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Question

(A) The electric flux $\phi$ through a surface is given by the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{S}$.$\phi

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$0$

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(A) The electric flux $\phi$ through a surface is given by 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,and the area vector $\vec{S}$ (which is perpendicular to the surface) is also in the plane of the paper (or perpendicular to it depending on orientation,but since the surface is in the plane of the paper,the area vector is perpendicular to the plane of the paper).Since the electric field $\vec{E}$ is in the plane of the paper and the area vector $\vec{S}$ is perpendicular to the plane of the paper,the angle $	heta$ between $\vec{E}$ and $\vec{S}$ is $90^\circ$.Therefore,$\phi = ES \cos 90^\circ = ES(0) = 0$.Thus,the electric flux associated with the surface is $0$.

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

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