Question

An arbitrary surface encloses an electric dipole. What is electric flux passing through the surface ?

Answer 1

This question utilizes the concept of Gauss’s theorem for determination of electric flux through an enclosed surface. The electric flux can be defined as the electric field multiplied by the area of the surface projected in a plane which is perpendicular to the field.

Formulae used:
$Electric{\kern 1pt} {\kern 1pt} Flux = \oint {\overrightarrow E \overrightarrow {ds} = \dfrac{{{q_T}}}{{{\varepsilon _o}}}}$
where $\oint {\overrightarrow E \overrightarrow {ds} }$ is the surface integral of the Electric field multiplied by small surface element $\overrightarrow {ds}$, ${q_T}$ is the total charge enclosed by the body and ${\varepsilon _o}$ is the electric constant.

Complete step by step answer:
Let the dipole be enclosed by an arbitrary surface $A$. We know that a dipole is a system of two equal and opposite charges $+ q$ and $- q$ , which are placed at a certain distance $d$ from each other. Thus, we can conclude that the total charge enclosed by the surface will be
${q_T} = q + ( - q) \\\ \Rightarrow {q_T} = q - q \\\ \Rightarrow {q_T} = 0 \\\$
Since total charge enclosed by the surface is zero, applying gauss’s law, we get electric flux as
$\text{Electric Flux} = \dfrac{{{q_T}}}{{{\varepsilon _o}}} \\\ \Rightarrow \text{Electric Flux} = \dfrac{0}{{{\varepsilon _o}}} \\\ \therefore \text{Electric Flux} = 0$

Therefore, electric flux through the arbitrary surface $A$ will be zero.

Note: Every electric charge, however small, gives out electric field lines. Plus charges give out electric lines whereas minus charges absorb electric field lines. In a dipole, all the field lines emanating from the plus charge gets absorbed in the minus charge, thus resulting in no stray electric field lines. Since no electric field lines are present, therefore, electric flux through the surface is non – existent.