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UY1: Electric field of a point charge

Derive the electric field of a point charge from Coulomb's law and solve direction-sensitive field calculations.

  • University Physics Year 1
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Learning objectives

  • Construct electric-field and potential models for discrete and continuous charge distributions.
Why this matters + quick links

This page gives the UY1 working model/result for Electric field of a point charge. You reuse it when you build fields/potentials by symmetry or superposition, and when you connect fields to forces, energy, and circuits.

1) At a glance

  • Prerequisites: field definition from The Electric Field, force law from Coulomb’s Law
  • Outcomes: compute vector E(vector r) for a point charge, state direction for q > 0 vs q < 0, and use scaling checks
  • Key result:
vector E(vector r) = (1/4πε₀)q/r²r hat
  • Common trap: calculating only the magnitude E = k|q|/r² and then guessing direction (sign is already in q)
  • Inverse-square scaling: doubling r reduces | vector E| by a factor of 4.
  • Direction: away from + q, toward -q.

Motivation / intuition

This result is the single “source block” you reuse everywhere: once you know the field of one point charge, more complicated fields follow by vector addition (superposition) or by turning many charges into an integral.

2) Setup

  • Source charge q fixed at the origin.
  • Field point P is distance r from source.
  • r hat is the unit vector from source to field point.
  • Use SI units: q in C, r in m, E in N C⁻¹.

3) Core derivation/explanation

Start from field definition and Coulomb force on test charge q₀:

vector E = (vector F)/q₀, vector F = (1/4πε₀)qq₀/r²r hat

Hence:

vector E = (1/4πε₀)q/r²r hat

Important sign point:

  • The equation already includes sign through q.
  • If q < 0, vector points opposite r hat.

Inverse-square meaning: if distance doubles, field magnitude falls by factor 4.

Quick checks (units + limits/sign)
  • Units: [vector E] = N/C; from kq/r² you get (N m² C⁻²)C/m² = N/C.
  • Limits/signs: as r → ∞, E → 0; if q < 0, the vector points toward the charge (opposite r hat).

4) Worked example(s)

A charge q = +5.0 nC is at the origin. Find the field at r = 0.20 m.

E = ((8.99 × 10⁹)(5.0 × 10⁻⁹))/(0.20)² = 1.12 × 10³ N C⁻¹

Direction is radially outward from the positive source.

If the source were -5.0 nC, magnitude is unchanged, but direction is radially inward.

5) Practice set (with hints + answers)

  1. A source charge is -2.0 nC. Find field magnitude at 0.10 m. Hint: use E = k|q|/r². Answer: 1.80 × 10³ N C⁻¹.

  2. For Q1, state field direction at that point. Hint: negative sources pull field lines inward. Answer: toward the source charge.

  3. Compare fields at r and 3r from the same point charge. Hint: inverse-square scaling. Answer: E(3r) = E(r)/9.

6) Summary + next steps

  • Point-charge fields are radial and inverse-square.
  • Sign errors are avoided by writing vectors, not just magnitudes.
  • This result is the building block for dipoles and continuous charge distributions.

Next: Electric Dipole Previous: The Electric Field As A Web Back To UY1: Electromagnetism