Past exam of the mathematics course of the University of Cambridge / 2021 / ib / Paper 2 / 16D (Electromagnetism)

WriteUsing the Taylor seriesand retaining terms through second order in $|\mathbf{y}|/|\mathbf{x}|$ givesThis is the beginning of the multipole expansion.

The method of images replaces the earthed plane by an image charge $-q$ at $(-a,0,0)$. The two Coulomb potentials cancel at $x=0$, so uniqueness for the Dirichlet problem makes the resulting field the physical field in $x>0$. With $\mathbf{r}=(x,y,z)$ and ${\mathbf{e}}_x=(1,0,0)$,and

The electric multipole expansion from part (a) gives, for $r\gg a$,Thus the leading field is that of an electric dipole.

On the plane,Taking the plane normal to be $+{\mathbf{e}}_x$ and using polar coordinates,With the outward normal of the region $x>0$, the sign is reversed. This is consistent with Gauss's law: all electric flux from the real charge terminates on the grounded conductor.

The electrostatic boundary condition gives the induced surface charge densityand its integral isIn the plane $z=0$, the field lines leave the positive charge, meet the conductor normally, and are the right-half-plane portions of the field lines joining the real charge to its negative image.

Successive reflections in the two grounded planes require four charges:Their total charge and electric dipole moment vanish. Applying the second-order expansion from part (a), the terms proportional to $a^2{x}^2$, $b^2{y}^2$, and $a^2+b^2$ also cancel, while the mixed terms add. ThusThis is an electric quadrupole potential.