Past exam of the electromagnetism and optics course of the University of Oxford / 2016 / 3

Take $d$ as the centre-to-centre separation and $ϵ={ϵ}_0{ϵ}_r$. Use the ideal lossless two-conductor transmission line model, with nonmagnetic surroundings and negligible internal-wire inductance. Opposite line charges $+{\lambda }_q$ and $-{\lambda }_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their electric fields, or adding their logarithmic electric potentials, gives the potential differenceThis is the two-wire capacitance at large separation. For opposite electric currents $+I$ and $-I$, the magnetic fields between the wires add. The linked magnetic flux per unit length is approximatelyThus the external inductance of a thin two-wire line is $L^{\prime }\simeq {\mu }_0\ln (d/a)/\pi$. The lossless telegrapher's equations give a forward wave with $V/I=\sqrt{L^{\prime }/C^{\prime }}$, so its characteristic impedance isThe product $L^{\prime }{C}^{\prime }={\mu }_0ϵ$ also gives the propagation speed $c/\sqrt{{ϵ}_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln (d/a)$ by $\mathrm{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite electrical conductivity and internal-wire magnetic energy require a more detailed line model.