The inward and outward electric flux for a closed surface in units of $N \cdot m^2/C$ are respectively $8 \times 10^3$ and $4 \times 10^3$. Then the total charge inside the surface is [where $\varepsilon_0$ = permittivity constant].

  • A
    $4 \times 10^3 \text{ C}$
  • B
    $-4 \times 10^3 \text{ C}$
  • C
    $\frac{-4 \times 10^3}{\varepsilon_0} \text{ C}$
  • D
    $-4 \times 10^3 \varepsilon_0 \text{ C}$

Explore More

Similar Questions

$A$ square Gaussian surface is placed in the $y-z$ plane. Its axis is along the $x-$axis and its center is at the origin. Two identical charges,each $Q$,are placed at points $(a, 0, 0)$ and $(-a, 0, 0)$. If each side length of the square is $2a$,then the electric flux passing through the square is:

In the figure,a point charge $+Q_1$ is at the centre of an imaginary spherical surface and another point charge $+Q_2$ is outside it. Point $P$ is on the surface of the sphere. Let $\Phi _s$ be the net electric flux through the sphere and $\vec E_p$ be the electric field at point $P$ on the sphere. Which of the following statements is $TRUE$?

An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

$A$ charge $Q$ is enclosed by a Gaussian surface of radius $R$. If the radius is doubled,then the outward electric flux will

If charge $q$ is placed on one of the vertex of a cube,then total electric flux passing through the cube is . . . . . . .

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