If the number of electric lines of force emer | vedclass

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

Question

(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q}{\varepsilon_0}$,where $q$ is the enclosed

Vedclass · Accepted Answer

$8.854 	imes 10^{-9}$

Vedclass · Answer

(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q}{\varepsilon_0}$,where $q$ is the enclosed charge and $\varepsilon_0$ is the permittivity of free space.Given that the number of electric lines of force (electric flux) $\phi = 1000$.The value of $\varepsilon_0 = 8.854 	imes 10^{-12} \, C^2 N^{-1} m^{-2}$.Substituting the values into the formula:$1000 = \frac{q}{8.854 	imes 10^{-12}}$$q = 1000 	imes 8.854 	imes 10^{-12}$$q = 8.854 	imes 10^{-9} \, C$.

If the number of electric lines of force emerging out of a closed surface is $1000$,then the charge enclosed by the surface is .......... $C$.

Similar Questions

The wrong statement about electric lines of force is:

The flux of the electric field $\overrightarrow{E} = 24 \hat{i} + 30 \hat{j} + 28 \hat{k} \text{ NC}^{-1}$ through an area of $20 \text{ m}^2$ on the $yz$ plane is

The $S.I.$ unit of electric flux is

Four closed surfaces $S_1, S_2, S_3$ and $S_4$ together with charges $+q, -q$ and $-2q$ are shown. Through which one of the surfaces is the net flux zero?

$A$ charge $q$ is placed at the corner of a cube of side $a$. The electric flux through the cube is

Vedclass Test Series

Exam Paper Generator

Online Exam Module