Figure shows the field produced by two point charges +q and -q of equal magnitude but opposite signs (an electric dipole). Find the electric flux through each of the closed surfaces A, B, C and D.

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Solution

Flux through closed surface A, $ϕ_{A} = \frac{Q_{i n}}{ϵ_{0}} = \frac{q}{ϵ_{0}}$
Flux through closed surface B, $ϕ_{B} = \frac{Q_{i n}}{ϵ_{0}} = - \frac{q}{ϵ_{0}}$
Flux through closed surface C, $ϕ_{C} = \frac{Q_{i n}}{ϵ_{0}} = \frac{q - q}{ϵ_{0}} = 0$
Flux through closed surface D, $ϕ_{D} = \frac{Q_{i n}}{ϵ_{0}} = 0$

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An electric dipole is an arrangement of two equal charges of equal magnitude but opposite nature, separated by some fixed distance.

Now consider a dipole enclosed inside a closed surface S. The flux through the surface S is zero.

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