An infinite,uniformly charged sheet with surface charge density $\sigma$ cuts through a spherical Gaussian surface of radius $R$ at a distance $x$ from its center,as shown in the figure. The electric flux $\Phi$ through the Gaussian surface is

  • A
    $\frac{{\pi {R^2}\sigma }}{{{\varepsilon _0}}}$
  • B
    $\frac{{2\pi {{\left( {{R^2} - {x^2}} \right)}^{}}\sigma }}{{{\varepsilon _0}}}$
  • C
    $\frac{{\pi {{\left( {R - x} \right)}^2}\sigma }}{{{\varepsilon _0}}}$
  • D
    $\frac{{\pi {{\left( {{R^2} - {x^2}} \right)}^{}}\sigma }}{{{\varepsilon _0}}}$

Explore More

Similar Questions

The total electric flux through a closed spherical surface of radius $r$ enclosing an electric dipole of dipole moment $2aq$ is (Give $\varepsilon_0=$ permittivity of free space)

$A$ charge is uniformly distributed on the surface of a spherical rubber balloon. As it is blown up,the total electric flux coming out of the surface

$A$ thin spherical shell encloses a concentric solid sphere. The radius of the shell is $(0.060)^{1/2} \ m$ and its surface charge density is $-10^{-5} \ C/m^2$. The radius of the solid sphere is $(0.01)^{1/3} \ m$ and its volumetric charge density is $3 \times 10^{-5} \ C/m^3$. $\varepsilon_0$ is the permittivity of free space in $C^2/Nm^2$. The electric flux through a spherical surface concentric with the spherical shell and of radius greater than that of the shell in $V-m$ is:

Four dipoles having charge $\pm e$ are placed inside a sphere. The total flux of $\vec{E}$ coming out of the sphere is

The dimensional formula for electric flux 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