Physics · Electrostatics

JEE Main 2026 — 24 January, Morning Shift — Question 42

There are three co-centric conducting spherical shells A,B\mathrm{A}, \mathrm{B} and C of radii a,b\mathrm{a}, \mathrm{b} and c respectively. The potential of the spheres A,B\mathrm{A}, \mathrm{B} and C respectively, are :

  1. Option A:

    14πϵ0(q1+q2+q3a),14πϵ0(q1+q2+q3 b),14πϵ0(q1+q2+q3c)\frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{a}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{~b}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{c}}\right)

  2. Option B:

    14πϵ0(q1+q2+q3a),14πϵ0(q1+q2 b+q3c),14πϵ0(q1a+q2 b+q3c)\frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{a}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{a}}+\frac{\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right)

  3. Option C:

    14πϵ0(q1a+q2 b+q3c),14πϵ0(q1+q2 b+q3c),14πϵ0(q1+q2+q3c)\frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{a}}+\frac{\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{c}}\right)

    Correct
  4. Option D:

    14πϵ0(q1a+q2 b+q3c),14πϵ0(q1+q2+q3 b),14πϵ0(q1+q2+q3c)\frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{a}}+\frac{\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{~b}}\right), \frac{1}{4 \pi \epsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{c}}\right)

Answer: C

Step-by-step solution

VA=Kq1a+Kq2 b+Kq3c=14πε0(q1a+q2 b+q3c)VB=Kq1 b+Kq2 b+Kq3c=14πε0(q1+q2 b+q3c)VC=Kq1c+Kq2c+Kq3c=14πε0(q1+q2+q3c)\begin{aligned} & \mathrm{V}_{\mathrm{A}}=\frac{\mathrm{Kq}_{1}}{\mathrm{a}}+\frac{\mathrm{Kq}_{2}}{\mathrm{~b}}+\frac{\mathrm{Kq}_{3}}{\mathrm{c}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1}}{\mathrm{a}}+\frac{\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right) & \mathrm{V}_{\mathrm{B}}=\frac{\mathrm{Kq}_{1}}{\mathrm{~b}}+\frac{\mathrm{Kq}_{2}}{\mathrm{~b}}+\frac{\mathrm{Kq}_{3}}{\mathrm{c}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}}{\mathrm{~b}}+\frac{\mathrm{q}_{3}}{\mathrm{c}}\right) & \mathrm{V}_{\mathrm{C}}=\frac{\mathrm{Kq}_{1}}{\mathrm{c}}+\frac{\mathrm{Kq}_{2}}{\mathrm{c}}+\frac{\mathrm{Kq}_{3}}{\mathrm{c}}=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{q}_{1}+\mathrm{q}_{2}+\mathrm{q}_{3}}{\mathrm{c}}\right) \end{aligned}
Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Conductors and Redistribution of Charge