If there is only one type of charge in the un | vedclass

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

Question

(D) According to Gauss's Law,the electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivi

Vedclass · Accepted Answer

$ \oint \vec{E} \cdot d\vec{s} = 0 $ if charge is outside,$ = \frac{q}{\varepsilon_{0}} $ if charge is inside

Vedclass · Answer

(D) According to Gauss's Law,the electric flux through any closed surface is equal to the net charge enclosed by the surface divided by the permittivity of free space $( \varepsilon_{0} )$.Mathematically,$ \oint \vec{E} \cdot d\vec{s} = \frac{q_{	ext{enclosed}}}{\varepsilon_{0}} $.If the charge is outside the Gaussian surface,the net charge enclosed $( q_{	ext{enclosed}} )$ is $ 0 $,so the flux $ \oint \vec{E} \cdot d\vec{s} = 0 $.If the charge is inside the Gaussian surface,the net charge enclosed is $ q $,so the flux $ \oint \vec{E} \cdot d\vec{s} = \frac{q}{\varepsilon_{0}} $.This law holds true regardless of whether there is only one type of charge or both types of charges in the universe.

If there is only one type of charge in the universe,then ($ \vec{E} $ = Electric field,$ \vec{d}s $ = Area vector):

Similar Questions

$A$ hollow cylinder has a charge $q$ inside it. If $\phi$ is the electric flux associated with the curved surface $B$,the flux linked with the plane surface $A$ will be

$A$ $6 \mu C$ charge is placed at the centre of a cube. What will be the electric flux through each face of the cube? (Take $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \ Nm^2 C^{-2}$)

$A$ hollow cylinder has a charge $q$ inside it. If $\phi$ is the electric flux in units of $\text{V}\cdot\text{m}$ associated with the curved surface $B$, what is the flux linked with the plane surface $A$ in units of $\text{V}\cdot\text{m}$?

The charge distribution is shown in the figure below. The electric flux through the surface $S$ due to these charges is .........

$A$ charge $q$ is placed at the centre of the base of a cubical box of side $a$ with the top open. The flux of the electric field through the open surface is not considered,but the flux through the remaining five surfaces of the cubical box is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module