Potential of a dipole and of many charges
Potential due to a dipole
An electric dipole is a pair of equal and opposite charges: −q at A and +q at B, a distance 2a apart. Its dipole moment p has size q × 2a, and points from −q to +q.
Take the centre of the dipole as the origin. We want the potential at a point P, a distance r from the centre, where the line OP makes an angle θ with p.
Potentials simply add. P is r₁ from +q and r₂ from −q, so V = (1/4πε₀)(q/r₁ − q/r₂).
Far from the dipole, r is much bigger than a. Then r₁ is about r − a cos θ, and r₂ is about r + a cos θ.
Work it through, keeping only the largest term: V = p cos θ/4πε₀r². In vector form, V = p·r̂/4πε₀r².
On the axis, on the side of +q, θ is zero and V = p/4πε₀r². On the other side, θ is 180°, and V has the same size but is negative.
On the equatorial line, θ is 90°, so V is zero everywhere along it. Every point there is equally far from +q and −q, so their potentials cancel exactly.
Notice two differences from a single charge. The dipole potential falls off as 1/r², not 1/r. And it depends on the direction θ, not only on r.
Worked example: q = 1 nC and 2a = 2 mm, so p = 2 × 10⁻¹² C m. At 10 cm along the axis, V = 9 × 10⁹ × 2 × 10⁻¹² ÷ 0.01, which is 1.8 V. At θ = 60°, it is half as much: 0.9 V.
Potential due to many charges
For any group of charges the rule is the same: the potential at a point is the sum of the potentials due to each charge on its own. This is the superposition principle.
V = (1/4πε₀)(q₁/r₁P + q₂/r₂P + … + qₙ/rₙP). Each r is the distance from that charge to the point P.
These are plain numbers with signs, not vectors. So the adding is easy: no components, no angles.
For a continuous charge, split it into tiny pieces and add their potentials the same way. For a uniformly charged spherical shell, the answer outside is V = q/4πε₀r, as if all the charge were at the centre.
Inside the shell, the field is zero, so moving a charge around in there takes no work. The potential is the same everywhere inside, equal to its value at the surface, q/4πε₀R.
Worked example: charges of 3 × 10⁻⁸ C and −2 × 10⁻⁸ C are 15 cm apart. Where on the line through them is the potential zero? Put the positive charge at the origin and the negative one at 15 cm.
Between them, at x cm from the origin: 3/x must equal 2/(15 − x). So 45 − 3x = 2x, and x = 9 cm.
Beyond the negative charge, at x: 3/x = 2/(x − 15). So 3x − 45 = 2x, and x = 45 cm.
To the left of the positive charge there is no such point: there the bigger charge is always nearer, so V stays positive. And at both zero-potential points, the field is not zero.
Try it: the dipole’s potential
- Move q₀ along the dashed white line halfway between the charges. V stays at zero, but the field arrow does not vanish.
- Move q₀ out along the axis to the right. Compare V with the dipole formula kp cos θ/r²: the agreement improves as r grows.
- Add a third charge and watch the map: the potentials just add up.
- Turn off the colour map to see the equipotential lines on their own.
On the equatorial line
A point P lies on the equatorial line of an electric dipole, 5 cm from its centre. Which is true at P?
- V = 0 and E = 0
- V = 0 but E ≠ 0
- V ≠ 0 but E = 0
- V ≠ 0 and E ≠ 0
Answer
B. P is equally far from +q and −q, so kq/r − kq/r = 0: the potentials cancel because they are scalars with opposite signs. The fields are vectors: their components along the line of the dipole add, giving E = kp/r³ antiparallel to p. A zero potential does not mean a zero field.
Four charges on a square
Four charges of +2 nC each sit at the corners of a square, each 10 cm from its centre. What is the potential at the centre?
- 0 V, because the charges cancel
- 180 V
- 720 V
- 509 V
Answer
C. Each charge gives kq/r = 9 × 10⁹ × 2 × 10⁻⁹ / 0.1 = 180 V. Potentials are scalars and all positive here, so they add: 4 × 180 = 720 V. It is the field at the centre that cancels to zero, not the potential. (509 V comes from using the side, 14.1 cm, instead of 10 cm.)
Summary
- Dipole (r ≫ a):
V = p cos θ/4πε₀r² = p·r̂/4πε₀r² - System of charges:
V = (1/4πε₀) Σ qᵢ/rᵢP - Charged shell, outside / inside:
V = q/4πε₀r / V = q/4πε₀R
- A dipole’s potential depends on r and θ, and falls off as 1/r².
- On the axis V = ±p/4πε₀r²; on the equatorial line V = 0, though E is not zero there.
- By superposition, the potential of many charges is the algebraic sum of their separate potentials.
- Inside a uniformly charged shell the potential is constant and equals its surface value.