{\mathbf{k}})\times10^{-3}

    Correct
  3. Option C:

    (3i^+6j^+6k^)×10−3(3\hat{\mathbf{i}}+6\hat{\mathbf{j}}+6\hat{\mathbf{k}})\times10^{-3}

  4. Option D:

    (−4i^−8j^−8k^)×10−3(-4\hat{\mathbf{i}}-8\hat{\mathbf{j}}-8\hat{\mathbf{k}})\times10^{-3}

Answer: B

Step-by-step solution

r⃗=r⃗2−r⃗1=(1−2)i^+(1−3)j^+(1−3)k^=−i^−2j^−2k^\vec{r} = \vec{r}_2 - \vec{r}_1 = (1-2)\hat{i} + (1-3)\hat{j} + (1-3)\hat{k} = -\hat{i} -2\hat{j} -2\hat{k}. r=1+4+4=3r = \sqrt{1+4+4}=3. Force F⃗=kq1q2r3r⃗=9×109×3×10−6×(−4)×10−627(−i^−2j^−2k^)=−108×10−327(−i^−2j^−2k^)=−4×10−3(−i^−2j^−2k^)=4×10−3(i^+2j^+2k^)\vec{F} = \frac{k q_1 q_2}{r^3}\vec{r} = \frac{9\times10^9 \times 3\times10^{-6} \times (-4)\times10^{-6}}{27} (-\hat{i}-2\hat{j}-2\hat{k}) = \frac{-108\times10^{-3}}{27} (-\hat{i}-2\hat{j}-2\hat{k}) = -4\times10^{-3} (-\hat{i}-2\hat{j}-2\hat{k}) = 4\times10^{-3}(\hat{i}+2\hat{j}+2\hat{k}).

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Electric Charge and Coulomb's Law