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(D) The electric flux $\phi$ through a surface is defined as the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{S}$.$\ph

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$0$

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(D) The electric flux $\phi$ through a surface is defined as the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{S}$.$\phi = \vec{E} \cdot \vec{S} = ES \cos 	heta$Here,the electric field $\vec{E}$ lies in the plane of the paper.The area vector $\vec{S}$ for a surface in the plane of the paper is perpendicular to the plane of the paper.Therefore,the angle $	heta$ between the electric field vector $\vec{E}$ and the area vector $\vec{S}$ is $90^{\circ}$.Since $\cos 90^{\circ} = 0$,the electric flux $\phi$ associated with the surface is:$\phi = ES \cos 90^{\circ} = ES(0) = 0$.

$A$ square surface of side $L$ metres is in the plane of the paper. $A$ uniform electric field $\vec{E} \text{ (V/m)}$,also in the plane of the paper,is limited only to the lower half of the square surface,(see figure). The electric flux in $SI$ units associated with the surface is

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