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For the cylinder-plane image charges geometry, the attractive force is the displayed quantity per unit length. One derivation uses the effective energy including the voltage source: at fixed cylinder voltage $V$, the force toward increasing centre height is $F_d=V^2(d{C}^{\prime }/dd)/2$. Since $C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$ decreases with $d$, this force points toward the plane. For $d\gg a$, $\lambda \simeq 2\pi {\epsilon }_{0}V/\ln (2d/a)$ and $f\simeq \pi {\epsilon }_0{V}^2/[d{\ln }^2(2d/a)]$. A total force additionally requires the cable length.

In $\mathrm{Re}z>0$ outside the circle $|z-d|=R$ with $d>R>0$, the displayed bounded harmonic function is zero on the imaginary axis and $-V_0$ on the circle. Put ${\tau }_0=\mathrm{arcosh}(d/R)$. The circle is the Apollonius circle $|(z-a)/(z+a)|=e^{-{\tau }_0}$; its centre and radius are $a\coth {\tau }_0=d$ and $a/\sinh {\tau }_0=R$. The point $a$ lies inside the excluded disk and $-a$ outside the half-plane, so the logarithmic modulus is harmonic throughout the physical domain. Local branches of the complex logarithm suffice; its real part is globally single-valued even though a global complex logarithm need not exist. The potential tends to zero at infinity. A boundedness or far-field condition is needed to infer uniqueness from the finite boundary data. This is also the geometry of cylinder-plane image charges.

For a circle in the right half-plane use $0<\lambda <1$. Its centre and radius from part (ii) satisfy $d^2-R^2=a^2$. Hence choosePut ${\tau }_0=\mathrm{arcosh}(d/R)$. Then $\sinh {\tau }_0=a/R$ andIndeed the Apollonius circle has centre $a\coth {\tau }_0=d$ and radius $a/\sinh {\tau }_0=R$. Its logarithmic modulus is therefore $-{\tau }_0$, while on $x=0$ the modulus is one. The logarithmic Dirichlet potential between a circle and a line isThe logarithmic singularity at $a$ lies inside the excluded disk, since $d-a<R$. The other singularity, at $-a$, lies outside the right half-plane. Thus the real logarithmic modulus is harmonic throughout the required region and has exactly the two specified boundary values.

Local branches of the complex logarithm justify harmonicity; its real part is globally single-valued even though the complex logarithm need not be single-valued around the excluded disk. The displayed solution is bounded and tends to zero at infinity. A boundedness or corresponding far-field condition supplies uniqueness in this unbounded domain; finite boundary values alone do not rule out unbounded additional harmonic solutions. The same geometry appears in cylinder-plane image charges.

Images and uniqueness. The method of images replaces a conducting boundary by fictitious charges outside the physical field region. Choose them so that the resulting electric potential satisfies the required conductor potentials. The Uniqueness of the Dirichlet problem then guarantees the same physical exterior field. For example, a point charge $q$ at height $h$ above a grounded plane has an image $-q$ at height $-h$: the plane is an equipotential at zero, and the charge is attracted toward it by $q^2/(16\pi {\epsilon }_0{h}^2)$. The image is a computational device; the physical conductor carries a continuous induced surface charge density.

Potential of a line. Surround an infinite line charge $\lambda$ by a coaxial Gaussian cylinder of radius $r$ and length $L$. Gauss's law gives $E_r(2\pi rL)=\lambda L/{\epsilon }_0$, hence $E_r=\lambda /(2\pi {\epsilon }_{0}r)$. Integrating $E_r=-dV/dr$ givesWriting an additive constant instead is equivalent. The reference length makes the logarithm dimensionless; an isolated infinite line cannot be assigned a finite zero potential at infinity.

Locating the two lines. Put the grounded plane at $x=0$ and the physical cylinder at $(x,y)=(d,0)$, with axis along $z$ and radius $a<d$. Set $c=\sqrt{d^2-a^2}$, and place line charges $+\lambda$ and $-\lambda$ at $(c,0)$ and $(-c,0)$. If $R_{+}$ and $R_{-}$ are their distances from an observation point, the cylinder-plane image charges giveOn $x=0$, $R_{-}=R_{+}$, so $V=0$. To check the cylinder, define $q=(d+c)/a$. The level set $R_{-}=q{R}_{+}$ is a circle: completing the square givesSince $d^2-c^2=a^2$, its centre is $d$ and radius is $a$. The positive line lies inside the cylinder; the negative line lies behind the plane. There is consequently no fictitious charge in the exterior physical region. A Gaussian contour around the cylinder encloses $+\lambda$ only, so the required physical charge per unit length is also correct. These boundary conditions and the far-field behaviour determine the physical solution by Uniqueness of the Dirichlet problem.

Cylinder voltage. The constant distance ratio on the cylinder immediately givesIts capacitance per unit length is $C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$. In the thin-cable limit, $\mathrm{arcosh}(d/a)\simeq \ln (2d/a)$.

Cable attraction. Because the cable is idealized as infinite, the meaningful answer is force per unit length. For $a=0.020\mathrm{m}$, centre height $d\simeq 10\mathrm{m}$ and $V_c=1.1\times 10^4\mathrm{V}$,The electrostatic attraction per unit length of a cylinder above a plane isFor example, this follows without any ambiguity about moving image positions from $F_d=\frac{1}{2}{V}_c^{2}d{C}^{\prime }/dd$ at fixed voltage, including the work of the voltage source. Its sign is negative because capacitance decreases with height. A cable length $L$ experiences approximately $fL$ downward when end effects are negligible. This is the electrostatic contribution only.