Electric Field from Coulomb's Law

Derivation

Electric Field from Coulomb's Law

D-039 Home PU-102 Threads force · fields Depends on superposition-of-forces

Statement

The electrostatic field E(r) at a field point r is defined as the electric force on a stationary test charge q₀ placed there, divided by q₀, in the limit q₀ → 0. Starting from Coulomb's law and the superposition of forces, this yields for a source distribution the field of a point charge E = (1/4πε₀) q r̂/r², and, by superposition, the volume integral E(r) = (1/4πε₀) ∫ ρ(r′) (r−r′)/|r−r′|³ d³r′ over a continuous charge density.

Why it matters

The field concept replaces action-at-a-distance with a local quantity defined at every point of space, whether or not a test charge is present. Once E(r) is known, the force on any charge q is simply F = qE, decoupling "what the sources do" from "what a probe feels."

This is the entry point to all of electrostatics: Gauss's law, the potential, boundary-value problems, and ultimately Maxwell's equations are all statements about the field E rather than about pairwise forces. The superposition integral derived here is the direct, brute-force solution that those later tools are designed to circumvent.

Assumptions

Coulomb's law holds for each source pair.If the inverse-square, charge-proportional force law failed, neither the point-charge field nor the integral would take this form; empirically the exponent is 2 to better than one part in 10¹⁵.

Forces superpose linearly (prior result).If the force from one source depended on the presence of others, we could not sum contributions independently and no single field could be assigned to the configuration.

All charges are static.Moving charges radiate and produce magnetic forces; dropping this brings in the full Maxwell dynamics and retardation, and E would no longer be purely radial from the instantaneous positions.

The test charge is vanishingly small.A finite q₀ polarises or displaces the sources (e.g. on a conductor), altering the very field it measures; the q₀ → 0 limit removes this back-reaction.

The charge density ρ is a well-behaved (integrable) function.If ρ contains non-integrable singularities the volume integral need not converge; idealised point and surface charges require separate treatment as limits or distributions.

Derivation

F₀ = (1/4πε₀) q q₀ r̂/r²

Coulomb's law for the force on test charge q₀ at displacement r = r r̂ from a source q at the origin. A

E(r) ≡ lim_q₀→0 F₀/q₀

Definition of the field: force per unit test charge, taking q₀ → 0 so the probe does not perturb the source. Symbolic division precedes any numbers. A

E(r) = (1/4πε₀) q r̂/r²

Substituting step 1 into step 2, the factor q₀ cancels exactly and the limit is trivial (nothing else depends on q₀). Field of a single point charge. A

E(r) = (1/4πε₀) q (r−r′)/|r−r′|³

Placing the source at a general point r′ rather than the origin: the separation vector becomes r−r′ with unit vector (r−r′)/|r−r′|, so r̂/r² = (r−r′)/|r−r′|³. B

E(r) = ∑_i (1/4πε₀) q_i (r−r_i)/|r−r_i|³

Superposition of forces (prior result), divided through by q₀: the total field is the vector sum of the point-charge fields of N discrete sources. B

q_i → ρ(r′) d³r′, ∑_i → ∫_V

Continuum limit: partition the source region into cells of volume d³r′ carrying charge ρ(r′) d³r′. As cell size → 0 the Riemann sum converges to an integral provided ρ is integrable. C

E(r) = (1/4πε₀) ∫_V ρ(r′) (r−r′)/|r−r′|³ d³r′

Result of the limit in step 6. The integrand is integrable even at r′ = r: the 1/|r−r′|² singularity is beaten by the d³r′ ∼ s²ds volume element in three dimensions. C

Result

E(r) = (1/4πε₀) ∫_V ρ(r′) (r−r′)/|r−r′|³ d³r′

Reading. Every source element ρ(r′)d³r′ contributes a field pointing along the separation vector r−r′ from source to field point, falling as the inverse square of distance; the total field is the vector sum (integral) of these contributions. The field exists at r whether or not any charge sits there. For a single point charge the integral collapses to (1/4πε₀)qr̂/r².

Units check. [ρ] = C m⁻³, [(r−r′)/|r−r′|³] = m⁻², [d³r′] = m³, and [1/4πε₀] = N m² C⁻². Product: (N m² C⁻²)(C m⁻³)(m⁻²)(m³) = N C⁻¹ = V m⁻¹, the correct units of electric field.

Limiting cases

Single point charge: ρ(r′) = q δ³(r′−r₀) recovers E = (1/4πε₀)q(r−r₀)/|r−r₀|³.
Far field of any bounded neutral-plus distribution: at |r| ≫ source size, E → (1/4πε₀)Q_totr̂/r², i.e. the distribution looks like a point charge Q_tot = ∫ρ d³r′.
Field at a symmetry point: where the distribution is symmetric under a reflection or rotation that reverses E, the integral vanishes (e.g. centre of a uniform sphere or ring).
Surface / line charge: take ρd³r′ → σdA′ or λdℓ′ as one or two dimensions collapse, giving the corresponding lower-dimensional integrals.

Breaks when

Charges move appreciably. For time-varying or fast sources the instantaneous Coulomb form fails; the field lags by the retarded time and acquires induction and radiation terms (Jefimenko / Liénard–Wiechert). The magnetic force qv×B also appears, so F = qE alone is incomplete.
Inside polarisable or conducting matter. The bare integral over free charge omits bound charge; one must use the full ρ including polarisation charge, or switch to D and ε. A finite test charge on a conductor also induces surface charge, violating the q₀→0 idealisation.
At the location of an idealised point charge. The self-field diverges; E is infinite at r = r₀ and the self-energy integral diverges, signalling the breakdown of the point-charge idealisation (needs QED / classical electron-radius cutoffs).
Relativistic or strong-field regimes. Near strong sources the vacuum is nonlinear (Euler–Heisenberg) and superposition itself fails at fields approaching the Schwinger limit ∼10¹⁸ V m⁻¹.

Failure modes

Cubing vs squaring. Writing the vector form with |r−r′|² in the denominator instead of the cube: the extra power of |r−r′| is what converts the unit vector into (r−r′) numerator; using square-with-vector-numerator gives a 1/r field, the wrong law.
Integrating over the field point. Integrating d³r instead of d³r′ — the primed variable ranges over sources; r is held fixed as the observation point.
Adding magnitudes not vectors. Summing |E_i| scalars instead of vector components; opposite contributions must cancel component-wise.
Keeping q₀ in the answer. Forgetting that the test charge cancels; the field is a property of the sources, independent of any probe.
Dropping the source's own displacement. Using r for the separation when the source is not at the origin, rather than r−r′.
Assuming E points radially from the origin. For an extended body E points along the local separation from each element, not from the coordinate origin.

Discussion

The definition E = lim_q₀→0 F/q₀ is doing real conceptual work: it strips the probe out of the description and assigns a vector to every point of empty space. This is the birth of the field as an independent physical entity — it carries energy (density ε₀E²/2), momentum, and angular momentum, and in the dynamical theory it propagates at c. Coulomb's law is merely the electrostatic shadow of this richer object.

The superposition integral is exact but often impractical: it is a vector integral with a directional kernel, and the (r−r′) numerator makes the components couple awkwardly. This is precisely why one usually detours through the scalar potential V(r) = (1/4πε₀)∫ρ/|r−r′| d³r′ and takes E = −∇V — a scalar integral is far easier — or exploits symmetry through Gauss's law.

The result is intimately tied to three dimensions. The 1/r² law is what makes ∇·E = ρ/ε₀ local (Gauss) and ∇×E = 0 (conservative), because (r−r′)/|r−r′|³ = −∇(1/|r−r′|) is a gradient with divergence 4πδ³. The whole differential structure of electrostatics is latent in this one integrand.

Rigorously, the field of a volume distribution is continuous everywhere, including inside the charge — the apparent 1/|r−r′|² singularity of the integrand is integrable because d³r′ = s² ds dΩ supplies two compensating powers of the separation s. What is not continuous is the normal component across a surface charge, which jumps by σ/ε₀; this discontinuity is the seed of the boundary conditions and, in the point limit, of the divergent self-energy that classical field theory cannot resolve.

Common misconceptions. The field is not "produced by" the test charge, nor does it require one to exist — the test charge only reveals it. And E is not the force: it is force per unit charge, with units V m⁻¹ = N C⁻¹, so a given field exerts different forces on different charges.

Worked examples

Field on the axis of a uniformly charged ring. Ring radius a, total charge Q, field at axial distance z from centre.

Set up: by symmetry the transverse components cancel, leaving only the axial (z) component. B

dE_z = (1/4πε₀) (dq/(a²+z²)) cosθ, cosθ = z/√(a²+z²)

Each element is at the same distance √(a²+z²); project onto the axis with cosθ. Symbols first. B

E_z = (1/4πε₀) z/(a²+z²)^3/2 ∫ dq = (1/4πε₀) Qz/(a²+z²)^3/2

The z-dependent factors are constant over the ring, so they come out of the integral; ∫dq = Q. B

Q = 2.0×10⁻⁸ C, a = 0.10 m, z = 0.20 m

Insert numbers. a²+z² = 0.01+0.04 = 0.05 m²; (0.05)^3/2 = 0.01118 m³. A

E_z = (8.988×10⁹)(2.0×10⁻⁸)(0.20)/0.01118 ≈ 3.2×10³ V m⁻¹

Reading. The field points away from the ring along the axis (for Q>0), magnitude ≈ 3.2 kV m⁻¹ directed in +z. It vanishes at the centre (z=0) and falls to the point-charge form for z≫a.

Net field from two point charges. q₁ = +3.0 nC at the origin, q₂ = −3.0 nC at (0.40 m, 0). Find E at P = (0.20 m, 0.20 m).

Setup: superpose the two point-charge fields as vectors (steps 3–5 of the derivation). A

r₁ = (0.20, 0.20), |r₁| = 0.283 m; r₂ = (−0.20, 0.20), |r₂| = 0.283 m

Separation vectors from each source to P, and their magnitudes (√0.08 = 0.283 m). A

E₁ = (1/4πε₀) q₁r₁/|r₁|³; |E₁| = (8.988×10⁹)(3.0×10⁻⁹)/0.08 = 337 V m⁻¹

Magnitude of each field; |r|² = 0.08 m². Directions: E₁ along +(0.20,0.20)/0.283, E₂ toward q₂ along −(−0.20,0.20)/0.283. B

E₁ = 337(0.707, 0.707) = (238, 238); E₂ = 337(0.707, −0.707) = (238, −238)

E₂ points from P toward the negative charge, i.e. along (0.20,−0.20)/0.283. Add components. B

E = E₁+E₂ = (476, 0) V m⁻¹, |E| ≈ 4.8×10² V m⁻¹ along +x

Reading. The y-components cancel and the x-components add, giving a field of about 480 V m⁻¹ pointing in +x — from the positive toward the negative charge, as expected for a dipole-like pair.

Problems

(A) A point charge q = 5.0 nC sits at the origin. Find the magnitude of E at r = 0.30 m.
Solution
E = (1/4πε₀)q/r² = (8.988×10⁹)(5.0×10⁻⁹)/(0.30)² = 44.94/0.09 = 4.99×10² V m⁻¹, directed radially outward.
(A) Two charges +2.0 nC and +2.0 nC lie on the x-axis at x = ±0.10 m. Find E at the origin.
Solution
By symmetry the two fields are equal in magnitude and opposite in direction along x, so they cancel: E = 0. (The origin is an unstable equilibrium point for a like test charge.)
(B) Find E on the axis of the ring in Worked Example 1 (Q = 2.0×10⁻⁸ C, a = 0.10 m) at the far distance z = 1.0 m, and compare with the point-charge value.
Solution
a²+z² = 0.01+1.00 = 1.01; (1.01)^3/2 = 1.015. E_z = (8.988×10⁹)(2.0×10⁻⁸)(1.0)/1.015 = 179.8/1.015 = 1.77×10² V m⁻¹. Point charge: kQ/z² = 179.8 V m⁻¹. They agree to ≈1.5%, confirming the ring looks point-like at z≫a.
(B) An electric dipole consists of +q and −q separated by d along z. Show that on the perpendicular bisector at distance r≫d the field magnitude is E ≈ (1/4πε₀)p/r³ with p = qd, and give the direction.
Solution
On the bisector each charge is at distance √(r²+(d/2)²). The r-components cancel; the z-components add. Each z-component is kq/(r²+d²/4) × (d/2)/√(r²+d²/4). Summing the two: E = kq d/(r²+d²/4)^3/2. For r≫d: E ≈ kqd/r³ = kp/r³, directed antiparallel to the dipole moment p (from + to −), i.e. in −z.
(C) A thin rod of length L carries uniform linear density λ. Find E at a point on its axis, a distance d beyond the near end. Evaluate for λ = 4.0×10⁻⁸ C m⁻¹, L = 0.50 m, d = 0.10 m.
Solution
Place the near end at distance d, element at distance x from the field point (d ≤ x ≤ d+L), charge dq = λ dx. E = kλ ∫_d^d+L dx/x² = kλ[1/d − 1/(d+L)] = kλL/[d(d+L)], along the axis away from the rod. Numbers: kλ = (8.988×10⁹)(4.0×10⁻⁸) = 359.5; L/[d(d+L)] = 0.50/[(0.10)(0.60)] = 0.50/0.06 = 8.333 m⁻¹. E = 359.5×8.333 ≈ 3.0×10³ V m⁻¹, pointing away from the rod along its axis.