Cylinder-plane image charges ($c=\sqrt{d^2-a^2},C^{\prime }=2\pi {\epsilon }_0/\mathrm{arcosh}(d/a)$)

An infinite conducting cylinder of radius $a$ at centre height $d>a$ above a grounded plane has the same exterior electric potential as opposite line charges $\pm \lambda$ at heights $\pm c$. If $R_{+}$ and $R_{-}$ are distances from the positive and negative lines, $V=\lambda \ln (R_{-}/R_{+})/(2\pi {\epsilon }_0)$. The plane has $R_{-}=R_{+}$; the cylinder has $R_{-}/R_{+}=(d+c)/a$. Its voltage is therefore $\lambda \mathrm{arcosh}(d/a)/(2\pi {\epsilon }_0)$, giving the displayed capacitance per unit length. The Uniqueness of the Dirichlet problem fixes the exterior field. A Gaussian contour surrounding the cylinder encloses the positive image line, so $\lambda$ is also its physical charge per unit length.

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.