12 min readConcept sheet3-mark derivations of the field on the axial and equatorial lines, 5-mark answers combining both, and 1-mark MCQ; also a JEE favourite

Electric dipole and the field due to a dipole

An electric dipole is two equal and opposite charges a small distance apart. Its field falls as 1/r³, faster than the field of a single charge.

Dipole momentp = q × 2a, directed from −q to +q; unit C m

Axial line (r ≫ a)E = (1/4πε₀) 2p/r³, in the direction of p

Equatorial line (r ≫ a)E = (1/4πε₀) p/r³, opposite to p

Key ratioAt the same large distance, E(axial) = 2 × E(equatorial)

What is an electric dipole?

An electric dipole is a pair of equal and opposite point charges, +q and −q, a small distance 2a apart. The line through them is the axis, and the midpoint is the centre. The dipole moment p = q × 2a is a vector pointing from −q to +q. The total charge is zero, but the field is not zero, because the two charges are at different places. Far away, the field falls as 1/r³, faster than the 1/r² of a single charge. Molecules such as H₂O and HCl are natural dipoles. An ideal (point) dipole has 2a → 0 and q → ∞, with p fixed.

Derivation: field on the axial line

−q is at A and +q at B, with AB = 2a and centre O. Point P is on the axis, at distance r from O, on the side of +q. Let k = 1/4πε₀.

1
Field at P due to +q (distance r − a): E₊ = kq/(r − a)², along p (away from B).
2
Field at P due to −q (distance r + a): E₋ = kq/(r + a)², opposite to p (towards A).
3
Net field: E = kq[1/(r − a)² − 1/(r + a)²], along p.
4
Simplify the bracket: [(r + a)² − (r − a)²]/(r² − a²)² = 4ar/(r² − a²)².
5
So E = k × 2(2aq)r/(r² − a²)² = (1/4πε₀) 2pr/(r² − a²)², along p.
6★
For a short dipole (r ≫ a), ignore a²: E = (1/4πε₀) 2p/r³, along p.

★ key step

Derivation: field on the equatorial line

Same dipole. Point P is on the perpendicular bisector of AB, at distance r from O.

1
Each charge is √(r² + a²) from P. So E₊ = E₋ = kq/(r² + a²).
2
E₊ points along BP, away from +q. E₋ points along PA, towards −q.
3
Resolve both. The components perpendicular to the axis are equal and opposite, so they cancel.
4
The components parallel to the axis add. Both point from +q towards −q, that is, opposite to p.
5
With cos θ = a/√(r² + a²): E = 2E₊ cos θ = 2kqa/(r² + a²)^(3/2), opposite to p.
6★
Using p = 2qa: E = (1/4πε₀) p/(r² + a²)^(3/2). For r ≫ a: E = (1/4πε₀) p/r³, opposite to p.

★ key step

How to draw the dipole figures

1
Draw a horizontal line. Mark −q at A (left) and +q at B (right). Mark O midway and write 2a for AB.
2
Draw an arrow p from −q to +q.
3
Axial point: mark P on the line beyond B, with OP = r. Draw a long arrow E₊ pointing away from B and a shorter arrow E₋ pointing back towards A.
4
Equatorial point: mark P above O on the perpendicular, with OP = r. Join AP and BP.
5
At P draw E₊ along BP extended (away from B) and E₋ along PA (towards A), of equal length. Show the resultant parallel to BA, opposite to p.

Worked example: field of a short dipole

Problem

Charges of +4 μC and −4 μC are 3 mm apart. Find (a) the dipole moment, and the field 20 cm from its centre (b) on the axis and (c) on the equatorial line.

Working

(a) p = q × 2a = 4 × 10⁻⁶ C × 3 × 10⁻³ m = 1.2 × 10⁻⁸ C m.
r = 0.20 m is much larger than a = 1.5 mm, so use the short-dipole formulas.
(b) Axial: E = 2kp/r³ = 2 × 9 × 10⁹ × 1.2 × 10⁻⁸ ÷ (0.20)³ = 216 ÷ 0.008 = 2.7 × 10⁴ N C⁻¹.
(c) Equatorial: E = kp/r³ = 108 ÷ 0.008 = 1.35 × 10⁴ N C⁻¹.

Answer

(a) 1.2 × 10⁻⁸ C m; (b) 2.7 × 10⁴ N C⁻¹ along p; (c) 1.35 × 10⁴ N C⁻¹ opposite to p

Point charge vs dipole

Single point chargeNet charge q. Field falls as 1/r². Same size at all points at the same distance (spherical symmetry).

Electric dipoleNet charge zero. Field falls as 1/r³. Depends on the direction from the dipole: on the axis it is twice the equatorial value.

In real life

A water molecule is a tiny dipole. A microwave oven works because its rapidly changing field keeps twisting the water dipoles in your rice or dal, and this jostling heats the food.

Exam angle

3 marks: “Derive an expression for the electric field at a point on the axial line of an electric dipole. Write it for a short dipole.” Fields of +q and −q: kq/(r − a)² and kq/(r + a)², in opposite directions. Net E = 2kpr/(r² − a²)². For r ≫ a, E = 2kp/r³, along p.

3 marks: “Derive an expression for the field at a point on the equatorial line of an electric dipole.” Each charge gives kq/(r² + a²). Perpendicular parts cancel; parallel parts add. E = kp/(r² + a²)^(3/2), opposite to p; for r ≫ a, E = kp/r³.

1 marks: “How does the field of a short dipole change when the distance from it is doubled?” E ∝ 1/r³, so it becomes one eighth.

Remember it

Dipole fields: axial is 2, equatorial is 1 (2p/r³ and p/r³). Axial points along p, equatorial against p.