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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 4 April · Evening Shift · Physics · Electrostatics

A charge q is placed at the center of one of the surface of a cube....

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

Options

  1. Aq2ϵ0\frac{q}{2 \epsilon_0}
  2. Bq8ϵ0\frac{q}{8 \epsilon_0}
  3. CZero
  4. Dq4ϵ0\frac{q}{4 \epsilon_0}

Correct answer

A. q2ϵ0\frac{q}{2 \epsilon_0}

Step-by-step solution

From A charge q is placed at the center of one of the surface of a cube. The flux linked with the cube is:
2ϕ=qϵ0ϕ=q2ϵ0\begin{aligned} & 2 \phi=\frac{\mathrm{q}}{\epsilon_0} \\ & \phi=\frac{\mathrm{q}}{2 \epsilon_0}\end{aligned}
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