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

By vector-adding the Coulomb fields of the two point charges of a dipole at an axial point, the exact field is $E = \dfrac{2kpr}{(r^2-a^2)^2}$, which reduces to $E = \dfrac{p}{2\pi\varepsilon_0 r^3}$ for a short dipole ($r\gg a$), directed along the dipole moment $\vec p$.

Setup

Consider an electric dipole consisting of charge $-q$ at point A and charge $+q$ at point B, separated by distance $2a$, with O as the midpoint. The dipole moment is $\vec p$, of magnitude $p = q(2a)$, directed from $-q$ to $+q$. Let P be a point on the axial line (the line through A, O, B extended) at distance $r$ from O, on the side of $+q$.

Distance of P from $+q$: $BP = r - a$

Distance of P from $-q$: $AP = r + a$

Step 1: Field due to $+q$ at P

$E_1 = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{q}{(r-a)^2}$

directed away from $+q$, i.e. along OP produced (same sense as $\vec p$).

Step 2: Field due to $-q$ at P

$E_2 = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{q}{(r+a)^2}$

directed towards $-q$, i.e. opposite to $\vec p$.

Step 3: Resultant field

Since $E_1$ and $E_2$ act along the same line but in opposite senses, and P is closer to $+q$ so $E_1 > E_2$, the resultant is

$E = E_1 - E_2 = \dfrac{q}{4\pi\varepsilon_0}\left[\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2}\right]$

directed along $\vec p$ (from $-q$ to $+q$ side).

Step 4: Simplify

$\dfrac{1}{(r-a)^2} - \dfrac{1}{(r+a)^2} = \dfrac{(r+a)^2-(r-a)^2}{(r^2-a^2)^2} = \dfrac{4ra}{(r^2-a^2)^2}$

so

$E = \dfrac{q}{4\pi\varepsilon_0}\cdot\dfrac{4ra}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2(2qa)r}{(r^2-a^2)^2} = \dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{2pr}{(r^2-a^2)^2}$ …