Question

Derive an expression for electric field intensity due to an electric dipole at a point on its axial line.

Answer 1

Electric field due to dipole on axial line
Explanation
Electric field due to +q charge
$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r-d{)}^2}$ along BP

Electric field due to -q charge

$E_2=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r+d{)}^2}$ along PA

$E=E_1+(-E_2)$ or $E=E_1-E_2$

$E=\frac{q}{4\pi {\epsilon }_0}[\frac{1}{(r-d{)}^2}-\frac{1}{(r+d{)}^2}]$

$E=\frac{q}{4\pi {\epsilon }_0}\left[\frac{4rd}{(r^2-d^2{)}^2}\right]$ along BP

$p=q2d$

When $r>>d$, $d^2$ can be neglected.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{2P}{r^3}$

E acts in the direction of dipole moment.