Result

Continuous distribution

(E1.5.1) E1.5 · Electric Field of a Continuous Charge Distribution →

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

Statement

The point-charge result, summed over infinitely many small pieces.

Symbols and units

$d q$	one small piece of charge	C
$λ, σ, ρ$	charge per length / area / volume	C/m, C/m², C/m³
$r$	distance from dq to the field point	m
$k_{e}$	Coulomb constant	8.99×10⁹ N·m²/C² (=1/4πε₀)
$E$	electric field at the field point	N/C

When it applies

Static charge, and r together with r̂ expressible in one integration variable.

Don't use when

Requires r and r̂ to be expressible in a single integration variable — an arbitrarily shaped distribution without symmetry may not reduce cleanly to this form.

Derivation

Each piece dq is small enough to count as a point charge, so it makes a field

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

Superposition, in the limit of infinitely many pieces, turns the sum into an integral:

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

Express dq through the charge density so you can integrate over geometry, not over charge:

d q = λ d x or σ d A or ρ d V

Worked examples

Ring of charge, on its axis

Every element lies at the same distance from the field point:

r = x^{2} + a^{2}

Perpendicular components cancel in pairs; the axial part remains:

d E_{x} = k_{e} \frac{d q}{x ^{2} + a ^{2}} \cdot \frac{x}{x ^{2} + a ^{2}}

All factors except dq are constant, and the integral of dq is Q:

E = \frac{k _{e} Q x}{( x ^{2} + a ^{2} ) ^{3/2}}

Checks: E = 0 at the centre; E → k_eQ/x² far away. The maximum lies at x = a/√2.

Watch out for

Use symmetry first: any component that cancels never needs integrating. Then write r and r̂ in the same variable as dq before you start.

Connects to

Continuous distribution Long line of charge