In a uniform electric field,a cube of side $1 \ cm$ is placed. The total energy stored in the cube is $8.85 \ \mu J$. The electric field is parallel to four of the faces of the cube. The electric flux through any one of the remaining two faces is

  • A
    $\frac{1}{5\sqrt{2}} \ V-m$
  • B
    $100\sqrt{2} \ V-m$
  • C
    $5\sqrt{2} \ V-m$
  • D
    $10\sqrt{2} \ V-m$

Explore More

Similar Questions

Why do electric field lines not form closed loops?

$A$ point charge causes an electric flux of $-2 \times 10^4 \ Nm^2 C^{-1}$ to pass through a spherical Gaussian surface of $8.0 \ cm$ radius,centred on the charge. The value of the point charge is: (Given $\epsilon_0 = 8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}$)

$A$ 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:

The figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. The total flux of the electric field through a closed surface $S$,due to all charges $q_1, q_2, q_3$ and $q_4$ is:

What is called a Gaussian surface?

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