Worked exercise — electric dipole on the axis

Problem

Two charges $+q$ and $-q$ separated by a small distance $d$. (a) Magnitude of the electric field on the dipole axis at distance $r \gg d$ from the centre. (b) Comparison with the field of a single charge.

Solution

(a) Field on the axis. Place the centre of the dipole at the origin and consider the point $P$ on the axis at distance $r$ from the centre. The point $P$ is at distance $r-d/2$ from the positive charge and $r+d/2$ from the negative one. The two contributions, both directed from $+q$ towards $-q$ (i.e. along the axis), are opposite in direction and so subtract in magnitude: $E = kq\left[\frac{1}{(r-d/2)^2} - \frac{1}{(r+d/2)^2}\right].$ Expanding to first order for $d/r \ll 1$ (with $(1\mp x)^{-2}\approx 1\pm 2x$): $E \approx kq\cdot\frac{2rd}{r^4} = \frac{2kqd}{r^3} = \ev{E = \frac{2kp}{r^3}},$ where $p = qd$ is the dipole moment.

(b) Comparison. A single charge generates a field that decays as $1/r^2$; the dipole decays as $1/r^3$, i.e. more rapidly. The physical reason: at large distances the two opposite charges appear almost coincident and their fields nearly cancel out, leaving only a residue that falls off faster. This is the same reason why, from afar, water appears electrically neutral even though its molecules possess a non-zero dipole moment.

Connections

Topics: Electric field and potential Concepts: Electric dipole · Electric field Skills: Superposition principle Methods: Gradient method Objects: Charged sphere