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(C) The electric field is uniform and given by $\vec{E} = 3 	imes 10^3 \hat{i} 	ext{ N/C}$.For a closed surface (cube) in a uniform electric field,t

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(C) The electric field is uniform and given by $\vec{E} = 3 	imes 10^3 \hat{i} 	ext{ N/C}$.For a closed surface (cube) in a uniform electric field,the net electric flux is given by Gauss's Law: $\phi_{	ext{net}} = \oint \vec{E} \cdot d\vec{A}$.Since the electric field is uniform,the flux entering one face is equal to the flux leaving the opposite face,but with an opposite sign.Specifically,for the two faces perpendicular to the $x$-axis,the area vectors are $\vec{A}_1 = -A \hat{i}$ and $\vec{A}_2 = A \hat{i}$.The flux through these faces is $\phi_1 = \vec{E} \cdot \vec{A}_1 = (3 	imes 10^3 \hat{i}) \cdot (-A \hat{i}) = -EA$ and $\phi_2 = \vec{E} \cdot \vec{A}_2 = (3 	imes 10^3 \hat{i}) \cdot (A \hat{i}) = EA$.For the other four faces,the area vectors are perpendicular to the electric field $(\vec{E} \cdot d\vec{A} = 0)$,so the flux through them is $0$.The net flux is $\phi_{	ext{net}} = \phi_1 + \phi_2 + 0 + 0 + 0 + 0 = -EA + EA = 0$.

The net electric flux due to a uniform electric field of $3 \times 10^3 \hat{i} \text{ NC}^{-1}$ through a cube of side $20 \text{ cm}$ oriented such that its faces are parallel to the coordinate planes is

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