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(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{net}}}{\varepsilon_0}$,where $

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(D) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{net}}}{\varepsilon_0}$,where $q_{	ext{net}}$ is the total charge enclosed within the surface.An electric dipole consists of two equal and opposite charges,$+e$ and $-e$. The net charge of a single dipole is $q_{	ext{dipole}} = (+e) + (-e) = 0$.Since there are eight such dipoles inside the cube,the total net charge enclosed is $q_{	ext{net}} = 8 	imes 0 = 0$.Therefore,the total electric flux coming out of the cube is $\phi = \frac{0}{\varepsilon_0} = 0$.

Eight dipoles of charges of magnitude $e$ are placed inside a cube. The total electric flux coming out of the cube will be

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