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(D) The electric flux $\phi_E$ through a surface area $A$ is defined by the dot product of the electric field vector $\vec{E}$ and the area vector $\v

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zero

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(D) The electric flux $\phi_E$ through a surface area $A$ is defined by the dot product of the electric field vector $\vec{E}$ and the area vector $\vec{A}$,given by $\phi_E = \vec{E} \cdot \vec{A} = EA \cos \alpha$,where $\alpha$ is the angle between the electric field vector $\vec{E}$ and the area vector $\vec{A}$ (the normal to the surface).In this problem,the square surface lies in the plane of the paper. The area vector $\vec{A}$ is perpendicular to the plane of the paper.The electric field $\vec{E}$ also lies in the plane of the paper.Since the area vector $\vec{A}$ is perpendicular to the plane of the paper and the electric field $\vec{E}$ lies within the plane of the paper,the angle $\alpha$ between $\vec{E}$ and $\vec{A}$ is $90^\circ$.Therefore,the electric flux is $\phi_E = EA \cos(90^\circ) = EA(0) = 0$.Thus,no electric field lines pass through the surface,and the electric flux is zero.

$A$ square surface of side $L \; m$ in the plane of the paper is placed in a uniform electric field $E \; (V/m)$ acting along the same plane at an angle $\theta$ with the horizontal side of the square as shown in the figure. The electric flux linked to the surface,in units of $V \cdot m$,is:

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