Potential of a point charge

Potential difference is defined by work: V(b) − V(a) = −∫ E ⋅ dl along a path from a to b, which is the work per unit charge that must be done against the field on the way, and minus the work the field itself does. For one point charge the field is E = (kq/r²) r̂. Choose a at infinity, where the field has died away, and walk straight in along a radius, so each step is dl = dr r̂ and E ⋅ dl = (kq/r²) dr. Then V(r) = −∫ kq/r² dr from ∞ to r = −[−kq/r] from ∞ to r = kq/r − 0 = kq/r. The whole result is that one integral. The minus sign in the definition cancels against the minus that comes from integrating r⁻², and the lower limit contributes nothing precisely because the field dies as 1/r². Notice also what the integration did to the exponent: a field going as r⁻² produced a potential going as r⁻¹.

Nothing in that derivation needed a radial path. Break an arbitrary path into small steps and split each step into a piece along r̂ and a piece perpendicular to it. The field is purely radial, so the perpendicular piece contributes E ⋅ dl = 0 — sliding around a sphere of fixed radius is free — and only the dr part survives. Two different routes between the same pair of points therefore accumulate the same integral, which is exactly the property that lets a potential exist at all. Choosing infinity as the zero is a separate matter, and it is available here because ∫ kq/r² dr converges as the far limit runs away: the field dies fast enough that the total work is finite. Move the zero somewhere else and every V picks up the same additive constant, which then cancels out of every potential difference.

In V = kq/r the distance r is positive and k is positive, so the sign of V is entirely the sign of q. Around a positive charge V is positive everywhere and decays toward zero far away; around a negative charge V is negative everywhere and climbs toward zero. Take q = −2.0 nC: at 4.0 cm, V = (8.99 × 10⁹)(−2.0 × 10⁻⁹)/(0.040) = −4.5 × 10² V, and that is the value of the point whether or not anything is sitting there. A second sign enters only when a charge is placed, through U = q₀V. Put an electron at that point and U = (−1.602 × 10⁻¹⁹ C)(−449.5 V) = +7.2 × 10⁻¹⁷ J, positive, because two negative charges repel and the pair would rather be further apart.

Potential and field are not two names for the same fall-off. Around a 1.0 nC charge, V(1.0 cm) = 899 V and V(2.0 cm) = 449.5 V, so doubling the distance halves the potential. Over the same step the field drops from 8.99 × 10⁴ V m⁻¹ to 2.25 × 10⁴ V m⁻¹, a factor of four. The link is one power of r: V/Eᵣ = r exactly for a point charge, and calculus moves between the two in either direction. Differentiate the potential and you get −d(kq/r)/dr = kq/r², the Coulomb field. Integrate the field in from infinity and the extra power of r appears from the integration. That identity is also the fastest unit check on the page, since V m⁻¹ multiplied by m is V, and it catches the commonest slip in this topic in a single line.

Because V depends on r alone, the surfaces of constant potential are spheres centred on the charge, and the field, always perpendicular to them, runs along the radii. Invert the formula as r = kq/V to see how those spheres are spaced. For q = 1.0 nC, kq = 8.99 V m, so the 100 V sphere sits at r = 9.0 cm, the 200 V sphere at 4.5 cm, the 300 V sphere at 3.0 cm and the 400 V sphere at 2.2 cm. Equal 100 V steps are 4.5 cm apart between the 100 V and 200 V shells, but only 0.75 cm apart between the 300 V and 400 V shells. The shells crowd where the field is strong, because a large E needs less distance to accumulate the same ΔV — the drawn version of Eᵣ = −dV/dr.

V = kq/r has no finite value at r = 0. That divergence belongs to the idealisation rather than to real charge: a metal sphere of radius R carrying q reaches only V = kq/R at its surface, so nothing physical ever climbs to infinity. Outside any spherically symmetric distribution, on the other hand, the point formula is exact — a 4.0 nC sphere of radius 1.0 cm and a 4.0 nC point are indistinguishable at 5.0 cm, where both give 7.2 × 10² V. The other limit is the reference. A distribution whose field dies no faster than 1/r, such as an infinite line of charge, makes the integral diverge at the far end, so V = 0 at infinity is simply unavailable and the zero has to be planted at a finite distance instead.