Review / Chapter 1 — Electrostatics

Fields of Continuous Distributions

Chapter 1 — Electrostatics

Real objects aren't made of a handful of point charges — they're continuous blobs of charge. Conceptually nothing changes: we slice the object into infinitesimal pieces, each of which acts like a point charge, and add up the contributions as an integral.

Charge densities

Depending on the geometry you pick one of three densities:

The integral

Let \(\vec r\) be the field point and \(\vec r\,'\) run over the source. Define \(\vec s = \vec r - \vec r\,'\). Then the field is

\[\vec E(\vec r) = \frac{1}{4\pi\epsilon_0}\int \frac{dq'}{s^2}\,\hat s = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec r\,')}{s^3}\,\vec s\, dV'.\]

Two different coordinates: \(\vec r\,'\) labels the source (what you integrate over), \(\vec r\) is where you're asking for the field (fixed during the integral).

Strategy: exploit symmetry

Direct integration of the full 3D integral is usually painful. In this course you'll always have enough symmetry to reduce the work:

  1. Pick coordinates aligned with the symmetry (cylindrical for rings/rods, spherical for shells/balls).
  2. Identify which components of \(\vec E\) must vanish by symmetry — only integrate the surviving component.
  3. Write \(dq\) in the natural coordinates and plug in.

For highly symmetric distributions (sphere, infinite line, infinite plane), Gauss's law is dramatically easier than brute-force integration.

Worked example: ring of charge

A thin ring of radius \(R\) carries uniform linear density \(\lambda\). On the axis at height \(z\) above the ring's center, each element contributes a field at distance \(s = \sqrt{R^2 + z^2}\). By symmetry only the \(z\)-component survives. With \(dq = \lambda\, R\, d\phi\),

\[E_z = \frac{1}{4\pi\epsilon_0}\int_0^{2\pi}\frac{\lambda R\,d\phi}{R^2 + z^2}\cdot \frac{z}{\sqrt{R^2+z^2}} = \frac{1}{4\pi\epsilon_0}\frac{Q z}{(R^2+z^2)^{3/2}},\]

where \(Q = 2\pi R\lambda\) is the total charge. As \(z\to\infty\) this reduces to \(kQ/z^2\), as it must.

Schematic of line, surface, and volume charge distributions.
Three flavors of continuous charge distribution: line, surface, volume.

Practice Problems

Problem 1easy
A uniformly charged ring (total charge \(Q\), radius \(R\)) lies in the \(xy\)-plane. What is \(\vec E\) at the center of the ring?
Hint
Symmetry argument.
Solution

Every element at angle \(\phi\) is matched by one at \(\phi + \pi\) on the opposite side — their fields cancel at the center.

Answer: \(\vec E = 0\).

Problem 2medium
For the ring above, find the \(z\) where \(E_z\) is maximum (on the axis, \(z>0\)).
Hint
Take \(dE_z/dz = 0\) for \(E_z = kQz/(R^2+z^2)^{3/2}\).
Solution

\(dE_z/dz \propto (R^2+z^2) - 3z^2 = 0\), so \(z^2 = R^2/2\), i.e. \(z = R/\sqrt 2\).

Answer: \(z = R/\sqrt 2\).

Problem 3medium
A rod of length \(L\) with uniform linear density \(\lambda\) lies along the \(x\)-axis from \(0\) to \(L\). Find \(E_x\) at a point \((x_0, 0)\) on the axis with \(x_0 > L\).
Hint
\(dE_x = k\,\lambda\,dx'/(x_0 - x')^2\). Integrate.
Solution

\(E_x = k\lambda \int_0^L dx'/(x_0-x')^2 = k\lambda \left[\tfrac{1}{x_0-x'}\right]_0^L = k\lambda\left(\tfrac{1}{x_0-L} - \tfrac{1}{x_0}\right) = \tfrac{k\lambda L}{x_0(x_0-L)}\).

Answer: \(E_x = \dfrac{kQ}{x_0(x_0-L)}\) with \(Q = \lambda L\), pointing in \(+\hat x\).

Problem 4hard
A uniformly charged disk of radius \(R\) and surface density \(\sigma\) lies in the \(xy\)-plane. Find \(E_z\) on the axis at height \(z > 0\).
Hint
Slice the disk into rings of radius \(r\), thickness \(dr\). Use the ring result with \(dq = \sigma\cdot 2\pi r\,dr\).
Solution

\(E_z = \int_0^R \frac{k\,z\,(2\pi\sigma r\, dr)}{(r^2+z^2)^{3/2}} = 2\pi k\sigma z\left[-\tfrac{1}{\sqrt{r^2+z^2}}\right]_0^R = \frac{\sigma}{2\epsilon_0}\left(1 - \frac{z}{\sqrt{R^2+z^2}}\right)\).

Answer: \(E_z = \dfrac{\sigma}{2\epsilon_0}\left(1 - \dfrac{z}{\sqrt{R^2+z^2}}\right)\). As \(R\to\infty\) this tends to \(\sigma/(2\epsilon_0)\) (the infinite-plane result).