Section 2.4: Continuous Charge Distributions

When charge is distributed continuously along a line, surface, or volume, the total electric field is found using integration.

General Formula:

\[ \vec{E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\vec{r'}) \, dV' \, (\vec{r} - \vec{r'})}{|\vec{r} - \vec{r'}|^3} \]

where \( \rho \) is the volume charge density, \( \vec{r'} \) the source point, and \( \vec{r} \) the field point.

Example 1

Find the electric field at a point along the axis of a uniformly charged rod of length \( L \) and total charge \( Q \).

Linear charge density: \( \lambda = \frac{Q}{L} \).

\( dE = \frac{1}{4\pi \epsilon_0} \frac{\lambda dx}{r^2} \).

Integrate from \( 0 \) to \( L \), considering geometry.

Result: \[ E = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q}{x\sqrt{x^2 + L^2}} \] along the axis, directed away from the rod if \( Q > 0 \).

Practice Problems

  1. Find the field at a point along the axis of a ring with total charge \( Q \).
  2. Calculate the field at the center of a uniformly charged arc of angle \( \theta \).
  3. A thin rod of charge lies on the \( y \)-axis from \( y=-a \) to \( y=+a \). Find the field on the \( x \)-axis.
  4. Find the net field due to an infinite line of charge using integration (and compare with Gauss’s law).
  5. Conceptual: Why does symmetry simplify integrals for continuous charge distributions?