Electricity and Magnetism · Problem 10.38

MAGNETIC FIELDS IN MATTER — Edward Purcell & David Morin

Physics — E&M · uniform-field-superposition-polarized-sphere

Problem

In Section 10.9, the fact that the electric field is uniform inside the polarized sphere was deduced from the form of the potential on the boundary. You can also prove it by superposing the internal fields of two balls of charge whose centers are separated. (a) Show that, inside a spherical uniform charge distribution, E\mathbf{E} is proportional to r\mathbf{r}. (b) Now take two spherical distributions with density ρ\rho and −ρ-\rho, centers at C1C_1 and C2C_2, and show that the resultant field is constant and parallel to the line from C1C_1 to C2C_2. Verify that the field can be written as −P/3ϵ0-\mathbf{P}/3\epsilon_0. (c) Analyze in the same way the field of a long cylindrical rod that is polarized perpendicular to its axis.
Intuitive explanation

The key insight is that superposing fields from displaced charge distributions—one positive, one negative—creates a uniform field in their overlap region. For part (a), you'll use Gauss's law inside a uniform sphere to show the field grows linearly with distance from center. Then in (b), offsetting two opposite spheres by a small displacement d\mathbf{d} produces a nearly uniform field in the overlap; the field magnitude depends only on the dipole moment P=ρd\mathbf{P} = \rho \mathbf{d}, not position. Part (c) extends this elegant superposition trick to cylindrical geometry, where the same principle yields a uniform transverse field.

Intuitive explanation

The key insight is that superposing fields from displaced charge distributions—one positive, one negative—creates a uniform field in their overlap region. For part (a), you'll use Gauss's law inside a uniform sphere to show the field grows linearly with distance from center. Then in (b), offsetting two opposite spheres by a small displacement d\mathbf{d} produces a nearly uniform field in the overlap; the field magnitude depends only on the dipole moment P=ρd\mathbf{P} = \rho \mathbf{d}, not position. Part (c) extends this elegant superposition trick to cylindrical geometry, where the same principle yields a uniform transverse field.

Related theory

Browse the problem library