Coulomb's law and electric field intensity

Why it matters

Coulomb's law is the experimental starting point of electrostatics: every capacitance, every electrostatic sensor and every insulation calculation ultimately rests on it. The electric field intensity E that follows from it lets you describe the effect of a charge distribution at a point without knowing what test charge will be placed there, which is how field problems are set up in the rest of the subject.

Key ideas

Coulomb's law. Two point charges Q1 and Q2 separated by a distance R in a homogeneous medium exert equal and opposite forces on each other, directed along the line joining them. The force is proportional to the product of the charges and inversely proportional to R². Like charges repel; unlike charges attract.

Assumptions and limits. The charges must be point charges (sizes much smaller than R), stationary, and in a linear, homogeneous, isotropic medium. In a medium of relative permittivity εr the force is reduced by the factor εr compared with free space.

Vector form. Write the force on Q2 as F12 = Q1Q2 (r2 − r1) / (4πε |r2 − r1|³). The vector R12 = r2 − r1 points from Q1 to Q2; with this form the sign of the product Q1Q2 automatically gives repulsion (positive) or attraction (negative). Do not combine a magnitude formula with a guessed direction.

Superposition. Electrostatic forces and fields are linear: the field of many charges is the vector sum of the fields each would produce alone. For continuous distributions, replace Q by ρL dl (line), ρS dS (surface) or ρv dv (volume) and integrate — always as a vector, resolving into constant unit vectors (Cartesian) before integrating.

Electric field intensity. E is the force per unit positive test charge: E = F / q as q → 0 (so the test charge does not disturb the source). Unit: N/C, which equals V/m. For a point charge E points radially away from a positive charge and towards a negative one, and falls as 1/r².

Standard results obtained by integrating Coulomb's law:

• Infinite uniform line charge: E falls as 1/ρ and points radially from the line.
• Infinite uniform sheet of charge: E is uniform, ρS/(2ε), independent of distance, normal to the sheet.
• On the axis of a ring of charge: E rises from zero at the centre to a maximum and then falls.

Field lines start on positive charges and end on negative charges (or at infinity); their density shows field strength; they never cross because E has one direction at each point; and they are tangent to the force on a positive test charge — not necessarily its trajectory.

Formulas

• F = Q1Q2 / (4πε₀εr R²) — magnitude of force (N); Q in coulombs (C), R in metres (m); ε₀ = 8.854 × 10⁻¹² F/m; εr = relative permittivity (1 for vacuum, about 1 for air).
• k = 1 / (4πε₀) ≈ 8.988 × 10⁹ N·m²/C² (often rounded to 9 × 10⁹).
• F12 = Q1Q2 (r2 − r1) / (4πε|r2 − r1|³) — vector force on Q2 due to Q1 (N), ε = ε₀εr.
• E = F / q — field intensity (N/C or V/m).
• E = Q / (4πεR²) a_R — field of a point charge; a_R is the unit vector from the charge to the field point.
• E = Σ Qk (r − rk) / (4πε|r − rk|³) — superposition for several point charges.
• E = ρL / (2περ) a_ρ — infinite line charge, ρL in C/m, ρ = perpendicular distance (m).
• E = ρS / (2ε) a_n — infinite sheet, ρS in C/m², a_n normal pointing away from the sheet.
• E = ρL a h / (2ε (h² + a²)^(3/2)) a_z — on the axis of a ring of radius a at height h.

Worked examples

Example 1 (standard). Charges Q1 = 2 μC and Q2 = −3 μC are 0.5 m apart in vacuum. Find the force between them and the field due to Q1 alone at a point 0.5 m from it.

1. F = k |Q1Q2| / R² = 8.988 × 10⁹ × (2 × 10⁻⁶ × 3 × 10⁻⁶) / (0.5)² = 8.988 × 10⁹ × 6 × 10⁻¹² / 0.25.
2. F = 0.05393 / 0.25 = 0.216 N, attractive (the charges have opposite signs).
3. E = k |Q1| / R² = 8.988 × 10⁹ × 2 × 10⁻⁶ / 0.25 = 71.9 kV/m (= 71.9 kN/C), pointing away from Q1. Note the scale: microcoulomb charges a fraction of a metre apart give forces of a fraction of a newton.

Example 2 (GATE level, vector superposition). Q1 = 1 nC at the origin and Q2 = −2 nC at (3, 0, 0) m, in free space. Find E at P(0, 4, 0) m.

1. From Q1: R1 = (0, 4, 0), |R1| = 4 m. E1 = kQ1 R1 / |R1|³ = 8.988 × 10⁹ × 10⁻⁹ × 4 / 64 a_y = 0.562 a_y V/m.
2. From Q2: R2 = P − (3, 0, 0) = (−3, 4, 0), |R2| = 5 m. E2 = kQ2 R2 / |R2|³ = 8.988 × 10⁹ × (−2 × 10⁻⁹) × (−3a_x + 4a_y) / 125 = 0.431 a_x − 0.575 a_y V/m.
3. Total: E = E1 + E2 = 0.431 a_x − 0.0135 a_y V/m, magnitude √(0.431² + 0.0135²) = 0.432 V/m. The y-components nearly cancel; a magnitude-only approach would have missed this completely.

Example 3 (GATE level, null point). +4 μC sits at x = 0 and −1 μC at x = 1 m on the x-axis. Where (other than at infinity) is E = 0?

1. Between the charges the two fields both point in +x (away from + and towards −), so they cannot cancel there.
2. The null must lie outside, nearer the smaller charge: x > 1 m. Equate magnitudes: 4/x² = 1/(x − 1)².
3. Taking square roots: 2(x − 1) = x, so x = 2 m. Check: k·4 × 10⁻⁶/4 = k·1 × 10⁻⁶/1. ✓

Common mistakes

• Forgetting the μ or n prefix (10⁻⁶, 10⁻⁹) — the most common source of answers off by 10⁶ or 10⁹.
• Adding field magnitudes instead of vectors. Fields from charges in different directions must be resolved into components.
• At the midpoint of +Q and −Q the fields add (both point towards −Q); they cancel only for two equal charges of the same sign.
• Using the magnitude formula and then guessing the direction; use the vector form R = r_field − r_source.
• Writing the medium's effect as kQ/(εR²) — the correct reduction is by εr: E = Q/(4πε₀εrR²).
• Squaring R in the vector form only once: the denominator is |R|³ because R itself is in the numerator.

For GATE IN

Typical items: force or field from two or three point charges (often at vertices of a square or triangle) requiring vector addition; location of a null point on a line; field of a line or sheet charge at a given distance; the effect of a dielectric medium. NAT answers are frequently asked in kV/m or mN, so watch prefixes. Practise setting up R = r − r′ quickly and checking symmetry to kill components before calculating.

Quick check

1. What happens to the force between two charges if their separation is halved?
2. What is the SI unit of E, apart from N/C?
3. Field of +Q and −Q at their midpoint: zero or not?
4. How does the field of an infinite line charge vary with distance?
5. What is the field of an infinite sheet with ρS = 8.854 nC/m² in free space?

Answers: 1. It becomes four times larger. 2. V/m. 3. Not zero — the two fields add, pointing towards −Q. 4. As 1/ρ. 5. 500 V/m.