Gauss's law is given by {epsilon _0}oint {vec | vedclass

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

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The number of electric field lines entering the surface must be equal to the number of electric field lines leaving it.

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(B) According to Gauss's Law,the total electric flux $\Phi_E$ through a closed surface is given by $\Phi_E = \oint {\vec E \cdot d\vec s} = \frac{q_{enclosed}}{\epsilon_0}$.If the net charge enclosed $q_{enclosed} = 0$,then the total electric flux $\Phi_E = 0$.Electric flux represents the net number of electric field lines passing through the surface.$A$ flux of zero implies that the number of electric field lines entering the surface is exactly equal to the number of electric field lines leaving the surface.Therefore,option $B$ is correct.

Gauss's law is given by ${\epsilon _0}\oint {\vec E \cdot d\vec s} = q$. If the net charge enclosed by a Gaussian surface is zero,then .......

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