Published by:
CGP EDU Academic Team
Published on: September 12, 2026
A charge q is placed at the center of a cube. Then the flux passing through one face of cube will be
$(a) \frac{q}{e_0} (b) \frac{q}{2 e_0} (c) \frac{q}{4 e_0} (d) \frac{q}{6 e_0}$
Text Solution
Verified by ExpertsThe correct answer is:
D
Net flux through the cube $\phi_{net} = \frac{Q}{\varepsilon_0} ;$ so flux through one face $\phi_{face} = \frac{q}{6 \epsilon_0}$
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A cylinder of radius R and length L is placed in a uniform electric field E parallel to the cylinde…
Electric field at a point varies as $r_{0}$ for
An electric charge q is placed at the center of a cube of side α. The electric flux on one of its f…
Total electric flux coming out of a unit positive charge put in air is
For a given surface the Gauss's law is stated as $\oint \mathbf{E} \cdot d\mathbf{s} = 0$ . From th…
A cube of side l is placed in a uniform field E, where $\mathbf{E} = \mathbf{E} \hat{i}$ . The net …