Question

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

Answer 1

Electric field due to an electric dipole at a point on its axial line: $AB$ is an electric dipole of two point charges $-q$ and $+q$ separated by small distance $2d$. $P$ is a point along the axial line of the dipole at a distance $r$ from the midpoint $O$ of the electric dipole.

The electric fiedl at the point $P$ due to $+q$ placed at $B$ is,

$E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{{(r-d)}^2}$ (along $BP$)

The electric field at the point $P$ due to $-q$ placed at $A$ is,

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

Therefore, the magnitude of resultant electric field $\left(E\right)$ acts in the direction of the vector with a greater, magnitude. The resultant electric field at $P$ is

$E=E_1+(-E_2)$

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

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

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

If the point $P$ is far away from the dipole, then $d\ll r$

$\therefore E=\frac{q}{4\pi {\epsilon }_0}-\frac{4rd}{r^4}=\frac{q}{4\pi {\epsilon }_0}\frac{4d}{r^3}$

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

[$\because$ Electric dipole moment $p=q\times 2d$]

$E$ acts in the direction of dipole moment.