Capacitors
Capacitance, series and parallel combinations, dielectrics and energy stored.
Capacitance and combinations
A capacitor is a pair of conductors separated by an insulator, used to store charge and energy. If the plates carry +Q and –Q with potential difference V, the capacitance is C = Q/V. Unit: farad (F) = C/V. Practical values are in μF (10⁻⁶ F) or pF (10⁻¹² F).
Parallel plate capacitor: C = ε₀A/d in vacuum, where A is plate area and d is separation. With a dielectric filling the gap, C = Kε₀A/d, so the capacitance increases K times.
Combinations:
- Series: 1/C = 1/C₁ + 1/C₂ + …; the same charge Q on each; the total is smaller than the smallest.
- Parallel: C = C₁ + C₂ + …; the same voltage across each.
Energy stored: U = ½CV² = Q²/(2C) = ½QV. The energy is stored in the electric field between the plates; energy density u = ½ε₀E².
Worked examples:
- A = 0.02 m², d = 1 mm: C = 8.854 × 10⁻¹² × 0.02 / 10⁻³ ≈ 1.77 × 10⁻¹⁰ F (177 pF).
- 2 μF and 3 μF in series: C = (2 × 3)/(2 + 3) = 1.2 μF; in parallel: 5 μF.
- 10 μF at 100 V: U = ½ × 10⁻⁵ × 10⁴ = 0.05 J.
Key points
- C = Q/V; C = Kε₀A/d.
- Series: reciprocals add. Parallel: values add.
- U = ½CV² = Q²/2C.
- C depends only on geometry and the medium, not on Q or V.
Exam tips
- For two capacitors in series use the shortcut C₁C₂/(C₁ + C₂).
- If a battery stays connected, V is fixed; if disconnected, Q is fixed. Decide this first in dielectric problems.
Common mistakes
- Using resistor rules for capacitors (they are reversed).
- Forgetting to convert mm to m or μF to F.
- Thinking a larger charge increases capacitance.
Practice questions
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