Past exam of the physics B course of the University of Cambridge / 2013 / Paper 1 / Section B

The image charge construction replaces conductor-induced charge by fictitious sources outside the physical domain. Match the conductor potentials and the far-field condition; uniqueness then fixes the potential in the physical region. For a grounded plane, reflect a charge and reverse its sign, giving attraction $q^2/(16\pi {ϵ}_0{h}^2)$. For a grounded sphere of radius $a$, a charge at distance $R>a$ has image $-qa/R$ at radial distance $a^2/R$. These are examples with prescribed zero potential, not arbitrary isolated conductors.^[ref]

Use horizontal $x$ and upward $y$ in the upper right-angle wedge. The sheets meet electrically, so take their common potential as zero and assume no externally imposed field. For a real charge at $(0,y)$, the image charges in a right-angle conducting wedge are $-q$ at $(y,0)$ and $(-y,0)$, and $+q$ at $(0,-y)$. The first two contributions have cancelling horizontal components. With $K_e=1/(4\pi {ϵ}_0)$,Thus the force magnitude at the original height is $\beta /y_0^2$, directed down toward the intersection. The sphere is treated as a point charge, with size much smaller than its distances from the sheets.

Since gravity is already balanced in the uncharged equilibrium, the charged vertical equilibrium satisfies $k(y_e-y_0)+\beta /y_e^2=0$. For $\beta /(k{y}_0^3)\ll 1$, $y_e\simeq y_0-\beta /(k{y}_0^2)$. Differentiating the force givesThe electrostatic softening above a conducting wedge reduces the restoring force; the small-charge vertical equilibrium remains stable.

For the lateral stability of a spring-supported charge in a wedge, the off-axis images of $(x,y)$ are $-q$ at $(y,x)$ and $(-y,-x)$ and $+q$ at $(-x,-y)$. Expansion near the bisector givesSo the electrostatic force is laterally destabilizing: moving toward either sheet strengthens its attraction. In the simplified model where the spring supplies only a vertical force, the overall equilibrium is a saddle and is unstable.

A three-dimensional spring suspension also has geometrical transverse stiffness. If its charged-equilibrium tension is $T$ and length is $\ell$, a small sideways displacement gives spring force $-(T/\ell )x$, with $T=mg+\beta /y_e^2$. The full stability conditions in this suspension model areMotion parallel to the intersection line has no electrostatic stiffness in the ideal infinite-wedge geometry and is stabilized by the positive spring tension. Thus lateral electrostatic destabilization alone does not prove instability of a real hanging spring. At fixed finite support geometry, a sufficiently small charge preserves stability because $D_x=O(q^2)$ while $T/\ell \to mg/\ell >0$. The given vertical-displacement approximation does not specify the attachment length or by itself establish the transverse inequality for an arbitrary charge. The general criterion above states what can be concluded; the saddle verdict applies to the distinct idealization with zero lateral support stiffness.

For distributed inductance $L$ and capacitance $C$, a small segment obeys $V_z=-L{I}_t$ and $I_z=-C{V}_t$. Differentiation gives $V_{zz}=LC{V}_{tt}$ and $I_{zz}=LC{I}_{tt}$. For a forward wave,A backward voltage wave has current $I^{-}=-V^{-}/Z$. This sign matters at an impedance discontinuity.

The source figure is the stacked-strip parallel-plate transmission line geometry, despite its use of the word coplanar. Gauss's law gives $C={ϵ}_0{ϵ}_{r}a/d$. Opposite surface currents give magnetic field $B={\mu }_{0}I/a$ between the strips by Ampère's law; the magnetic flux per unit length is $Bd$, hence $L=Bd/I={\mu }_{0}d/a$. ThereforeNo numerical $a/d$ is given; the impedance is appropriately left in terms of that ratio. The line is taken to be nonmagnetic and lossless.

Increasing the gap by $\alpha$ changes $L$ to $\alpha L$ and $C$ to $C/\alpha$, so $Z_2=\alpha Z_1$ while $v$ is unchanged. voltage and current continuity at the step giveConsequently the voltage and energy-flux transmission coefficients areHere amplitudes are peak voltage amplitudes, with mean power $|V{|}^2/(2Z)$. The reciprocal gaps have $r_V(1/\alpha )=-r_V(\alpha )$: equal reflected energy but opposite voltage phase. A voltage transmission coefficient above one does not imply energy gain.

For a finite impedance section of a transmission line, put $\ell =l_2-l_1$ and $\delta =\omega \ell /v$. Multiple reflections inside the altered section interfere. Summing their geometric series, relative to the first boundary,Transmission is unity at $\delta =n\pi$ and minimal at $\delta =(n+1/2)\pi$, with minima $4{\alpha }^2/(1+{\alpha }^2{)}^2$. Resonances are separated by $\Delta f=v/(2\ell )$; the long-wavelength limit also transmits fully. Matching the far termination removes end echoes, not the two internal reflections.

For a moving loop, Faraday's law of induction is $\mathcal{E}=\oint (\mathbf{E}+\mathbf{u}\times \mathbf{B})\cdot d\mathbf{l}=-d{\Phi }_B/dt$, where $\mathbf{u}$ is the local wire velocity, ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$, and the oriented boundary and surface follow the right-hand rule. For a fixed loop the motional term vanishes. Lenz's law is the minus sign: the induced response opposes the change of linked flux, not necessarily the applied field itself.

Let $s$ increase downhill and choose positive current $I$ to produce an uphill magnetic force. The normal component of the imposed field is $B\cos \theta$, so the changing external flux is $Bl\cos \theta s$. With initially zero current and no resistance, conservation of flux linkage gives $LI=Bl\cos \theta s$. The rod feels gravity vertically downward, a rail reaction normal to the slope, and magnetic force of magnitude $IlB$ horizontally uphill. Its downhill component is $-IlB\cos \theta$; its normal component presses the rod into the rails, giving $N=mg\cos \theta +IlB\sin \theta$.

The inductive oscillation of a sliding rod on inclined rails follows fromThe magnetic force is not itself parallel to the rails. Defining ${\omega }_B=Bl\cos \theta /\sqrt{mL}$ and $s_e=mg\sin \theta /(m{\omega }_B^2)$, release from rest givesThe first stop occurs at $t=\pi /{\omega }_B$; the rod then returns and oscillates between $0$ and $2s_e$. At $s_e$ the force is zero but the speed is maximal. Mechanical potential energy is reversibly exchanged with kinetic energy and $L{I}^2/2$; a perfectly conducting circuit provides no damping.

For the resistive long-time drift of an inductive sliding rod, with loop resistance $R>0$, useJoule heating damps the transient oscillation. On sufficiently long rails the asymptotic motion is steady sliding, not permanent rest:The zero-resistance model has persistent oscillations; its late-time limit cannot be interchanged with taking $R\to 0$. These results assume $B\cos \theta \ne 0$, smooth contacts, negligible rod self-inductance and constant geometry.

For cold, collisionless, unmagnetized electrons with stationary heavy ions, $-m_e{\omega }^2\mathbf{x}=-e\mathbf{E}$. Therefore $\mathbf{P}=-ne\mathbf{x}=-n{e}^2\mathbf{E}/(m_e{\omega }^2)$ andThis is the cold collisionless plasma dielectric response. For a propagating transverse wave, $\mathbf{B}=\mathbf{k}\times \mathbf{E}/\omega$, so $E/B=c$ in vacuum and $E/B=c/n_p$ in plasma, where $n_p=\sqrt{{ϵ}_r}$. Equivalently $E/H=Z_0$ and $Z_0/n_p$ for $\mu ={\mu }_0$.

Converting the density gives $n=10^{18}{\mathrm{m}}^{-3}$, so ${\omega }_p=5.64\times 10^{10}{\mathrm{s}}^{-1}$ and $f_p=8.98\mathrm{GHz}$. At $10\mathrm{GHz}$,At an interface without independently imposed surface charge or sheet current, tangential $\mathbf{E},\mathbf{H}$ and normal $\mathbf{B},\mathbf{D}$ are continuous, with the electron response included in the effective permittivity. Normal $E$ need not be continuous; the induced electron charge is represented by that response. In a convention treating it explicitly as a surface charge, the normal vacuum-displacement jump equals that charge instead.

For the downward transmitted wave, transversality gives $k_x{E}_x+k_z{E}_z=0$. The measured field therefore has $k_x/(-k_z)=3$, so $\tan {\theta }_t=3$ and $\sin {\theta }_t=3/\sqrt{10}$. Tangential phase matching givesFor inferring incident power from a transmitted plasma field, choose ${\mathbf{E}}_i=E_i(\cos {\theta }_i,0,\sin {\theta }_i)$ and ${\mathbf{E}}_r=E_r(\cos {\theta }_i,0,-\sin {\theta }_i)$. Their magnetic fields have opposite $y$ directions. The two tangential conditions areHence $E_i\simeq 1.2464\mathrm{V}{\mathrm{m}}^{-1}$ and $E_r\simeq -0.1458\mathrm{V}{\mathrm{m}}^{-1}$. Since these are peak phasor amplitudes,This is incident power per area perpendicular to its propagation. Per area of the interface the incoming normal flux is $⟨S_i⟩\cos {\theta }_i\simeq 1.87\times 10^{-3}\mathrm{W}{\mathrm{m}}^{-2}$. The reflected sign is a phase reversal, not negative reflected energy.

At $1\mathrm{GHz}<f_p$, the relative permittivity is negative and there is no propagating transverse bulk wave. The field decays evanescently into the plasma and the ideal collisionless half-space reflects all incoming normal power. Collisions would allow absorption; a complex refractive index here does not by itself imply dissipative heating.