For a closed surface oint vec{E} cdot dvec{s} | vedclass

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(D) According to Gauss's Law,the net electric flux $\phi_E$ through a closed surface is given by $\phi_E = \oint \vec{E} \cdot d\vec{s} = \frac{q_{	e

Vedclass · Accepted Answer

The number of electric field lines entering the surface is equal to the number of electric field lines exiting the surface.

Vedclass · Answer

(D) According to Gauss's Law,the net electric flux $\phi_E$ through a closed surface is given by $\phi_E = \oint \vec{E} \cdot d\vec{s} = \frac{q_{	ext{enclosed}}}{\epsilon_0}$.Given that $\oint \vec{E} \cdot d\vec{s} = 0$,it implies that the net charge enclosed by the surface is zero $(q_{	ext{enclosed}} = 0)$.Electric flux represents the net number of electric field lines passing through a surface.Since the net flux is zero,the number of electric field lines entering the surface must be exactly equal to the number of electric field lines exiting the surface.

For a closed surface $\oint \vec{E} \cdot d\vec{s} = 0$,then:

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