Axial: E = 2kp/r³. Equatorial: E = kp/r³. Direction analysis.

Electric Dipole — Field on Axial and Equatorial Points

Question

A short electric dipole has dipole moment p. Derive expressions for the electric field intensity at a point on (a) the axial line and (b) the equatorial line, both at distance r from the centre. Compare their magnitudes and directions.

Solution — Step by Step

Place the dipole along the x-axis: charge $-q$ at $-a$ and $+q$ at $+a$ , so dipole moment $p = q \cdot 2a$ pointing from $-q$ to $+q$ (i.e., in the $+x$ direction).

We treat the “short dipole” approximation throughout: $r \gg a$ .

Point P is on the axis, at distance $r$ from centre. Distances from each charge:

E_+ = \frac{kq}{(r-a)^2} \quad \text{(towards P, i.e. away from } +q\text{)}

E_- = \frac{kq}{(r+a)^2} \quad \text{(towards } -q\text{, i.e. away from P)}

Net field along axial direction (both components point in same direction — towards $+x$ ):

E_{axial} = kq\left[\frac{1}{(r-a)^2} - \frac{1}{(r+a)^2}\right]

Expanding and using $r \gg a$ :

E_{axial} = kq \cdot \frac{4ar}{(r^2 - a^2)^2} \approx \frac{4kqa}{r^3} = \frac{2kp}{r^3}

Direction: along $\hat{p}$ (same as dipole moment)

Point Q is on the perpendicular bisector at distance $r$ . Distance from each charge: $\sqrt{r^2 + a^2}$ .

Both charges contribute equal magnitude $E_+= E_- = \frac{kq}{r^2 + a^2}$ .

The components along the bisector cancel. Only the components along the axis (anti-parallel to $\hat{p}$ ) survive:

E_{equatorial} = 2 \cdot \frac{kq}{r^2+a^2} \cdot \frac{a}{\sqrt{r^2+a^2}} = \frac{kqa}{(r^2+a^2)^{3/2}}

For $r \gg a$ :

\boxed{E_{equatorial} = \frac{kp}{r^3}}

Direction: anti-parallel to $\hat{p}$ (opposite to dipole moment)

Point	Field magnitude	Direction
Axial	$\dfrac{2kp}{r^3}$	Along $\hat{p}$
Equatorial	$\dfrac{kp}{r^3}$	Opposite to $\hat{p}$

Axial field is exactly twice the equatorial field at the same distance. This factor of 2 is a standard board/JEE numerical — memorise it, but also know why.

Why This Works

On the axial line, both charges pull/push the test charge in the same net direction, so their contributions add constructively. On the equatorial line, the perpendicular components cancel and only the weaker axial components survive — this is why the equatorial field is half.

The $1/r^3$ dependence (vs. $1/r^2$ for a single charge) is the key signature of a dipole. The two opposite charges partially cancel each other at large distances, so the field falls off faster.

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

E_{equatorial} = \frac{kp}{r^3} = \frac{p}{4\pi\varepsilon_0 r^3}

Both valid for $r \gg 2a$ (short dipole approximation).

Alternative Method — Using Potential Approach

We can derive the general field using the dipole potential $V = \frac{kp\cos\theta}{r^2}$ and then find $E_r$ and $E_\theta$ by differentiation:

E_r = -\frac{\partial V}{\partial r} = \frac{2kp\cos\theta}{r^3}

E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = \frac{kp\sin\theta}{r^3}

For the axial point: $\theta = 0°$ , so $E_\theta = 0$ and $E_r = \frac{2kp}{r^3}$ . ✓

For the equatorial point: $\theta = 90°$ , so $E_r = 0$ and $E_\theta = \frac{kp}{r^3}$ , directed opposite to $\hat{p}$ . ✓

This method is faster once you’re comfortable with potential — worth using in JEE where time is tight.

Common Mistake

Most students write the equatorial field direction as “along $\hat{p}$ ” — copying the axial result. The equatorial field points anti-parallel to $\hat{p}$ . In JEE Main 2024 Shift 1, the direction question specifically tested this. The $+q$ side pulls the test charge towards $-q$ (anti- $\hat{p}$ ), and the $-q$ side pushes the same way. Both agree: direction is $-\hat{p}$ .

Quick memory trick: “Axial = Alike” — field direction same as $\hat{p}$ . “Equatorial = Enemy” — field opposes $\hat{p}$ . The factor of 2 difference in magnitude is non-negotiable — it appears in numericals where both fields are given and you must identify which is which.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Potential Approach

Common Mistake

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