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Two point dipoles separated by $\mathbf{r}$ have interaction proportional to

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.

Take the wire at $(a,0)$ to carry $+\lambda$ and that at $(-a,0)$ to carry $-\lambda$. Ifthen, up to one additive constant,Thus an equipotential with $k=2\pi {\epsilon }_0\varphi /\lambda$ satisfies $r_{-}/r_{+}=e^k$. For $k\ne 0$, completing the square gives the Apollonius circleIts centre is $(a\coth k,0)$ and its radius is $a/|\sinh k|$. Positive and negative values give nested circles around the positive and negative wires. The electric field is orthogonal to these circles and points from the positive wire toward the negative wire. For $\varphi =0$, the limiting equipotential is the straight line $x=0$.

Direct superposition givesIn the limit $a\to 0$ with $\lambda a=p$, the potential and field become those of a two-dimensional electric dipole:and

The even, spherically symmetric hydrogen ground state has no permanent electric dipole, so a uniform electric field gives no first-order energy shift. Odd-parity excited states mix in at second order, yielding a negative shift quadratic in the field.