If the radius of the spherical Gaussian surfa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enc}}{\epsilon_0}$,where $q_{enc}$ i

Vedclass · Accepted Answer

remains unchanged.

Vedclass · Answer

(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enc}}{\epsilon_0}$,where $q_{enc}$ is the total charge enclosed by the surface and $\epsilon_0$ is the permittivity of free space.Since the electric flux depends only on the charge enclosed by the surface and not on the size or shape of the Gaussian surface,increasing the radius of the spherical Gaussian surface does not change the amount of charge enclosed.Therefore,the electric flux remains unchanged.

If the radius of the spherical Gaussian surface is increased,then the electric flux due to a point charge enclosed by the surface:

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 $:-$

The adjoining diagram shows the electric lines of force emerging from a charged body. If the electric fields at $A$ and $B$ are $E_A$ and $E_B$ respectively and the distance between them is $r$,then

If electric flux is coming out of a Gaussian surface,what is associated with the surface?

$A$ point charge $+q$ is placed at the centre of a cube of side $L$. The electric flux emerging from the cube is

Which of the following is a correct statement?

Vedclass Test Series

Exam Paper Generator

Online Exam Module