Question

Derive an expression for intensity of electric field at a point broadside position or an equatorial line of an electric dipole.

Answer 1

Now, suppose that the point P is situated on the right-bisector of the dipole AB at a distance r metre from its mid-point 0 (Fig. (a)]

Again Let $E_1$ and $E_2$ be the magnitudes of the intensities of the electric field at P due to the charges $+q$ and $-q$ of the dipole respectively. The distance of P from each charge is $\sqrt{r^2+l^2}$. Therefore,

and $E_1=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r^2+l^2)}$ away from $+q$

The magnitudes of $E_1$ and $E_2$ are equal (but directions are different). On resolving $E_1$ and $E_2$ into two components parallel and perpendicular to AB, the components perpendicular to AB ($E_{1}sin\theta and{E}_{2}sin\theta$) cancel each other (because they are equal and opposite), while the components parallel to AB ($E_{1}cos\theta and{E}_{2}cos\theta$ ), being in the same direction, add up [Fig. (b)]. Hence the resultant intensity of electric field at the point P is

$E=E_{1}cos\theta +E_{2}cos\theta$

$=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r^2+l^2)}cos\theta +\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r^2+l^2)}cos\theta$

$=\frac{1}{4\pi {\epsilon }_0}\frac{q}{(r^2+l^2)}2cos\theta$

But 2ql=p (moment of electric dipole)

$\therefore$ $=\frac{1}{4\pi {\epsilon }_0}\frac{p}{(r^2+l^2{)}^{32}}$

The direction of electric field E is 'antiparallel' to the dipole axis.

If $r$ is very large compared to $2l$ ($r>>2l$), then $l^2$ may be neglected in comparison to $r^2$.

$E=\frac{1}{4\pi {\epsilon }_0}\frac{p}{r^3}$Newton/Coulomb