ZIMSEC A Level · Physics · Form 6
Electric fields
- Coulomb's law: F = Qq/(4πε₀r²) (ε₀ = 8.85 × 10⁻¹² F m⁻¹).
- Field strength E = F/q; for a point charge E = Q/(4πε₀r²); between parallel plates E = V/d (uniform field).
- Electric potential V = Q/(4πε₀r); work done moving a charge = qΔV.
Worked example: Two plates 5.0 mm apart have a potential difference of 2 000 V. Find the force on an electron between them (e = 1.6 × 10⁻¹⁹ C).
E = 2 000 ÷ 0.0050 = 4.0 × 10⁵ V m⁻¹; F = eE = 1.6 × 10⁻¹⁹ × 4.0 × 10⁵ = 6.4 × 10⁻¹⁴ N
Charged particles in fields
An electron moving across a uniform field follows a parabolic path, like a projectile.
Capacitance
- C = Q/V (farad). Energy stored W = ½QV = ½CV².
- Capacitors in parallel: C = C₁ + C₂; in series: 1/C = 1/C₁ + 1/C₂.
- Discharge through a resistor: Q = Q₀e^(−t/RC); the time constant τ = RC is the time for charge to fall to about 37% of its initial value.
Worked example: A 470 µF capacitor is charged to 12 V. Find the energy stored.
W = ½ × 470 × 10⁻⁶ × 144 = 0.0338 J
Exam tip: Remember µF = 10⁻⁶ F – a frequent source of power-of-ten errors.
Check yourself
- Find the electric field 0.10 m from a charge of 2.0 × 10⁻⁹ C.
- Two 10 µF capacitors are connected in series. Find the total capacitance.
- A 100 µF capacitor discharges through a 10 kΩ resistor. Find the time constant.
Show answers
- E = 2.0 × 10⁻⁹ ÷ (4π × 8.85 × 10⁻¹² × 0.01) ≈ 1.8 × 10³ V m⁻¹
- 5 µF
- τ = RC = 10⁴ × 10⁻⁴ = 1.0 s
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