$A$ charge $Q$ is placed at a distance $a/2$ above the centre of the square surface of edge $a$ as shown in the figure. The electric flux through the square surface is

  • A
    $\frac{Q}{3\varepsilon_0}$
  • B
    $\frac{Q}{6\varepsilon_0}$
  • C
    $\frac{Q}{2\varepsilon_0}$
  • D
    $\frac{Q}{\varepsilon_0}$

Explore More

Similar Questions

$A$ hollow cylinder has a charge $q$ at its center. If $\phi$ is the electric flux associated with the curved surface $B,$ the flux linked with the plane surface $A$ will be $:-$

$A$ cubical Gaussian surface has a side of length $a = 10 \,cm$. Electric field lines are parallel to the $X$-axis as shown in the figure. The magnitudes of the electric fields through surfaces $ABCD$ and $EFGH$ are $6 \,kNC^{-1}$ and $9 \,kNC^{-1}$ respectively. Then,the total charge enclosed by the cube is (Take $\varepsilon_0 = 9 \times 10^{-12} \,Fm^{-1}$): (in $\,nC$)

The figure shows the electric field lines. The spacing between the lines is parallel to the paper at every point. If the magnitude of the field at $A$ is $40 \ N/C$,then the approximate magnitude of the field at $B$ is ....... $N/C$.

Consider the charges and the Gaussian surface shown in the figure. When calculating the electric flux through the spherical surface,the electric field is due to which of the following?

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