Physics · Electrostatics

JEE Main 2026 — 2 April, Morning Shift — Question 7

Two charged conducting spheres $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of radii $8\mathrm{cm}$ and $18\mathrm{cm}$ are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ spheres are $\mathrm{E}_{\mathrm{S1}}$ and $\mathrm{E}_{\mathrm{S2}}$ respectively. The value of $\frac{\mathrm{E}_{\mathrm{S1}}}{\mathrm{E}_{\mathrm{S2}}}$ is

Option A:
$\frac{3}{2}$
Option B:
$\frac{2}{3}$
Option C:
$\frac{4}{9}$
Option D:
$\frac{9}{4}$
Correct

Answer: D

Step-by-step solution

When connected, potentials equal: $\frac{kQ_1}{r_1}=\frac{kQ_2}{r_2}$ and $E = \frac{kQ}{r^2}$ → $\frac{E_1}{E_2}=\frac{r_2}{r_1}=\frac{18}{8}=\frac{9}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Physics
Chapter: Electrostatics
Topic: Properties of Conductors and Redistribution of Charge

Two short electric dipoles A and B having dipole moment p 1 and p 2 respectively are placed with their axis mutually perpendicular as shown…

Answer: D

Step-by-step solution

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