Two equations are given below:A. oint E cdot | vedclass

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

Question

(B) Equation $A$ is $\oint E \cdot dA = \frac{Q}{\varepsilon_0}$,which is the Gauss law for electrostatics. It states that the total electric flux thr

Vedclass · Accepted Answer

$A$. Gauss law for electric fields,$B$. Gauss law for magnetic fields

Vedclass · Answer

(B) Equation $A$ is $\oint E \cdot dA = \frac{Q}{\varepsilon_0}$,which is the Gauss law for electrostatics. It states that the total electric flux through a closed surface is equal to the enclosed charge divided by the permittivity of free space.Equation $B$ is $\oint B \cdot dA = 0$,which is the Gauss law for magnetism. It implies that magnetic monopoles do not exist,meaning the net magnetic flux through any closed surface is always zero.

Two equations are given below:
$A. \oint E \cdot dA = \frac{Q}{\varepsilon_0}$
$B. \oint B \cdot dA = 0$
They are:

Similar Questions

State Gauss's law in electrostatics and provide its mathematical expression.

The linear charge density of a wire is $8.85\,\mu C/m$. The radius and height of the cylinder are $3\,m$ and $4\,m$ respectively. Find the electric flux passing through the cylinder.

Electric flux through a surface of area $100 \ m^2$ lying in the $xy$ plane is (in $V-m$) if $\vec E = \hat i + \sqrt 2 \hat j + \sqrt 3 \hat k$.

The electric flux coming out from a unit positive charge kept in air is:

The black shapes in the figure below are closed surfaces. The electric field lines are shown by dashed arrows. For which case,the net flux through the surfaces is non-zero?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Two equations are given below:$A. \oint E \cdot dA = \frac{Q}{\varepsilon_0}$$B. \oint B \cdot dA = 0$They are: