Field on Axial Line of Dipole

Field at distance r from centre along the dipole axis. Direction is along p̂. The approximation holds when r ≫ a (half-length of dipole).

Setup

Place the dipole along the x-axis: $-q$ at $x = -a$, $+q$ at $x = +a$. Centre at origin. Dipole length $2a$, so $p = 2qa$.

Consider a point P at distance $r$ from the centre along the axis ($r > a$).

Fields from each charge

Field at P due to $+q$ (distance $r - a$, pointing away from $+q$, i.e., in $+x$ direction):

$$E_+ = \frac{q}{4\pi\varepsilon_0(r-a)^2}$$

Field at P due to $-q$ (distance $r + a$, pointing toward $-q$, i.e., in $-x$ direction):

$$E_- = \frac{q}{4\pi\varepsilon_0(r+a)^2}$$

Net field

Both contributions are along the x-axis. $E_+$ dominates (closer charge):

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

Expanding:

$$E = \frac{q}{4\pi\varepsilon_0} \cdot \frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2} = \frac{q}{4\pi\varepsilon_0} \cdot \frac{4ar}{(r^2-a^2)^2}$$

Since $p = 2qa$:

$$\boxed{E = \frac{2pr}{4\pi\varepsilon_0(r^2 - a^2)^2}}$$

Direction: along $\vec{p}$ (in $+x$ direction).

Far-field approximation ($r \gg a$)

$(r^2 - a^2)^2 \approx r^4$:

$$E \approx \frac{2p}{4\pi\varepsilon_0 r^3}$$