Result

Point charge

E = k_{e} \frac{q}{r ^{2}} r

Statement

The field of a single point charge: radially out from a positive charge, radially in towards a negative one.

Symbols and units

$E$	field of the charge	N/C
$q$	source charge, with sign	C
$r$	distance from the charge	m
$r$	unit vector pointing away from it	—
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)

When it applies

A point charge — or anywhere outside a spherically symmetric distribution.

Don't use when

Diverges as r → 0 — the point-charge idealization breaks down at very small separations where the charge's real size matters.

Derivation

Start from the force a test charge q₀ would feel at distance r:

F_{e} = k_{e} \frac{q q _{0}}{r ^{2}} r

Divide by q₀, following the definition of E. The test charge cancels:

E = \frac{F _{e}}{q _{0}} = k_{e} \frac{q}{r ^{2}} r

Worked examples

Field near a small charge

q = 2.0 nC at r = 5.0 cm:

E = \frac{( 8.99 \times 1 0 ^{9} ) ( 2 \times 1 0 ^{- 9} )}{( 0.050 ) ^{2}} \approx 7.2 \times 1 0^{3} N/C

Field at the electron orbit in hydrogen

E = \frac{( 8.99 \times 1 0 ^{9} ) ( 1.6 \times 1 0 ^{- 19} )}{( 5.3 \times 1 0 ^{- 11} ) ^{2}} \approx 5.1 \times 1 0^{11} N/C

Air breaks down near 3 × 10⁶ N/C; atomic fields exceed that by five orders of magnitude.

Two charges: the dipole field far away

+q at +a and −q at −a, evaluated on the axis at x ≫ a:

E = k_{e} q [\frac{1}{( x - a ) ^{2}} - \frac{1}{( x + a ) ^{2}}] \approx \frac{4 k _{e} q a}{x ^{3}}

The dipole field falls as 1/x³ because the two contributions almost cancel.

Watch out for

r runs from the charge to the point you care about, not from the origin. And note this diverges as r → 0 — a sign that the point-charge idea is an idealisation.

Connects to

Coulomb's law Point charge Uniformly charged sphere