Start free

Physics · Electrostatics

JEE Main 2024 — 6 April, Shift 2 — Question 38

Two identical conducting spheres PP and SS with charge Q on each, repel each other with a force 16N. A third identical uncharged conducting sphere R is successively brought in contact with the two spheres. The new force of repulsion between P and S is :

  1. Option A:

    4 N

  2. Option B:

    6 N

    Correct
  3. Option C:

    1 N

  4. Option D:

    12 N

Answer: B

Step-by-step solution

New force between P&SP \& S is :

FPS∝Q2×3Q4\mathrm{F}_{\mathrm{PS}} \propto \frac{\mathrm{Q}}{2} \times \frac{3 \mathrm{Q}}{4}

FPS∝3Q28=38×16=6\mathrm{F}_{\mathrm{PS}} \propto \frac{3 \mathrm{Q}^{2}}{8}=\frac{3}{8} \times 16=6

Solution figure

Answer key and solution verified before publishing.

Practise Electrostatics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free
Exam
JEE Main 2024
Paper
6 April, Shift 2
Subject
Physics
Chapter
Electrostatics
Topic
Properties of Conductors and Redistribution of Charge
Two identical conducting spheres P and S with charge Q on each, repel… | JEE Main 2024 PYQ with Solution · DhiX AI