Past exam of the physics B course of the University of Cambridge / 2010 / Paper 2B / Section B

For steady fields in two leaky dielectric layers, set $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$. Then $E_1={\sigma }_{2}V/\mathcal{D}$, $E_2={\sigma }_{1}V/\mathcal{D}$, $J={\sigma }_1{\sigma }_{2}V/\mathcal{D}$. The interfacial free and total charges are $({ϵ}_2{\sigma }_1-{ϵ}_1{\sigma }_2)V/\mathcal{D}$ and ${ϵ}_0({\sigma }_1-{\sigma }_2)V/\mathcal{D}$, respectively.

Full derivation: solution.

During partial entry, resistive magnetic braking gives $v=g\tau (1-e^{-t/\tau })$ with $\tau =mR/(B_{0}a{)}^2$. The lossless inductive limit gives $x=g(1-\cos \omega t)/{\omega }^2$, $v=g\sin \omega t/\omega$, with $\omega =B_{0}a/\sqrt{mL}$. Complete immersion changes the equations; the solution below distinguishes reversal from full entry.

Full derivation: solution.

The parallel-plate transmission line has per-length $C=ϵb/a$ and $L=\mu a/b$. The lossless telegrapher's equations imply wave speed $1/\sqrt{LC}=1/\sqrt{\mu ϵ}$, independent of the much slower electron drift velocity.

Full derivation: solution.

The image charge for a grounded conducting sphere is $-qa/d$ at radius $a^2/d$. With symmetric real charges, their exact force-balance equation has a solution close to $d/a=8$. Holding potential $V$ adds force $qaV/d^2$ to the grounded result, approximately $qV/(64a)$ at that separation.

Full derivation: solution.