Question:
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
Solution:
From Gauss' law, the electric flux through an enclosed surface is given by
$\int_{S} E \cdot d s=\frac{q}{\varepsilon_{o}}$
Where, q is the net charge enclose by the surface
Now total charge enclose by the dipole = (‒ q + q) = 0
Therefore, electric flux through a surface enclosing a dipole = 0.
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