In sufficiently symmetric geometries, the method of images provides a way to solve Poisson’s equation \(|\partial_{\textbf x}|^2\phi=-\rho/\varepsilon_0\) in a domain \(V\) subject to either Dirichlet or Neumann boundary conditions (required for the uniqueness theorem to hold) by strategically placing charges in the “unphysical region” \(\textbf R^3-V\) such as to ensure the boundary conditions are met. It works because of linearity and the fact by placing image charges outside the physical region \(V\), one isn’t tampering with \(\rho\) in that region so Poisson’s equation truly is solved.
In the following problems, the goal is to compute (in the suggested order):
- The electrostatic potential \(\phi(\textbf x)\) everywhere (i.e. both in regions of free space and inside materials).
- The electrostatic field \(\textbf E(\textbf x)=-\frac{\partial\phi}{\partial\textbf x}\)
- The induced charge density \(\sigma\) on any conducting surfaces, along with the total charge \(Q\) on such surfaces.
- The force \(\textbf F\) between any conductors.
- The internal fields (\(\textbf D,\textbf E,\textbf P,\phi\)) and bound charge distributions \(\rho_b,\sigma_b\) for any dielectrics.
- The resistance/self-capacitance/self-inductance/mutual capacitance/mutual inductance of any conductors? (although that isn’t really electrostatics anymore…)
Problem: Consider placing a point charge \(q\) at the point \((0,0,z)\) a distance \(z\) from an infinite planar conductor at \(z=0\).
Solution: Place an image point charge \(-q\) at \((0,0,-z)\).
Problem: Now instead of a point charge, consider a line charge with linear charge density \(\chi\).
Solution:
Problem: Instead of a line charge, place a line “cylinder” of charge of radius \(a\).
Solution: Applying the cosine law:
\[\rho_1^2=R^2+(d-\sqrt{d^2-R^2})^2-2R(d-\sqrt{d^2-R^2})\cos\phi\]
\[\rho_2^2=R^2+(d+\sqrt{d^2+R^2})^2-2a(d+\sqrt{d^2-R^2})\cos\phi\]
So eliminating the \(\cos\phi\), one finds that it is indeed possible to isolate solely for the ratio \(\rho_1/\rho_2\) as a function of constant parameters, confirming that it is an equipotential surface as required.
Aside: this is nothing more than Apollonius’s construction of a circle as the set of all points whose distances \(\rho,\rho’\) from \(2\) “foci” are in a fixed ratio \(\rho’/\rho\). Indeed, if the two foci are separated by a “semi-axis” \(a\) (thus their full separation is \(2a\)), then the distance \(d\) from the midpoint of the two foci to the center of the Apollonian circle and its radius \(R\) satisfy (using the extreme points on the circle):
\[\frac{\rho’}{\rho}=\frac{d+(a-R)}{a+R-d}=\frac{a+R+d}{d-(a-R)}\]
So:
\[d^2=a^2+R^2\]
\[\frac{\rho’}{\rho}=\frac{d+a}{R}=\frac{a}{R}+\sqrt{1+\left(\frac{a}{R}\right)^2}\]
or inverting these relations:
\[d=\]
\[R=\]
Problem: Consider a magnetic dipole \(\boldsymbol{\mu}\) suspended above a superconducting surface so that on this surface all magnetic fields are expelled.
Problem: An electrostatic dipole \(\boldsymbol{\pi}\) a distance from a conducting plane?
Problem: Consider a conducting sphere in an asymptotically uniform background electrostatic field \(\textbf E_0\).
Problem: Replace the conducting sphere by an insulating sphere (aka a linear dielectric sphere) of permittivity \(\varepsilon\) (comment on how this relates to the Clausius-Mossotti relation).
Problem: Instead of linear dielectric sphere, consider linear diamagnetic sphere in a uniform magnetic field \(\textbf B_0\).
Problem: Consider an \(N\)-gon of conducting sheets (quadrupole, octupole, etc.)
Problem: A point charge in a conducting spherical cavity (Green’s function for that domain).
Problem: A point charge outside the sphere.
Problem: (example with infinitely many image point charges?)
These ideas extend immediately to potential flows in fluid mechanics…describe all the analogous situations and analogous results without doing all the work again. Similarly for steady-state temperature distributions, and anywhere that Laplace’s equation with suitable boundary conditions shows up.