Result

Continuous distribution

(E3.5.1) E3.5 · Electric Potential Due to Continuous Charge Distributions →

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

Statement

Add the potential contributed by each small piece of charge.

Symbols and units

$V$	potential	V
$d q$	one small piece of charge	C
$r$	distance from dq to the point	m
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)

When it applies

Static charge distribution, V = 0 at infinity.

Don't use when

Requires a static charge distribution with V = 0 fixed at infinity — a distribution extending to infinity itself (e.g. an infinite line or plane of charge) makes this reference divergent and undefined.

Derivation

Each piece acts as a point charge:

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

Add over the whole distribution — a plain scalar integral, no components:

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

If you need the field afterwards, differentiate rather than integrating again:

E_{x} = - \frac{d V}{d x}

Worked examples

Ring of charge, on its axis

All elements lie at the same distance:

V = \frac{k _{e} Q}{x ^{2} + a ^{2}}

Differentiating recovers the field found by vector integration:

E_{x} = - \frac{d V}{d x} = \frac{k _{e} Q x}{( x ^{2} + a ^{2} ) ^{3/2}}

Watch out for

It is 1/r here, not 1/r², and there is no unit vector. That missing vector is the whole reason this route is usually less work.

Connects to

Continuous distribution Field from potential