$A$ charge '$Q$ $\mu C$' is placed at the centre of a cube. The flux through one face and two opposite faces of the cube is respectively:

  • A
    $\frac{Q}{6 \epsilon_{0}} \mu Vm, \quad \frac{Q}{3 \epsilon_{0}} \mu Vm$
  • B
    $\frac{Q}{12 \epsilon_{0}} \mu Vm, \quad \frac{Q}{\epsilon_{0}} \mu Vm$
  • C
    $\frac{Q}{6 \epsilon_{0}} \mu Vm, \quad \frac{Q}{2 \epsilon_{0}} \mu Vm$
  • D
    $\frac{Q}{12 \epsilon_{0}} \mu Vm, \quad \frac{Q}{3 \epsilon_{0}} \mu Vm$

Explore More

Similar Questions

If electric flux through a cubic Gaussian surface is $1.9 \times 10^5 \text{ Nm}^2 \text{C}^{-1}$,then the electric charge at its center is . . . . . . . (Edge length of cube = $9.0 \text{ cm}$).

Four charges $2 \mu C, -3 \mu C, 4 \mu C, -4 \mu C$ and $-1 \mu C$ are enclosed by a Gaussian surface of radius $2 \ m$. The net outward flux through the Gaussian surface is (in $\mu V-m$):

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

$A$ positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is the flux through a closed Gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is the flux through a closed Gaussian surface $S_2$ whose radius is just greater than $R_1$,then:

Gauss's law should be invalid if

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo