Question

An electric dipole of dipole moment \(2 \times 10^{-6}\) cm is kept inside a closed surface. What will be the net electric flux coming out from the surface?

Hint:

Electric flux depends on enclosed net charge; dipoles have zero net charge so zero net flux.

Answer 1

According to Gauss's law, the net electric flux \( \Phi_E \) through any closed surface is directly proportional to the net charge \( Q_{\text{enc}} \) enclosed within that surface. Mathematically, it is expressed as: \[ \Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}, \] where \( \vec{E} \) is the electric field, \( d\vec{A} \) is the differential area vector on the closed surface, and \( \varepsilon_0 \) is the permittivity of free space.
An electric dipole consists of two equal and opposite charges, \( +q \) and \( -q \), separated by a small distance. Since these charges are equal in magnitude but opposite in sign, the total net charge enclosed by any closed surface that contains the dipole is: \[ Q_{\text{enc}} = +q + (-q) = 0. \] Therefore, by Gauss's law, the net electric flux through any closed surface enclosing an electric dipole is: \[ \Phi_E = \frac{0}{\varepsilon_0} = 0. \] This means that despite the presence of strong electric fields around the dipole, the total outward electric flux through a closed surface around it is zero because the positive and negative charges contribute equal and opposite fluxes that cancel each other out.
Hence, the net electric flux through a closed surface enclosing an electric dipole is always zero.