Question

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

Answer 1

Gauss Theorem:

• According to Gauss's Law, the total electric flux out of a closed surface equals the charge contained divided by the permittivity.
• The electric flux in an area is calculated by multiplying the electric field by the area of the surface projected in a plane and perpendicular to the field.

${\mathrm{\varphi }}_{\mathrm{E}}=\frac{\mathrm{Q}}{{\mathrm{\epsilon }}_0}$

• Where, ${\mathrm{\varphi }}_{\mathrm{E}}$ is electric flux through a closed surface $\mathrm{S}$ enclosing any volume V
• $\mathrm{Q}$ is total charge enclosed within $\mathrm{V}$
• ${\mathrm{\epsilon }}_0$ is dielectric constant of free space
• If any arbitrary surface encloses a dipole, the net charge is zero.
• As the dipole is made up of two equal and opposing charges.
• The electric flux is also zero as:

${\mathrm{\varphi }}_{\mathrm{E}}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}=\frac{0}{{\mathrm{\epsilon }}_0}=0$

Hence, the electric flux to the surface is zero.