Question

Derive expression for the electric field intensity due to an electric dipole at a point on the axis:

1- On the axial line 2- On equatorial line.

And also its cases.

Answer 1

Answer:-

1. For axial line

as we can see in the figure

electric field at P (EB) due to +q

$E_B=\frac{1}{4\pi \epsilon_o}.\frac{q}{(BP)^2} \ \ \ along \ BP\\ \ \ \ \ \ = \frac{1}{4\pi \epsilon_o}.\frac{q}{(r-a)^2}$

Electric field at P due to -q (EA)

$E_A=\frac{1}{4\pi \epsilon_o}.\frac{q}{(AP)^2} \ \ \ along \ PA\\ \ \ \ \ \ = \frac{1}{4\pi \epsilon_o}.\frac{q}{(r+a)^2}$

Net field at P is given by

$E_P=E_B-E_A$

=$\frac{1}{4\pi\epsilon_o}[\frac{q}{(r-a)^2}-\frac{q}{(r+a)^2}]$

simplifying , we get

$E_p= \frac{2qa}{4\pi\epsilon_o}.\frac{2r}{(r^2-a^2)^2}$

2.for equatorial line

as we see in the figure

$E_A=\frac{1}{4\pi\epsilon_o}\frac{q}{(AP)^2} \ \ \ \ \ \ along \ PA \\ E_A=\frac{1}{4\pi\epsilon_o}.\frac{q}{(r^2+a^2)} \\$

$E_B=\frac{1}{4\pi\epsilon_o}\frac{q}{(BP)^2} \ \ \ \ \ \ along \ PD \\ E_B=\frac{1}{4\pi\epsilon_o}.\frac{q}{(r^2+a^2)} \\$

the resultant intensity ios te vector sum of the intestines along PA and PB. EA and EB can be resolved into vertical and horizontal components . The vertical components of EA and EB cancel each other as they are equal and oppositely directed . It is the horizontal components which add up to give the resultant field .

$E_A=E_B=\frac{1}{4\pi\epsilon_o}.\frac{q}{(\sqrt{r^2+a^2})^2}=\frac{1}{4\pi\epsilon_o}.\frac{q}{({r^2+a^2})}$

E=2EAcos q

Substituting , $cos\theta=\frac{a}{(r^2+a^2)^\frac{1}{2}}$ in the the above equation

we get

$E=2E_acos\theta=\frac{2}{4\pi\epsilon_o}.\frac{q}{(r^2+a^2)}.\frac{a}{(r^2+a^2)^{\frac{1}{2}}}$