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

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

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

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(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:

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