Answer to Question #89925 in Electricity and Magnetism for Shivam Nishad

Question

Derive an expression for the electric field at a point :
(i) along the axis of a dipole, and
(ii) on the perpendicular bisector of the
dipole axis

Answer 1

(i)



$$E=E_1+E_2 (1)$$

where

$$E_1=\frac { 1}{4\pi ɛ_0}\frac { q}{(r-a) ^2}$$

$$E_2=\frac { 1}{4\pi ɛ_0}\frac { -q}{(r+a) ^2}$$

Using (1) we got:

$$E=\frac { 1}{4\pi ɛ_0} 4aq \frac { r}{(r^2-a^2) ^2} (2)$$

The dipole moment (p) is given by formula

$$p=2qa (3)$$

We put (3) in (2):

$$E=\frac { 1}{2\pi ɛ_0} \frac { pr}{(r^2-a^2) ^2} (4)$$

Consider the far field case (i.e., the case where r >> a)

Then one can see that r² >> a², which allows us to simplify the equation in the far field.

$$E=\frac { 1}{2\pi ɛ_0} \frac { p}{r^3}$$

(ii)

$$E=2E_1 \ cos θ (1)$$

where

$$E_1=\frac { 1}{4\pi ɛ_0} \frac { q}{a ^2+x^2}$$

$$\ cos θ=\frac { a}{\sqrt { a ^2+x^2}}$$

Using (1) we got:

$$E=\frac { 1}{4\pi ɛ_0} \frac { 2qa}{ \sqrt[3]{ a ^2+x^2}}$$

By inspection we can conclude that when x>>a then this approaches

$$E=\frac { 1}{4\pi ɛ_0} \frac { 2qa}{ x^3}$$