Capacitors and capacitance
Section: Capacitance | Syllabus: Cambridge A Level Physics (9702)
The Parallel Plate Capacitor
A capacitor is a device designed to store electric charge and electric potential energy. A parallel plate capacitor consists of two metal plates separated by an insulator known as a dielectric.
- Dielectric Materials: Polyester, polycarbonate, polythene, mica, or air.
- Construction: To save space, many capacitors use metal foil plates rolled up with a thin dielectric into a cylinder.
The electrical symbol consists of two parallel lines of equal length.
Charge Storage
When a capacitor is connected to a cell:
- Electrons flow from the negative terminal onto one plate, making it negatively charged ( -Q ).
- Electrons flow off the other plate towards the positive terminal, leaving it positively charged ( +Q ).
- The dielectric prevents electrons from passing through the capacitor gap.
- Flow stops when the potential difference V across the capacitor equals the e.m.f. of the cell.
The overall charge is zero (+Q + (-Q) = 0), but we say the capacitor "stores charge Q".
Defining Capacitance (C)
Charge Q is directly proportional to potential difference V. The constant of proportionality is capacitance.
C = (Q/V) Examiner DefinitionCapacitance is the charge stored per unit potential difference (P.D.).
Unit: Farad (F). 1 F = 1 C V^-1.
Tolerance: Capacitors often have a tolerance (percentage uncertainty) of ± 10\% to ± 20\%.
Factors Affecting Capacitance
For a parallel-plate capacitor, the capacitance depends on the physical dimensions and the dielectric material:
C = (_0 _r A/d)Where A is the area of overlap of the plates, d is the separation, _0 is the permittivity of free space, and _r is the relative permittivity of the dielectric.
Key PointThe capacitance of a parallel-plate capacitor increases with the area of the metal plates and decreases with their separation.
Capacitance of an Isolated Conductor
Any isolated conductor can store charge. For an isolated conducting sphere of radius R:
1. Potential at surface V = (Q/4π_0 R)
2. Since C = Q / V:
C = 4π_0 RWorked Examples
Worked Example: Basic Calculations1. Find P.D. for 1.2~ storing 10~:
V = Q / C = (10 × 10^-6) / (1.2 × 10^-6) = 8.33 V.
2. Find max charge for 10~ with 12 V rating:
Q = CV = (10 × 10^-6) × 12 = 1.2 × 10^-4 C (120~).
Question: Determine the capacitance of the Earth (R = 6371 km, _0 = 8.85 × 10^-12 F m^-1).
SolutionC = 4π × (8.85 × 10^-12) × (6371 × 10^3)
C ≈ 7.1 × 10^-4 F (710 ~).
Interactive revision notes, videos and practice questions load below.