Uniform electric fields
Section: Electric Fields | Syllabus: Cambridge AS 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 Derivation 1.
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. = 1/2mv^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 = 1/2at^2 = 1/2(qE/m)t^2 - Vertical velocity: v_v = at Worked Examples Worked Example 1: Oil Drop Suspension Question: 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. Solution 1. 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 Deflection Question: 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. Solution 1. 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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