An arbitrary surface enclosed a dipole . what is the electric flux through this surface ?

Answer 1

Answer :

If any arbitrary surface encloses a dipole, the net charge is zero.

Dipole consists of two equal and opposite charges.

$E= \large\frac{1}{4 \pi \in_0}.\frac{q}{r^2}$

q is the charge , r is the distance from the charge .

$q=0$

$E= 0$

Electric flux $\Delta \phi =E. \Delta S$

$\therefore \Delta \phi =0$

Comments

Wrong process
Here E is not equal to 0,
Only q=0
See this https://www.physicsforums.com/threads/electric-dipole-and-gauss-law.161268/