Uniform electric fields
Section: Electric Fields | Syllabus: Cambridge A Level Physics (9702)
Field Between Parallel Plates
A uniform electric field is created by two parallel metal plates with a potential difference (V or Δ V) across them. The field lines are parallel and equally spaced.
E = (V/d) Derivation1. Work done to move charge q between plates: W = Fd.
2. Since F = Eq, W = (Eq)d.
3. By definition, V = W/q, so V = (Eqd) / q = Ed E = V/d.
The Electron Gun
An electron gun uses a heated cathode to emit electrons via thermionic emission. These electrons are accelerated towards an anode by a potential difference V.
Gain in K.E. = (1/2)mv^2 = eV The Electronvolt (eV)One electronvolt is the energy gained by an electron when it is accelerated through a potential difference of 1 Volt.
1 eV = 1.60 × 10^-19 J
Motion of Charged Particles
When entering a field perpendicularly, a charged particle follow a parabolic path. This is because it has constant horizontal velocity (v_H) and constant vertical acceleration (a = qE/m).
Trajectory Equations- Time to cross: t = x / v_H
- Vertical deflection: y = (1/2)at^2 = (1/2)((qE/m))t^2
- Vertical velocity: v_v = at
Worked Examples
Worked Example 1: Oil Drop SuspensionQuestion: An oil drop with one excess electron is suspended between plates 2.0 cm apart at 320 V. Calculate the mass of the oil drop.
Solution1. For equilibrium, electric force = weight: Eq = mg.
2. E = V / d = 320 / 0.02 = 16000 V m^-1.
3. m = (Eq) / g = (16000 × 1.6 × 10^-19) / 9.81 = 2.6 × 10^-16 kg.
Worked Example 2: Particle DeflectionQuestion: An electron (v_H = 2.0 × 10^7 m s^-1) enters a field (E = 3000 V m^-1) of length 10 cm. Find the deflection y.
Solution1. t = x / v_H = 0.10 / (2.0 × 10^7) = 5.0 × 10^-9 s.
2. a = eE / m = (1.6 × 10^-19 × 3000) / (9.11 × 10^-31) = 5.27 × 10^14 m s^-2.
3. y = 0.5 a t^2 = 0.5 × (5.27 × 10^14) × (5.0 × 10^-9)^2 = 6.6 × 10^-3 m (or 6.6 mm).
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