Result

Point charge

(E3.3.1) E3.3 · Electric Potential and Potential Energy Due to Point Charges →

V = k_{e} \frac{q}{r}

Statement

The potential around a single charge, taking V = 0 infinitely far away.

Symbols and units

$V$	potential, zero at infinity	V
$q$	charge, with sign	C
$r$	distance from the charge	m
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)

When it applies

Point charges, with V = 0 taken infinitely far away.

Don't use when

The reference V = 0 at infinity is a convention specific to an isolated point charge — a problem with a fixed boundary or ground reference elsewhere requires re-deriving the potential relative to that reference instead.

Derivation

Start from the definition and walk in radially from infinity:

V (r) - V (\infty) = - \int_{\infty}^{r} E \cdot d s

Moving radially, E·ds = E dr, with the point-charge field:

V (r) = - \int_{\infty}^{r} k_{e} \frac{q}{r ^{'2}} d r^{'}

Integrate — the antiderivative of 1/r'² is −1/r':

V (r) = - k_{e} q [- \frac{1}{r ^{'}}]_{\infty}^{r} = k_{e} \frac{q}{r}

For several charges, add the scalars directly:

V = k_{e} i \sum \frac{q _{i}}{r _{i}}

Worked examples

Midway between +q and −q

V = k_{e} \frac{q}{d} + k_{e} \frac{- q}{d} = 0, E = \frac{2 k _{e} q}{d ^{2}}

V = 0 where E is largest. Between two equal positive charges the reverse holds: E = 0 and V is largest.

Potential of a small charged sphere

q = 1 nC on a sphere of radius 1 cm:

V = \frac{( 8.99 \times 1 0 ^{9} ) ( 1 0 ^{- 9} )}{0.010} = 900 V

Static discharges involve kilovolts but negligible charge.

Watch out for

Keep the signs — this is an algebraic sum. Note V ~ 1/r while E ~ 1/r²: the extra power comes from the integration.

Connects to

Point charge Energy of a pair Charged sphere