Electric dipole potential derivation and formula

Key concept: This derivation is one of those places where the geometry does most of the work. Once you write the two potentials carefully and use the far-field approximation, the dipole form drops out cleanly.

Step by step

For a dipole made of charges $+q$ and $-q$, the electric potential at a point $M$ is the scalar sum of the two contributions:

$V_M = V_A(-q)+V_A(+q).$

Using the point-charge potential,

$V_M=\frac{Kq}{r_1}-\frac{Kq}{r_2}=Kq\,\frac{r_2-r_1}{r_1r_2}.$

Far-field approximation

If the observation point is far from the dipole compared with the separation $a$, then $a\ll r$. In that case the two distances are almost equal, and the field lines appear nearly parallel.

We use the standard approximations:

$r_1r_2\approx r^2,\qquad r_2-r_1\approx a\cos\theta.$

Substituting gives

$V_M\approx Kq\,\frac{a\cos\theta}{r^2}.$

Since the dipole moment magnitude is

$p=qa,$

we obtain

$V_M=K\,\frac{p\cos\theta}{r^2}.$

Using vectors, because

$\vec p\cdot\vec r=pr\cos\theta,$

we can also write

$V_M=K\,\frac{\vec p\cdot\vec r}{r^3}.$

That is the standard potential of an electric dipole in the far field.

Pitfall alert

A common mistake is to think the dipole potential behaves like the electric field and falls as $1/r^3$. The potential itself is proportional to $1/r^2$; the $1/r^3$ appears only after rewriting it with the vector dot product.

Try different conditions

If the point is not far from the dipole, the approximation $r_1r_2\approx r^2$ is no longer safe, and you must keep the exact expression $V=Kq(1/r_1-1/r_2)$. If the dipole is oriented perpendicular to the observation direction, then $\cos\theta=0$ and the potential vanishes in this approximation.

Further reading

dipole moment, electric potential, far-field approximation