Electric dipole — field along axial and equatorial line derivation

Question

An electric dipole consists of charges $+q$ and $-q$ separated by distance $2a$ . Derive the electric field at a point (a) on the axial line at distance $r$ from the centre, and (b) on the equatorial line at distance $r$ from the centre. Assume $r \gg a$ .

(NCERT Class 12, Chapter 1 — board exam favourite and JEE Main regular)

Solution — Step by Step

The electric dipole moment is $\vec{p} = q \cdot 2a\,\hat{p}$ , directed from the negative to the positive charge. We place the dipole along the x-axis with its centre at the origin.

A point P is on the axial line at distance $r$ from the centre. The distances from P to the $+q$ and $-q$ charges are $(r-a)$ and $(r+a)$ respectively.

E_{axial} = \frac{kq}{(r-a)^2} - \frac{kq}{(r+a)^2} = kq\left[\frac{(r+a)^2 - (r-a)^2}{(r^2-a^2)^2}\right]

Numerator: $(r+a)^2 - (r-a)^2 = 4ra$

E_{axial} = \frac{4kqra}{(r^2 - a^2)^2}

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

\boxed{E_{axial} = \frac{2kp}{r^3} = \frac{1}{4\pi\varepsilon_0}\frac{2p}{r^3}}

Direction: along $\vec{p}$ (from $-q$ to $+q$ ).

A point Q is on the perpendicular bisector at distance $r$ from the centre. Each charge is at distance $\sqrt{r^2 + a^2}$ from Q.

The fields from $+q$ and $-q$ have equal magnitudes: $E_0 = \frac{kq}{r^2 + a^2}$ .

The horizontal components (along the dipole axis) add up while the vertical components cancel by symmetry.

E_{eq} = 2E_0 \cos\alpha = 2 \cdot \frac{kq}{r^2+a^2} \cdot \frac{a}{\sqrt{r^2+a^2}} = \frac{2kqa}{(r^2+a^2)^{3/2}}

For $r \gg a$ : $(r^2 + a^2)^{3/2} \approx r^3$

\boxed{E_{eq} = \frac{kp}{r^3} = \frac{1}{4\pi\varepsilon_0}\frac{p}{r^3}}

Direction: antiparallel to $\vec{p}$ (from $+q$ to $-q$ ).

E_{axial} = 2 \times E_{equatorial}

At the same distance from the dipole, the axial field is exactly twice the equatorial field. Both fall off as $1/r^3$ — much faster than the $1/r^2$ field of a point charge.

Why This Works

A dipole has zero net charge, so at large distances the fields from the two charges nearly cancel. What remains is the “leftover” from the slight imbalance — and this residual field decays as $1/r^3$ . The axial field is stronger because both charge contributions point in the same direction, while on the equatorial line only the components along the dipole axis survive.

The $1/r^3$ dependence is the signature of a dipole field. Quadrupoles fall off as $1/r^4$ , octupoles as $1/r^5$ , and so on.

Alternative Method

Using the general formula for a dipole: $E_r = \frac{2kp\cos\theta}{r^3}$ and $E_\theta = \frac{kp\sin\theta}{r^3}$ . At the axial line $\theta = 0$ : $E = 2kp/r^3$ . At the equatorial line $\theta = 90°$ : $E = kp/r^3$ .

For CBSE boards, the derivation is worth 5 marks. Make sure to clearly state the approximation $r \gg a$ and show the simplification step. For JEE, remember the ratio $E_{axial}/E_{eq} = 2$ — this one-liner solves many MCQs instantly.

Common Mistake

The direction of the equatorial field is the most common error. Students write it as parallel to $\vec{p}$ — it is actually antiparallel (opposite to $\vec{p}$ ). On the equatorial line, the field points from the positive charge toward the negative charge, which is opposite to the dipole moment direction. Mixing this up flips the sign in vector problems.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next