91Ó°ÊÓ

The distance of point P from the center of the dipole $\left(\mathrm{r}\right)$is very large as compared to the distance between the two charges of the dipole $\left(\mathrm{d}\right)$.

$\left(\mathrm{r}≫\mathrm{d}\right)$

A combination of two equal and opposite charges placed very close to each other, is known as a dipole. If the two charges having magnitude $\left(q\right)$are placed d distance apart, then the dipole moment $\left(p\right)$is given as-

$\stackrel{\mathbf{→}}{\mathbf{p}}\mathbf{=}\mathbf{q}\stackrel{\mathbf{→}}{\mathbf{d}}$

Using the concept of the electric dipole, the expression for the electric field vector can be given. Again, from this expression, the direction of the electric field can be calculated as given.

The electric field of an electric dipole along the equatorial axis,

$\begin{array}{rcl}\mathbf{E}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{o}}}\frac{\mathbf{qd}}{{\mathbf{r}}^{\mathbf{3}}}\\ & \mathbf{=}& \frac{\mathbf{1}}{\mathbf{4}{\mathbf{Ï€Î\mu }}_{\mathbf{o}}}\frac{\mathbf{p}}{{\mathbf{r}}^{\mathbf{3}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right)\end{array}$

Considering the figure below, the magnitude of the net electric field at point P can be given using equation (i) as:

$\begin{array}{rcl}\left|{\stackrel{→}{\mathrm{E}}}_{\mathrm{net}}\right|& =& 2{\mathrm{E}}_1\sin \left(\mathrm{θ}\right)\\ & =& 2\left(\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{\mathrm{qd}}{\left[{\left(\mathrm{d}/2\right)}^2+{\mathrm{r}}^2\right]}\right)\frac{\mathrm{d}/2}{{\left[{\left(\mathrm{d}/2\right)}^2+{\mathrm{r}}^2\right]}^{1/2}}\\ & =& \frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{\mathrm{qd}}{{\left[{\left(\mathrm{d}/2\right)}^2+{\mathrm{r}}^2\right]}^{3/2}}\end{array}$

Here E₁ is the magnitude of the electric field due to each charge. The electric field due to charge +q is directed away from charge q and the field due to -q is directed towards the charge -q.

For r>>d, we can substitute the above equation with ${\left[{\left(\mathrm{d}/2\right)}^2+{\mathrm{r}}^2\right]}^{3/2}≈{\mathrm{r}}^3$.

Thus, the expression above reduces to

$\left|\stackrel{→}{{\mathrm{E}}_{\mathrm{net}}}\right|≈\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{\mathrm{qd}}{{\mathrm{r}}^3}$

Hence, the magnitude of the net dipole’s electric field is$\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{\mathrm{qd}}{{\mathrm{r}}^3}$

From the above figure, it is clear that the net electric field at point P is directed along $-\hat{\mathrm{j}}$.