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Physics · Electrostatics

JEE Main 2024 — 4 April, Shift 2 — Question 47

A charge qq is placed at the center of one of the surface of a cube. The flux linked with the cube is :-

  1. Option A:

    q4ϵ0\frac{q}{4 \epsilon_{0}}

  2. Option B:

    q2ϵ0\frac{q}{2 \epsilon_{0}}

    Correct
  3. Option C:

    q8ϵ0\frac{q}{8 \epsilon_{0}}

  4. Option D:

    Zero

Answer: B

Step-by-step solution

2ϕ=qϵ02 \phi=\frac{q}{\epsilon_{0}}

ϕ=q2ϵ0\phi=\frac{\mathrm{q}}{2 \epsilon_{0}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Electrostatics
Topic
Electric flux and Gauss's Law
A charge q is placed at the center of one of the surface of a cube.… | JEE Main 2024 PYQ with Solution · DhiX AI