3.7.2.1FieldsCore

Coulomb’s Law and Electric Field Strength

Swap mass for charge in Newton’s law of gravitation and you get $Coulomb’s law$ — the electric force between two charges, following exactly the same inverse-square pattern. But unlike gravity, this force can push as well as pull, which changes everything about how electric fields behave.

40 min Video by Physeeks 3.7.2 Electric fields

Coulomb’s Law and Electric Field Strength — A Level PhysicsWatch the full walkthrough before the notes below.

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What you'll be able to do

State and use Coulomb’s law for the force between two point charges
Define electric field strength and relate it to force
Calculate electric field strength for a point charge (radial field)
Calculate electric field strength between parallel plates (uniform field)
Compare electric and gravitational fields

Coulomb’s law

$Coulomb’s law$ states that the electrostatic force between two point charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. Unlike gravity, this force can be $attractive$ (opposite charges) or $repulsive$ (like charges) — the sign of the product $Qq$ tells you which.

F = \frac{Qq}{4 π ε _{0} r ^{2}}

Coulomb’s law; ε₀ = permittivity of free space (8.85 × 10⁻¹² F m⁻¹).

1Use

F = \frac{Qq}{4 π ε _{0} r ^{2}}

F = \frac{3.0 \times 1 0 ^{- 6} \times 2.0 \times 1 0 ^{- 6}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 0.5 0 ^{2}}

3Numerator:

6.0 \times 1 0^{- 12}

. Denominator:

4 π \times 8.85 \times 1 0^{- 12} \times 0.25 \approx 2.78 \times 1 0^{- 11}

Answer

F \approx 0.216

N, attractive (opposite charges).

Tip — The 4πε₀ combination appears so often it’s worth treating as a single constant, $k = \frac{1}{4 π ε _{0}} \approx 8.99 \times 1 0^{9}$ N m² C⁻², giving the more compact form F = kQq/r².

Electric field strength — radial fields

An $electric field$ is a region where a charge experiences a force. $Electric field strength$ , $E$ , is the force per unit (positive) charge at a point, a vector pointing in the direction a small positive test charge would be pushed. Around an isolated point charge, the field is $radial$ : pointing directly away from a positive charge (or directly toward a negative one), weakening with distance according to an inverse-square law.

E = \frac{F}{Q} = \frac{Q}{4 π ε _{0} r ^{2}}

Electric field strength (radial field around a point charge Q).

1Use

E = \frac{Q}{4 π ε _{0} r ^{2}}

E = \frac{5.0 \times 1 0 ^{- 9}}{4 π \times 8.85 \times 1 0 ^{- 12} \times 0.2 0 ^{2}}

3Denominator:

4 π \times 8.85 \times 1 0^{- 12} \times 0.040 \approx 4.45 \times 1 0^{- 12}

Answer

E \approx 1.12 \times 1 0^{3}

N C⁻¹, directed away from the charge.

Electric field strength — uniform fields

Between two flat, parallel plates connected to a p.d., the field is $uniform$ : field lines are parallel, equally spaced, and point from the positive plate to the negative plate. Field strength here depends only on the p.d. across the plates and their separation, not on distance from either plate.

E = \frac{V}{d}

Uniform field between parallel plates (V = p.d. between plates, d = separation).

1Convert separation to metres:

d = 5.0 \times 1 0^{- 3}

2Use

E = \frac{V}{d} = \frac{300}{5.0 \times 1 0 ^{- 3}}

Answer

E = 6.0 \times 1 0^{4}

V m⁻¹

Tip — A uniform field has the same strength everywhere between the plates — no inverse-square weakening — which is why parallel-plate setups are used whenever a constant force on a charge is needed, e.g. deflecting a beam of charged particles.

Comparing electric and gravitational fields

Electric and gravitational fields share the same mathematical form (both inverse-square for a point source), but differ in two crucial ways: gravitational force is always attractive, while electric force can be attractive or repulsive; and while there is only one "sign" of mass, charge comes in two signs, so electric fields can be created, cancelled and shielded in ways gravitational fields cannot.

Tip — A helpful analogy for translating between the two topics: charge ↔ mass, electric field strength E ↔ gravitational field strength g, Coulomb’s constant 1/(4πε₀) ↔ Newton’s constant G — but always remember the sign difference.

Equation recap

F = \frac{Qq}{4 π ε _{0} r ^{2}}

Coulomb’s law.

E = \frac{F}{Q} = \frac{Q}{4 π ε _{0} r ^{2}}

Electric field strength (radial, point charge).

E = \frac{V}{d}

Electric field strength (uniform, parallel plates).

Common mistakes to avoid

Forgetting electric force can be repulsive as well as attractive.

Like charges repel, unlike charges attract — always check the signs of both charges before deciding the direction of the force.

Using E = Q/(4πε₀r²) for the field between parallel plates.

That formula is for a RADIAL field around a point charge. Between parallel plates, use the uniform-field formula E = V/d instead.

Thinking electric field strength weakens with distance between parallel plates, like it does around a point charge.

A uniform field (parallel plates) has the same strength everywhere between the plates — there is no distance-dependence, unlike a radial field.

Key takeaways

Coulomb’s law: F = Qq/(4πε₀r²) — attractive for opposite charges, repulsive for like charges.
Electric field strength E = F/Q; around a point charge, E = Q/(4πε₀r²) (radial, inverse-square).
Between parallel plates, the field is uniform: E = V/d, independent of position between the plates.
Electric and gravitational fields share the same inverse-square mathematical form, but electric force can be attractive or repulsive while gravity is always attractive.

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Coulomb’s Law and Electric Field Strength

Swap mass for charge in Newton’s law of gravitation and you get

Coulomb’s law

— the electric force between two charges, following exactly the same inverse-square pattern. But unlike gravity, this force can push as well as pull, which changes everything about how electric fields behave.

40 min Video by Physeeks 3.7.2 Electric fields