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Two equal and opposite charges add up to zero charge, yet they still produce a field. The electric dipole is one of the first results in Class 12 electrostatics, and both JEE and NEET test it with formulas, ratios and directions. This post derives the field on the two special lines, checks the numbers on a worked example, and shows when the familiar short-dipole formulas stop working.
What do the syllabus and NCERT cover?
"Electric dipole, electric field due to a dipole" appears in the electrostatics unit of the JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus. Section 1.10 of the NCERT Class 12 chapter on Electric Charges and Fields defines a dipole as charges q and −q separated by 2a, with dipole moment p = q × 2a pointing from −q to q. It then derives the field at a point on the axis and at a point on the equatorial plane.
What is the field on the axis?
Put P on the axis at distance r from the centre, on the side of +q. The +q charge is at r − a and the −q charge at r + a. Both fields lie along the axis, in opposite directions:
E = kq[1/(r − a)² − 1/(r + a)²] = kq × 4ar/(r² − a²)².
That is NCERT's exact result. For r much larger than a, the denominator becomes r⁴ and
E_axis ≈ 2kp/r³, pointing along p.
What is the field on the equatorial line?
Now put P on the perpendicular bisector at distance r from the centre. Each charge is √(r² + a²) away, so the two fields have equal size. Their components perpendicular to the axis cancel, and the components along the axis add, pointing from +q towards −q. NCERT's exact result is
E = kq × 2a/(r² + a²)^(3/2), and for r much larger than a,
E_equatorial ≈ kp/r³, pointing opposite to p.
Why does the field fall as 1/r³?
Far away, the two charges look almost the same distance off, so their fields nearly cancel. What is left depends on the small difference in distance, 2a, compared with r. That extra factor of a/r turns 1/r² into 1/r³. NCERT also notes that the far field depends only on the product qa, which is why p is defined at all.
Worked example: numbers on both lines
Charges ±2.0 nC are 4.0 mm apart. Find the field 10 cm from the centre (a) on the axis and (b) on the equatorial line. Take k = 9 × 10⁹ N m²/C².
a = 2.0 mm, so p = 2.0 × 10⁻⁹ × 4.0 × 10⁻³ = 8.0 × 10⁻¹² C m, and kp = 0.072 N m²/C.
r³ = (0.10)³ = 1.0 × 10⁻³ m³.
(a) Axis: 2kp/r³ = 2 × 0.072 / 1.0 × 10⁻³ = 144 N/C, along p.
(b) Equatorial: kp/r³ = 72 N/C, opposite to p.
Exact check for (a): kq × 4ar/(r² − a²)² = 18 × (4 × 0.002 × 0.10)/(0.009996)² ≈ 144.1 N/C. The approximation is off by less than 0.1%, because r is 50 times a.
When does the short-dipole formula fail?
Compare exact and approximate values as r shrinks. On the axis, exact ÷ approximate = r⁴/(r² − a²)².
At r = 10a: (100/99)² ≈ 1.02, so the approximation is 2% low.
At r = 2a: 16/9 ≈ 1.78, so the approximation is 44% low compared with the true value.
On the equatorial line at r = 10a, the exact value is 1/(1.01)^(3/2) ≈ 0.985 of the approximation, so it is about 1.5% high. If a question gives r only a few times a, use the exact formulas.
What about a general point?
For a point at angle θ from the axis, far from a short dipole, the field magnitude is (kp/r³)√(3cos²θ + 1). At θ = 0 this gives 2kp/r³, and at θ = 90° it gives kp/r³, matching both results above. JEE problems sometimes ask for this, and it follows from adding the two components.
For the potential and the torque on a dipole in a field, see our post on electric potential and dipoles. For continuous charges instead of two points, see the field of a finite charged rod.
Which mistakes cost marks?
Equatorial direction. It points opposite to p, from +q towards −q. A sign slip here is the most common error.
Using q instead of p. In the short-dipole formulas, p = q × 2a, not q × a.
Distance from the wrong point. r is measured from the centre of the dipole.
Applying 1/r³ when r is close to a. Check the ratio r/a first.
Unit slips. Convert mm and cm to metres before cubing.
A practice routine
Derive both exact results from Coulomb's law, from a blank page.
Solve NCERT Example 1.9 and confirm the 2:1 ratio of its two answers.
Make a short table, in your notebook, of the axial error at r = 2a, 5a and 10a.
Solve two problems asking for the force on a charge placed on each line.
When you need to revise vector addition of fields first, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, so you can fix the gap before returning to dipoles.
Tonight, redo the worked example with the point at 2.0 cm and compare the exact and approximate answers.
