Capacitors and capacitance
Section: Capacitance | Syllabus: Cambridge AS 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. Figure 19.1: Capacitor Construction & Symbol 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 Definition Capacitance 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 = ε_0 ε_r A/d Where A is the area of overlap of the plates, d is the separation, ε_0 is the permittivity of free space, and ε_r is the relative permittivity of the dielectric.
Key Point The 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/4πε_0 R 2. Since C = Q / V: C = 4πε_0 R Worked Examples Worked Example: Basic Calculations 1. Find P.D. for 1.2~μF storing 10~μC: V = Q / C = (10 × 10^-6) / (1.2 × 10^-6) = 8.33 V.
2. Find max charge for 10~μF with 12 V rating: Q = CV = (10 × 10^-6) × 12 = 1.2 × 10^-4 C (120~μC). Worked Example: Capacitance of the Earth Question: Determine the capacitance of the Earth (R = 6371 km, ε_0 = 8.85 × 10^-12 F m^-1).
Solution C = 4π × (8.85 × 10^-12) × (6371 × 10^3) C 7.1 × 10^-4 F (710 ~μF).
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