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The free-space Green function of the Laplacian for the Poisson equation gives the electric potentialFor $r=|\mathbf{x}|\gg R$, its electric multipole expansion begins withDefine the total electric charge and electric dipole moment byThen

The analogous solution for the magnetic vector potential isIts apparent $r^{-1}$ coefficient is ${\mu }_0(4\pi {)}^{-1}\int \mathbf{J}{d}^3{x}^{\prime }$. For each component, the supplied identity and the divergence-free vector field condition giveThe divergence theorem turns this into a surface integral outside the compact support of the current density, where $\mathbf{J}=0$. Hence $\int \mathbf{J}{d}^3{x}^{\prime }=0$, so the $1/r$ term vanishes.

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.

Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.