Review / Chapter 3 — Conductors

Capacitance

Chapter 3 — Conductors

For a conductor, charge and potential are linearly related: doubling the charge doubles the field everywhere, which doubles the potential. That linear ratio is the capacitance — a property of geometry alone.

Definitions

For an isolated conductor at potential \(V\) (with respect to infinity) carrying charge \(Q\):

\[C = \frac{Q}{V}.\]

For a pair of conductors (a "capacitor") carrying \(+Q\) and \(-Q\) with potential difference \(\Delta V = V_+ - V_-\):

\[C = \frac{|Q|}{\Delta V}.\]

Units: \(1\,\mathrm{farad} = 1\,\mathrm{C/V}\). A farad is huge — most real capacitors are pF to mF.

Capacitance depends only on geometry

The linearity of Maxwell's equations in vacuum means if \(Q\) doubles, all fields and potentials double. So \(Q/V\) is independent of \(Q\). It's a geometric quantity.

Isolated sphere

A conducting sphere of radius \(R\) carrying charge \(Q\) has potential \(V = kQ/R = Q/(4\pi\epsilon_0 R)\). Hence

\[C_\text{sphere} = \frac{Q}{V} = 4\pi\epsilon_0 R.\]

Numerically: \(R = 1\,\mathrm{m}\) gives \(C \approx 111\,\mathrm{pF}\). To store 1 coulomb you'd need \(V \approx 9\times 10^9\,\mathrm{V}\). Farads are big.

Concentric spheres

Inner shell radius \(R_1\) with charge \(+Q\), outer shell \(R_2\) with \(-Q\). Between them \(E = kQ/r^2\), so

\[\Delta V = \int_{R_1}^{R_2} \frac{kQ}{r^2}\,dr = kQ\left(\frac{1}{R_1} - \frac{1}{R_2}\right)\]

and

\[C = \frac{Q}{\Delta V} = 4\pi\epsilon_0\,\frac{R_1 R_2}{R_2 - R_1}.\]

Limit \(R_2\to\infty\): \(C\to 4\pi\epsilon_0 R_1\), recovering the isolated-sphere answer.

Long coaxial cable

Inner cylinder radius \(a\), outer radius \(b\), length \(L\). Using the line-charge field \(E = \lambda/(2\pi\epsilon_0 r)\) with \(\lambda = Q/L\):

\[\Delta V = \frac{Q}{2\pi\epsilon_0 L}\ln(b/a)\quad\Rightarrow\quad C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}.\]

General recipe

  1. Place charge \(\pm Q\) on the two conductors (or \(Q\) on the isolated one).
  2. Find \(\vec E\) between them using Gauss or superposition.
  3. Integrate to get \(\Delta V\).
  4. Compute \(C = Q/\Delta V\). The \(Q\) cancels out, leaving only geometry.

Practice Problems

Problem 1easy
A conducting sphere with radius \(R = 5\,\mathrm{cm}\) is held at potential \(V = 1000\,\mathrm{V}\). Find the charge on it and its self-capacitance.
Hint
\(C = 4\pi\epsilon_0 R\), \(Q = CV\).
Solution

\(C = 4\pi\epsilon_0(0.05) = 5.56\,\mathrm{pF}\). \(Q = CV = (5.56\times 10^{-12})(1000) = 5.56\,\mathrm{nC}\).

Answer: \(C \approx 5.6\,\mathrm{pF}\), \(Q \approx 5.6\,\mathrm{nC}\).

Problem 2medium
Two concentric conducting shells have radii \(R_1\) and \(R_2 > R_1\), with charge \(+Q\) on the inner and \(-Q\) on the outer (the outer shell is grounded). Find the capacitance and check the limit \(R_2\to\infty\).
Hint
Field between shells is purely \(kQ/r^2\).
Solution

\(V_\text{inner} = kQ(1/R_1 - 1/R_2)\) (outer shell is grounded at \(V=0\)). So

\[C = \frac{Q}{V_\text{inner}} = 4\pi\epsilon_0\frac{R_1 R_2}{R_2 - R_1}.\]

Limit \(R_2\to\infty\): \(C\to 4\pi\epsilon_0 R_1\), recovering the isolated sphere. ✓

Answer: \(C = 4\pi\epsilon_0 R_1 R_2/(R_2 - R_1)\).

Problem 3medium
A coaxial cable has inner radius \(a = 1\,\mathrm{mm}\), outer radius \(b = 5\,\mathrm{mm}\). Find the capacitance per meter.
Hint
\(C/L = 2\pi\epsilon_0/\ln(b/a)\).
Solution

\(\ln(5/1) = \ln 5 \approx 1.609\). \(C/L = 2\pi(8.85\times 10^{-12})/1.609 \approx 3.5\times 10^{-11}\,\mathrm{F/m} = 35\,\mathrm{pF/m}\).

Answer: \(\approx 35\,\mathrm{pF/m}\).

Problem 4medium
Estimate Earth's self-capacitance (\(R \approx 6.4\times 10^6\,\mathrm{m}\)).
Hint
\(C = 4\pi\epsilon_0 R\).
Solution

\(C = 4\pi(8.85\times 10^{-12})(6.4\times 10^6) \approx 7.1\times 10^{-4}\,\mathrm{F} = 710\,\mu\mathrm{F}\).

Answer: \(\sim 700\,\mu\mathrm{F}\) — still less than one farad.