10 min readConcept sheet3-mark derivation of the dipole potential, and 2 or 3-mark numericals on several charges; also a JEE favourite

Potential due to a dipole and due to a system of charges

Potentials add as plain numbers with signs. A short dipole gives V = p cos θ/(4πε₀r²), which is zero on the equatorial line.

Short dipole (r ≫ a)V = (1/4πε₀) p cos θ/r², with θ measured from the direction of p

Special pointsAxial line: V = ±p/(4πε₀r²); equatorial line: V = 0

System of chargesV = (1/4πε₀)(q₁/r₁ + q₂/r₂ + ... + qₙ/rₙ), an algebraic sum

Thin charged shellOutside: V = q/(4πε₀r); inside: V = q/(4πε₀R), the same at every point

Derivation: potential due to an electric dipole

−q is at A and +q at B, with AB = 2a and centre O. P is at distance r from O, and OP makes an angle θ with p.

1
Let r₁ = BP (from +q) and r₂ = AP (from −q). Potentials add: V = (q/4πε₀)(1/r₁ − 1/r₂).
2
From the geometry: r₁² = r² + a² − 2ar cos θ and r₂² = r² + a² + 2ar cos θ.
3
For r ≫ a, drop a²/r²: r₁² ≈ r²(1 − 2a cos θ/r) and r₂² ≈ r²(1 + 2a cos θ/r).
4
Use the binomial approximation: 1/r₁ ≈ (1/r)(1 + a cos θ/r) and 1/r₂ ≈ (1/r)(1 − a cos θ/r).
5
Subtract: 1/r₁ − 1/r₂ ≈ 2a cos θ/r².
6★
So, V = (1/4πε₀)(2aq) cos θ/r² = (1/4πε₀) p cos θ/r², or V = p·r̂/(4πε₀r²).

★ key step

Dipole potential at special points

Position of P	Angle θ	Potential V
On the axis, on the +q side	0°	+p/(4πε₀r²)
On the axis, on the −q side	180°	−p/(4πε₀r²)
On the equatorial line	90°	0

Point charge vs dipole

Point chargeV = q/(4πε₀r). Falls as 1/r. The same in every direction (spherically symmetric).

Short dipoleV = p cos θ/(4πε₀r²). Falls as 1/r². Depends on the angle θ: it is zero on the equatorial line, even though E is not zero there.

Potential due to a system of charges

Potential is a scalar, so the potential of many charges is the algebraic sum of their separate potentials: V = (1/4πε₀) Σ qᵢ/rᵢ. Put in each charge with its sign. There are no directions to worry about, which makes potential easier to find than field. For a continuous charge, split it into small pieces and add (integrate). For a uniformly charged thin shell, the potential outside is q/(4πε₀r), as if the charge were at the centre. Inside, E = 0, so the potential is constant and equal to its value at the surface, q/(4πε₀R).

Worked example: four charges at a square

Problem

Charges of +5 nC, −3 nC, +4 nC and −2 nC sit at the corners of a square whose diagonal is 20 cm. Find the potential at the centre.

Working

Each corner is half a diagonal, 0.10 m, from the centre.
Since r is the same for all, V = k(q₁ + q₂ + q₃ + q₄)/r.
Sum of charges = 5 − 3 + 4 − 2 = +4 nC.
V = 9 × 10⁹ × 4 × 10⁻⁹ ÷ 0.10 = 36 ÷ 0.10 = 360 V.

Answer

360 V

Worked example: potential of a short dipole

Problem

A short dipole has p = 2 × 10⁻¹⁰ C m. Find the potential 10 cm from its centre (a) on its axis, on the +q side, (b) at θ = 60° to the axis, (c) on its equatorial line.

Working

(a) V = kp/r² = 9 × 10⁹ × 2 × 10⁻¹⁰ ÷ (0.10)² = 1.8 ÷ 0.01 = 180 V.
(b) V = kp cos 60°/r² = 180 × 0.5 = 90 V.
(c) cos 90° = 0, so V = 0.

Answer

(a) 180 V; (b) 90 V; (c) zero

In real life

An ECG in a hospital records tiny potential differences on the skin. The beating heart acts roughly like a changing electric dipole, so electrodes on the arms and legs pick up different potentials.

Exam angle

3 marks: “Derive an expression for the potential at a point due to a short electric dipole. Hence write the potential on its axial and equatorial lines.” V = (q/4πε₀)(1/r₁ − 1/r₂); for r ≫ a, 1/r₁ − 1/r₂ ≈ 2a cos θ/r², so V = p cos θ/(4πε₀r²). Axial: ±p/(4πε₀r²); equatorial: 0.

2 marks: “Charges of +3 nC, −1 nC, +4 nC and +2 nC sit at the corners of a square whose diagonal is 20 cm. Find the potential at its centre.” Each charge is 0.10 m away. Sum = +8 nC. V = 9 × 10⁹ × 8 × 10⁻⁹ ÷ 0.10 = 720 V.

1 marks: “What is the potential at a point on the equatorial line of an electric dipole?” Zero, since θ = 90° and cos 90° = 0.

Remember it

Potentials just add, with signs (no arrows). Dipole: V ∝ cos θ/r², zero on the equator.