define capacitance, as applied to both isolated spherical conductors and to parallel plate capacitors

Published by Patrick Mutisya · 14 days ago

Cambridge A-Level Physics 9702 – Capacitors and Capacitance

Capacitors and Capacitance

Learning Objective

Define capacitance and apply the definition to both isolated spherical conductors and parallel‑plate capacitors.

What is Capacitance?

Capacitance, \$C\$, is the ability of a system to store electric charge per unit potential difference. It is defined mathematically as

\$C = \frac{Q}{V}\$

where \$Q\$ is the magnitude of charge on one conductor and \$V\$ is the potential difference between the conductors (or between a conductor and infinity for an isolated body).

Isolated Spherical Conductor

An isolated conducting sphere of radius \$R\$ carries a charge \$Q\$. The potential of the sphere relative to infinity is

\$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}\$

Substituting into the definition of capacitance gives

\$C{\text{sphere}} = \frac{Q}{V} = 4\pi\varepsilon0 R\$

Key points:

  • The capacitance is directly proportional to the radius of the sphere.

  • It depends only on geometry and the permittivity of free space, \$\varepsilon_0\$.

Suggested diagram: An isolated spherical conductor of radius \$R\$ with charge \$+Q\$ on its surface and the electric field lines radiating outward.

Parallel‑Plate Capacitor

A parallel‑plate capacitor consists of two large, flat conducting plates of area \$A\$, separated by a small distance \$d\$ (with \$d \ll \sqrt{A}\$). The space between them may be filled with a dielectric of relative permittivity \$\kappa\$ (for vacuum or air, \$\kappa = 1\$).

The electric field between the plates is approximately uniform:

\$E = \frac{V}{d}\$

The charge on each plate is \$Q = \varepsilon_0 \kappa A E\$, so combining the expressions gives the capacitance:

\$C{\text{parallel}} = \frac{Q}{V} = \varepsilon0 \kappa \frac{A}{d}\$

Important observations:

  • Capacitance increases with larger plate area \$A\$.

  • Capacitance decreases as the plate separation \$d\$ increases.

  • Introducing a dielectric ( \$\kappa > 1\$ ) multiplies the capacitance by \$\kappa\$.

Suggested diagram: Two parallel plates of area \$A\$ separated by distance \$d\$, showing uniform electric field lines and the dielectric (if present).

Comparison of the Two Configurations

ConfigurationCapacitance FormulaKey Dependence
Isolated spherical conductor\$C = 4\pi\varepsilon_0 R\$Proportional to radius \$R\$; independent of surrounding medium (vacuum).
Parallel‑plate capacitor\$C = \varepsilon_0 \kappa \dfrac{A}{d}\$Proportional to plate area \$A\$, inversely proportional to separation \$d\$, multiplied by dielectric constant \$\kappa\$.

Worked Example: Spherical Conductor

  1. Given a metal sphere of radius \$R = 5\ \text{cm}\$, calculate its capacitance.

  2. Use \$\varepsilon_0 = 8.85\times10^{-12}\ \text{F m}^{-1}\$.

  3. \$C = 4\pi (8.85\times10^{-12}\ \text{F m}^{-1})(0.05\ \text{m}) \approx 5.6\times10^{-12}\ \text{F} = 5.6\ \text{pF}\$

Worked Example: Parallel‑Plate Capacitor

  1. Two plates each of area \$A = 0.02\ \text{m}^2\$ are separated by \$d = 1\ \text{mm}\$ and filled with a dielectric of \$\kappa = 2.5\$.

  2. \$C = (8.85\times10^{-12}\ \text{F m}^{-1})(2.5)\frac{0.02\ \text{m}^2}{1\times10^{-3}\ \text{m}} \approx 4.4\times10^{-10}\ \text{F} = 440\ \text{pF}\$

Key Take‑aways

  • Capacitance is a geometric property; it does not depend on the amount of charge stored.

  • For isolated conductors, the reference point is infinity; for capacitors, the reference is the opposite plate.

  • Understanding how \$C\$ varies with size, separation, and dielectric material is essential for designing circuits and selecting components.