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A Level Physics 9702 · A Level · Topic 18 · ~4 min read

Electric fields

Electric field strength, uniform fields between parallel plates, the motion of charged particles in a uniform field, Coulomb's law for point charges, and electric potential and its link to field strength.

Electric field linesField strength E = F/qUniform field E = ΔV/ΔdCharged particles between platesCoulomb's lawField of a point chargeElectric potentialPotential energy
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Paper tag = where it is examinedDATA & FORMULAE given in the examC1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • Electric field strength E = F/q: the force per unit positive charge. Its unit is N C⁻¹ = V m⁻¹.
  • Uniform field between parallel plates: E = ΔV/Δd. A charge in it feels a constant force, so it moves in a parabola (like a projectile).
  • Point charges: F = Qq/(4πε₀r²) and E = Q/(4πε₀r²). Both follow an inverse-square law.
  • Electric potential V = Q/(4πε₀r): work done per unit positive charge bringing it from infinity. Field strength is the negative potential gradient.

By the end you can

  • Understand electric fields and field lines, and define E = F/q.
  • Use E = ΔV/Δd for a uniform field and describe the motion of charged particles in it.
  • Recall and use Coulomb's law and E = Q/(4πε₀r²).
  • Define and use electric potential V = Q/(4πε₀r) and electric potential energy.

1Field strength

Definitions and uniform fields

Electric field strength — the force per unit positive charge acting on a stationary charge: E = F/q.

+positive: lines outwards+−dipole+ + + + +− − − − −parallel plates: uniform
Field lines go from + to −; closer lines mean a stronger field.
E = ΔV/Δd
uniform field between parallel plates
electronconstant force upwards (towards +): parabola, like a projectilestraight lineafter the plates
Between parallel plates the field is uniform: equally spaced parallel lines.
Worked example

Two parallel plates 2.0 cm apart have a p.d. of 500 V. Find the field strength and the force and acceleration of an electron between them.

  1. E = 500/0.020 = 2.5 × 10⁴ V m⁻¹ A1
  2. F = eE = 1.60 × 10⁻¹⁹ × 2.5 × 10⁴ = 4.0 × 10⁻¹⁵ N A1
  3. a = F/m = 4.0 × 10⁻¹⁵/9.11 × 10⁻³¹ = 4.4 × 10¹⁵ m s⁻² A1

Deflecting a beam

  • Horizontal: no force, constant velocity, so t = L/v. Vertical: constant acceleration a = eE/m, so y = ½at².
Worked example

An electron enters the field between plates of length 5.0 cm at 2.0 × 10⁷ m s⁻¹, parallel to them. The field is 1.0 × 10⁴ V m⁻¹. Find its vertical deflection on leaving.

  1. t = 0.050/2.0 × 10⁷ = 2.5 × 10⁻⁹ s C1
  2. a = eE/m = 1.76 × 10¹⁵ m s⁻² C1
  3. y = ½at² = 5.5 mm A1
Common mistakeAssuming the electron accelerates along the plates. The force is only perpendicular to the plates; its horizontal velocity is unchanged.

2Point charges

Coulomb's law

F = Qq/(4πε₀r²) ; E = Q/(4πε₀r²) ; V = Q/(4πε₀r) DATA & FORMULAE
1/(4πε₀) = 8.99 × 10⁹ N m² C⁻²

Electric potential — the work done per unit positive charge in bringing a small test charge from infinity to the point.

Worked example

Find the field strength and potential 0.10 m from a point charge of +4.0 nC.

  1. E = 8.99 × 10⁹ × 4.0 × 10⁻⁹/0.10² = 3.6 × 10³ N C⁻¹ A1
  2. V = 8.99 × 10⁹ × 4.0 × 10⁻⁹/0.10 = 360 V A1
  • Potential energy of charge q at potential V is qV = Qq/(4πε₀r). Unlike gravity, it can be positive (repulsion) or negative (attraction).
  • Field strength = −potential gradient: E = −dV/dr. Equipotentials are perpendicular to field lines.
Common mistakeSquaring r in the potential, or using the diameter instead of the radius of a charged sphere.
Linking questionThe same inverse-square ideas for masses are in Gravitational fields; storing charge is in Capacitance; and current and charge are in Electricity.

Test yourself — cover the answers first

Quick check

  1. Define electric field strength.
  2. Give two units for E.
  3. How does E vary with distance from a point charge?
  4. What shape path does an electron follow between parallel plates?
  5. Define electric potential.

Exam-style

  1. Two point charges of +2.0 μC are 0.30 m apart. Find the force between them.
  2. An oil drop of mass 3.2 × 10⁻¹⁵ kg is held stationary between plates 1.0 cm apart at 980 V. Find its charge.
  3. Find the work done moving +3.0 μC from a point at 200 V to a point at 50 V.
Show answers
  1. force per unit positive charge
  2. N C⁻¹ and V m⁻¹
  3. E ∝ 1/r²
  4. a parabola
  5. work done per unit positive charge bringing it from infinity
  6. 8.99 × 10⁹ × (2.0 × 10⁻⁶)²/0.09 = 0.40 N (repulsive)
  7. qV/d = mg ⇒ q = 3.2 × 10⁻¹⁵ × 9.81 × 0.010/980 = 3.2 × 10⁻¹⁹ C (2e)
  8. W = qΔV = 3.0 × 10⁻⁶ × (−150) = −4.5 × 10⁻⁴ J (the field does work)
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.