Question 278 of 311

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

A$\frac{q}{2 \epsilon_0}$B$\frac{q}{8 \epsilon_0}$CZeroD$\frac{q}{4 \epsilon_0}$

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Answer: (A) $\frac{q}{2 \epsilon_0}$

From
$\begin{aligned} & 2 \phi=\frac{\mathrm{q}}{\epsilon_0} \\ & \phi=\frac{\mathrm{q}}{2 \epsilon_0}\end{aligned}$

JEE Main Physics practice question on Electrostatics (Electric Flux and Gauss Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.