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Physics · Electrostatics

JEE Main 2025 — 24 January, Evening Shift — Question 62

A small uncharged conducting sphere is placed in contact with an identical sphere but having 4×10−84 \times 10^{-8} C charge and then removed to a distance such that the force of repulsion between them is 9×10−3 N9 \times 10^{-3} \mathrm{~N}. The distance between them is (Take 14πε0\frac{1}{4 \pi \varepsilon_{0}} as 9×1099 \times 10^{9} in SI units)

  1. Option A:

    2cm

    Correct
  2. Option B:

    3cm

  3. Option C:

    4cm

  4. Option D:

    1cm

Answer: A

Step-by-step solution

F=k(θ2)(θ2)r2\mathrm{F}=\frac{\mathrm{k}\left(\frac{\theta}{2}\right)\left(\frac{\theta}{2}\right)}{\mathrm{r}^{2}}

9×10−3=9×109×(4×10−8)×4×10−84×r29 \times 10^{-3}=\frac{9 \times 10^{9} \times\left(4 \times 10^{-8}\right) \times 4 \times 10^{-8}}{4 \times \mathrm{r}^{2}}

r2=9×109×16×10−164×9×10−3=4×10−4\mathrm{r}^{2}=\frac{9 \times 10^{9} \times 16 \times 10^{-16}}{4 \times 9 \times 10^{-3}}=4 \times 10^{-4}

r=2×10−2 m⇒2 cm\mathrm{r}=2 \times 10^{-2} \mathrm{~m} \Rightarrow 2 \mathrm{~cm}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
24 January, Evening Shift
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Electric Charge and Coulomb's Law

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