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A driven cylindrical conductor between two grounded coaxial conductors forms two coaxial cylindrical capacitors in parallel. For radii $a,\lambda a,2a$, the capacitance per unit length is the sum $C^{\prime }=2\pi {ϵ}_0[1/\log \lambda +1/\log (2/\lambda )]$. Writing $u=\log \lambda$ and $L=\log 2$, the function $1/u+1/(L-u)$ is strictly convex on $(0,L)$ and symmetric about $L/2$. Hence $C^{\prime }$ has its unique minimum at $\lambda =\sqrt{2}$, with value $8\pi {ϵ}_0/\log 2$. As $\lambda =1+\delta$ approaches the inner conductor, $C^{\prime }\sim 2\pi {ϵ}_0/\delta$, agreeing with the local parallel-plate capacitor approximation.

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.

Neglect end effects, so the electrostatic potential depends only on the cylindrical radius $r$. The radial Laplace equation is $(r{\varphi }^{\prime }{)}^{\prime }=0$, with general solution $A\log r+B$. Imposing the conductor boundary conditions givesThe radial electric field is $E_r=-{\varphi }^{\prime }$. Just inside the middle cylinder it is $-V/[\lambda a\log \lambda ]$, directed towards the inner cylinder; just outside it is $V/[\lambda a\log (2/\lambda )]$, directed towards the outer cylinder. By Gauss's law, its charge per unit length isHence the capacitance per unit length isThe two coaxial cylindrical capacitors contribute in parallel because both connect the middle cylinder to the same grounded potential. This is the coaxial capacitor with grounded inner and outer cylinders construction.

For $\lambda =1+\delta$ with $\delta \to 0^{+}$, $\log \lambda =\delta +O({\delta }^2)$ while $\log (2/\lambda )\to \log 2$. ThereforeThe inner gap has width $a\delta$ and area per unit length approximately $2\pi a$. Its parallel-plate capacitor approximation gives exactly the same leading result; the outer gap contributes only a bounded correction.

For the extremum, set $u=\log \lambda$, $L=\log 2$, so $0<u<L$. ThenThus the unique stationary point is $u=L/2$. Strict convexity and divergence at both endpoints show it is a global minimum:There is no maximum on $1<\lambda <2$.

Gauss's law in electrostatics statesCylindrical symmetry and a coaxial Gaussian cylinder giveThe field vanishes inside each perfect conductor, and outside the cable because the total enclosed charge per unit length is zero.

Choose the outer conductor's potential to be zero. Since $E=-\nabla V$,The capacitance per unit length is thereforeThe electrostatic energy per unit length isSubstitution of $C$ verifies $U=Q^2/(2C)$.

Introduce cylinder length $\ell$ and line charge density $\lambda =Q/\ell$; only charge per length matters in the long-cylinder approximation. A coaxial cylindrical Gaussian surface gives $E_{r}2\pi r\ell =Q/{ϵ}_0$. Choosing the outer cylinder's electric potential to be zero,The capacitance per unit length follows from $C=Q/[V(a)-V(b)]$:If $Q$ is intended as a line charge rather than a finite total charge, replace $\lambda$ by that $Q$. Neglecting end effects, equal opposite cylinder charges make the exterior electric field vanish; the potential difference between the conductors remains the quantity defining their capacitance.

Put line charges $+\lambda$ and $-\lambda$ on the two axes. For $d\gg a_1,a_2$, the surface potentials differ byThus the capacitance per unit length in vacuum isThe two logarithms belong to one potential difference, not two independent capacitances in parallel. A homogeneous dielectric replaces ${ϵ}_0$ by its absolute permittivity; finite radii and proximity corrections are neglected.

The characteristic impedance of a lossless transmission line is the ratio of the voltage to the current amplitude of a single forward-travelling wave. If $\mathcal{L}$ and $\mathcal{C}$ denote inductance and capacitance per unit length, the lossless telegrapher equations giveA backward wave has $I_{-}=-V_{-}/Z$ when current is positive along $+z$. Thus characteristic impedance is not a resistor's electrical resistance: it describes travelling-wave amplitudes and determines impedance matching, power transfer and reflections. A semi-infinite line, or a line terminated by a matching load, presents that impedance at its input.

At a junction with no lumped branch, voltage and electric current are continuous:Eliminating $V_t$ gives the voltage reflection coefficientFor real positive $Z_1$, incident and reflected waves occupy the same line, so their power ratio is $|r{|}^2$.

For the slightly mismatched coaxial cables, the change in characteristic impedance isConsequentlyThe requested power limit $0.01\%=10^{-4}$ implies $|\Delta b|/b<0.02\ln (b/a)$. Since $\ln (b/a)=2\pi nZ/Z_{\mathrm{vac}}=1.918$ for the stated line, the adjacent cables' outer radii must match within approximately $3.84\%$. If a tolerance is quoted separately as a symmetric manufacturing range $\pm \delta b$ and arbitrary worst-case pairs can differ by $2\delta b$, the corresponding individual tolerance is approximately $\pm 1.92\%$. The leading calculation is for small mismatches; the exact logarithmic expression gives slightly asymmetric positive and negative limits. The vacuum wave impedance used here is approximately $376.73\Omega$.^[ref]

A needle is a shunt electrical resistance $R_n$, so the load seen at its location is the continuation of the line in parallel with $R_n$: $Z_L=Z{R}_n/(Z+R_n)$. The local voltage reflection coefficient and transmission amplitude areFor periodic resistive loading of a transmission line, the reflection from the next needle travels an additional round trip $2L$ and crosses each previous needle twice. Taking the first needle as phase reference, the single-return approximation for periodic line reflections therefore sums toHere $v_p$ is the propagation phase velocity in the cable, $c/n$ for the stated nondispersive nonmagnetic filling. Squaring the amplitude gives the requested expressionFor a finite number $N$ of needles the amplitude has the additional numerator $1-[t^2{e}^{i\varphi }{]}^N$. The displayed semi-infinite result assumes sufficiently many needles, no terminal reflection, and neglects intrinsic propagation loss.

For real $r<0$, write $q=t^2$. The denominator of the power coefficient is $1+q^2-2q\cos \varphi$. It has constructive peaks at $\varphi =2\pi j$ and minima at $\varphi =(2j+1)\pi$, repeating with frequency spacing $\Delta \nu =v_p/(2L)$. Within this approximation,Weak individual shunts thus give narrow interference peaks, with phase width of order $|r|$.

The requested sum excludes paths that reflect repeatedly between different needles. It should not be read as an exact multiple-scattering result, particularly at the coherent peaks. For example the exact semi-infinite amplitude satisfiesAt $\varphi =2\pi j$ this equation reduces to $(R+1{)}^2=0$: the exact ideal infinite array has unit power reflection there, rather than the first-return approximation's value near $1/4$. This does not change the requested geometric-series derivation, but identifies its physical limitation.

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.

Replace two well-separated cylindrical conductors by opposite axial line charges. Their surface-to-surface potential difference is $\lambda [\ln (d/a_1)+\ln (d/a_2)]/(2\pi ϵ)$, giving the displayed capacitance per unit length. The approximation needs $d\gg a_1,a_2$ and a homogeneous surrounding permittivity.