If oint_s vec{E} cdot overrightarrow{d S}=0 o | vedclass

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

Question

(B) The given expression $\oint_s \vec{E} \cdot \overrightarrow{d S} = 0$ represents the total electric flux $\phi$ through a closed surface.According

Vedclass · Accepted Answer

the number of flux lines entering the surface must be equal to the number of flux lines leaving it.

Vedclass · Answer

(B) The given expression $\oint_s \vec{E} \cdot \overrightarrow{d S} = 0$ represents the total electric flux $\phi$ through a closed surface.According to Gauss's Law,the total flux through a closed surface is equal to the net charge enclosed divided by the permittivity of free space,$\phi = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Since the total flux is zero,the net charge enclosed by the surface must be zero $(q_{	ext{enclosed}} = 0)$.This implies that the number of electric field lines entering the surface is exactly equal to the number of electric field lines leaving the surface,resulting in a net flux of zero.

If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface,then:

Similar Questions

$A$ charge $q$ is placed inside a hollow cylinder. The electric flux associated with the curved surface $B$ is $\phi$. The flux associated with a plane surface (either $A$ or $C$) will be:

Explain the electric field lines and the magnitude of the electric field.

$A$ hemisphere of radius $R$ is placed in a uniform electric field $E$ so that its axis is parallel to the field. Which of the following statement$(s)$ is/are true?

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface,the electric field will be due to:

Three charges $q_1 = 1\,\mu C, q_2 = 2\,\mu C$,and $q_3 = -3\,\mu C$ and four surfaces $S_1, S_2, S_3$,and $S_4$ are shown in the figure. The electric flux emerging through surface $S_2$ in $N\cdot m^2/C$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module