= Solution
In a charge-free region of homogeneous <permittivity>, <Gauss's law> and $\mathbf E=-\nabla V$ give <Laplace's equation>, $\nabla^2V=0$. Prescribed surface <electric potentials> specify a <Dirichlet problem>. Its <Uniqueness of the Dirichlet problem> permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same <boundary conditions> is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. <Green's first identity> gives $\int|\nabla u|^2d^3x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal <electric field> instead determine a potential only up to an additive constant when the usual compatibility condition holds.
A cylindrical <Gaussian surface> around an infinite <line charge> gives $E_r=\lambda/(2\pi\epsilon_0r)$. The <electric potential difference> is therefore
$$
V(r_2)-V(r_1)=-\int_{r_1}^{r_2}E_r\,dr
=\frac{\lambda}{2\pi\epsilon_0}\ln\frac{r_1}{r_2}.
$$
An isolated infinite <line charge> has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to0$ there. Define $r_b^2=(x-b)^2+y^2$ and $r_c^2=(x-c)^2+y^2$. <superposition> gives
$$
V(x,y)=\frac{\lambda}{2\pi\epsilon_0}\ln\frac{r_c}{r_b}.
$$
On $x^2+y^2=bc$,
$$
r_c^2=c(b+c-2x),\qquad r_b^2=b(b+c-2x),
\qquad
\boxed{V=\frac{\lambda}{4\pi\epsilon_0}\ln\frac cb.}
$$
Thus the circle, and its cylindrical extension, is an <equipotential>. This is an <Apollonius circle>: a constant ratio of distances gives a circle rather than a constant distance from either charge.
For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The <cylinder-plane image charges> are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and <Uniqueness of the Dirichlet problem> applies. A <Gaussian surface> enclosing the physical cylinder confirms that its <charge per unit length> is indeed $\lambda$. Its <voltage> is
$$
V_c=\frac{\lambda}{2\pi\epsilon_0}\operatorname{arcosh}\frac Ha.
$$
The physical field <energy> per unit <length> at fixed cylinder charge is $U'=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:
$$
\frac{F_z}{\ell}=-\frac{\partial U'}{\partial H}
=-\frac{\lambda^2}{4\pi\epsilon_0\sqrt{H^2-a^2}},
\qquad
\boxed{\left.\frac{\mathbf F}{\ell}\right|_{H=2a}
=-\frac{\lambda^2}{4\pi\epsilon_0\sqrt3\,a}\,\widehat{\mathbf z}.}
$$
The <electrostatic attraction per unit length of a cylinder above a plane> is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2012/paper-1-cylinder-images.png]
{title=Original <cylinder-plane image charges> construction and exterior <equipotentials>}
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