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A non-ideal amplifier is not specified by infinite input impedance alone: its open-loop gain and output impedance must also be modeled. Take its internal controlled output to be $u=-A(s)v$, where $v$ is the inverting-input potential, and place a resistance $r_o$ between $u$ and the actual output $y=V_{\mathrm{out}}$. The non-inverting input remains grounded; assume no external load. This gives an explicit finite-gain operational-amplifier integrator model.

Writing Kirchhoff's current law at the input and output nodes givesThe second equation gives $y=(sC{r}_o-A)v/(1+sC{r}_o)$. Substitution in the first gives the general transfer functionFor sinusoidal phasors, set $s=i\omega$. The numerator and denominator are complex, so the magnitude gives the amplitude ratio and their arguments give the relative phase.

The simplest model takes $r_o=0$ and a constant finite gain $A_0>0$. It givesThus below ${\omega }_l=[(1+A_0)RC{]}^{-1}$ the amplifier has gain magnitude $A_0$ and inversion. Above that frequency it approaches an integrator with magnitude $A_0/[(1+A_0)\omega RC]\simeq 1/(\omega RC)$ and phase $+90^{\circ }$. Constant finite gain by itself predicts no further high-frequency turnover.

For a more realistic sketch, use the standard dominant-pole open-loop gain $A(s)=A_0/(1+s/{\omega }_a)$, still with $r_o=0$. ThenThe exact frequency response isAssuming $A_0\gg 1$ and $A_0{\omega }_{a}RC\gg 1$, the turnover frequencies are well separated: ${\omega }_l\simeq 1/(A_{0}RC)$ and ${\omega }_h\simeq A_0{\omega }_a$, the approximate unity-gain angular frequency of the amplifier. The regimes arewith magnitude slopes $0,-20,-40\mathrm{dB}$ per decade and smooth phase transitions. This is the solid curve in the figure; $180^{\circ }$ and $-180^{\circ }$ denote the same phase.

If finite $r_o$ is retained and $A(i\omega )\to 0$, the general expression instead approaches $r_o/(R+r_o)$ at very high frequency: the capacitor couples the input to the output through a passive voltage divider. The dashed curve illustrates that eventual feedthrough. Other amplifier poles or an external load would alter the response again. No particular values of $A_0,{\omega }_a,r_o$ are given in the paper, so the numerical sketch is illustrative and the general expression states the model dependence explicitly.

Temperature changes affect measurements through several routes. Thermal expansion changes lengths and alignments; temperature-dependent resistance and refractive index change electrical and optical calibrations; temperature differences between dissimilar junctions generate Seebeck effect voltages. These slow environmental changes are distinct from unavoidable equilibrium thermal fluctuations and Johnson-Nyquist noise. A useful strategy identifies the coupling to the measured signal before choosing a remedy.

Reduce environmental coupling by enclosing the apparatus against drafts, using thermal insulation and radiation shields, isolating heat sources, and minimizing dissipated power. A vacuum enclosure suppresses convection but leaves thermal conduction and thermal radiation paths to control. A large heat capacity reduces rapid temperature excursions. In a simple thermal model with heat capacity $C_T$ and thermal conductance $G_T$ to a reservoir, the relaxation time is $C_T/G_T$, so adequate settling time is necessary after changing the apparatus. A temperature sensor close to the sensitive element and stable negative feedback heating or cooling can regulate its temperature; poorly placed sensors or excessive feedback gain can leave gradients or cause oscillations.

Electrical thermal noise has the classical low-frequency one-sided voltage power spectral densityfor a resistor measured over a flat bandwidth $B$ without significant loading. More generally replace $B$ by the effective noise bandwidth. Lower temperature or narrower bandwidth reduces the noise, and resistance choices must account for source loading and the amplifier's own voltage and current noise. A lock-in amplifier uses modulation and narrowband detection to reject much low-frequency drift and out-of-band noise, while thermal noise inside the accepted band remains. Temperature stabilization does not remove equilibrium fluctuations.

Use a matched reference or differential bridge to reject common environmental changes. Keep junctions isothermal and choose matched materials to reduce thermoelectric offsets. For a resistance measurement, reversing a small test current separates a stable offset from the wanted voltage:Reversal must be fast enough that the offset stays approximately fixed, with enough settling time and sufficiently little self-heating. Recording the sample temperature permits a measured calibration slope $dS/dT$ to correct a signal to a reference temperature; the correction's calibration and thermometer uncertainties must be propagated. Independent white noise fluctuations decrease under averaging, whereas correlated temperature drift may not decrease as the inverse square root of the number of readings. Temperature records and repeated reference measurements help distinguish the two.

Digital acquisition converts a continuous voltage signal into timed numerical samples. A practical chain is a suitably buffered and amplified sensor signal, an analogue anti-aliasing filter, a sampling or sample-and-hold stage, an analog-to-digital converter, and a clocked data-storage system. Choose the gain so that the interesting variation uses the available input range without clipping. Finite input impedance can load a sensor, so an appropriate buffer or operational amplifier may be needed.

Sampling makes time discrete. If the conditioned signal is confined below frequency $f_{\max }$, choose a sampling rate $f_s>2f_{\max }$, allowing a transition band for the physical filter. The Nyquist–Shannon sampling theorem permits ideal reconstruction of a suitably bandlimited function from exact samples. Without filtering, components separated by integer multiples of $f_s$ have identical samples; in particular real sinusoids at $f$ and $f_s-f$ can be indistinguishable. This aliasing cannot generally be undone by later digital filtering. Oversampling can ease filter design, but does not replace the requirement to suppress unwanted high-frequency content before sampling.

Amplitude quantization is separate from sampling. A uniform $b$-bit converter covering voltage span $V_{\mathrm{span}}$ has step $q\simeq V_{\mathrm{span}}/2^b$. Rounding an in-range input gives the uniform-quantization error bound $|ϵ|\le q/2$. If the rounding errors can additionally be modeled as uniform and uncorrelated, their standard deviation is $q/\sqrt{12}$. This statistical estimate is not a guarantee for a constant input or a coherent periodic signal. More bits increase resolution, but accuracy also depends on voltage-reference calibration, offset, gain error, nonlinearity and electronic noise. Out-of-range signals clip and invalidate the rounding bound.

Clock quality matters too. Timing jitter $\delta t$ gives a voltage error approximately $(dV/dt)\delta t$; for a sinusoid its scale grows with $2\pi f{V}_p{\sigma }_t$. A clock calibration error biases measured frequencies and time intervals. Simultaneous multichannel signals need known relative timing, and the sampling and analogue settling times must be appropriate to the signal bandwidth.

For $n_c$ channels the raw recording rate is approximately $n_c{f}_{s}b$ bits per second, before overhead, so storage and sustained transfer speed must prevent missing data. Store units, channel identities, timestamps, sampling rate, input ranges and calibration information along with the readings. Preserve the original samples when applying digital filters or other processing. Averaging can improve precision for independent fluctuating errors, but cannot restore aliased frequencies, clipped peaks or a fixed rounding bias. Digital recording makes reproducible analysis and transmission possible when its sampling, amplitude and timing limitations are all controlled.

Take a p-n junction made from one semiconductor, with the p-type semiconductor on the left and the n-type semiconductor on the right. At thermal equilibrium the chemical potential $\mu =E_F$ is horizontal throughout. Before contact the chemical potential lies nearer the valence band on the p side and nearer the conduction band on the n side; after contact the band edges bend to permit one common value.

In this orientation both the conduction band edge $E_C$ and the valence band edge $E_V$ decrease from p to n by $e{V}_{\mathrm{bi}}$. For a homojunction their separation $E_g=E_C-E_V$ is constant. The acceptor impurity level $E_A$ lies just above $E_V$ where acceptors are present, and the donor impurity level $E_D$ lies just below $E_C$ where donors are present. These impurity levels follow the local band bending; they do not form one additional horizontal band across the device.

The shaded depletion region has few mobile electrons or electron holes and contains fixed negative ionized acceptor impurities on the p side and positive ionized donor impurities on the n side. Thus its electric field points from n to p. Its neutral outer regions have negligible electric field, so the band edges become flat there. In the depletion approximation, charge neutrality requires $N_A{x}_p=N_D{x}_n$; the symmetric drawing illustrates $N_A=N_D$ rather than assuming that equality is universal. Diffusion and field-driven drift current cancel at equilibrium: no applied voltage does not mean zero internal electric field or zero built-in electric potential difference.

A Fourier series separates a periodic physical signal into harmonics. For example, a periodic voltage applied to a linear circuit can be treated one frequency at a time; the harmonic responses then superpose to give the output waveform.

Write $w=\Delta T$ for the pulse width and $\delta =w/T$ for its duty cycle, with $0<w<T$. The waveform is a train of positive rectangular pulses centred on $t=jT$, each with area one. It is even about $t=0$, so every sine coefficient vanishes. Cosine orthogonality givesand, for $n\ge 1$,This is the Fourier series of a periodic rectangular pulse:The actual constant term in the series is $a_0/2=1/T$, not $a_0$. As usual for a jump discontinuity, the series converges to the average of the one-sided values at each pulse edge, namely $1/(2w)$, irrespective of the value assigned at that isolated endpoint. Elsewhere it converges to the rectangular waveform.

Counting the constant as the first term, the first three terms areso their sum isIf “first three” instead means the first three nonconstant harmonics, add $a_3\cos (6\pi t/T)$; the original plot shows that alternative partial sum as well. The three-term approximation rounds the narrow peaks and ripples between them; many harmonics are needed to represent a sharp pulse.

The original sketches use the illustrative duty cycle $\delta =0.15$ and cover $-T\le t\le T$. The coefficient panel also displays the first spectral zero.

For $n\delta \ll 1$, expand the sinc function:Thus the low-index coefficients, including $a_0$, have nearly equal values, although their constant-series contribution still has the factor $1/2$. The first sinc zero occurs at $n\delta =1$, or $n=T/w$. If this is not an integer, the closest integer coefficient is near zero rather than exactly zero: writing $n\delta =1+\epsilon$ gives $a_n\simeq -(2/T)\epsilon$. This pulse-width scale of Fourier coefficients means that a narrower pulse requires significant harmonics extending to proportionally larger indices, the reciprocal time-width/frequency-width relation.

In a charge-free region of homogeneous permittivity, Gauss's law and $\mathbf{E}=-\nabla V$ give Laplace's equation, ${\nabla }^{2}V=0$. Prescribed surface electric potentials specify a Dirichlet problem. Its Uniqueness of the Dirichlet problem permits replacing conductors by fictitious charges outside the physical solution region: a harmonic candidate with the same boundary conditions is the physical solution. For an exterior problem its behaviour at infinity must also be prescribed. For a proof, subtract two candidate potentials to obtain $u$ with zero boundary values. Green's first identity gives $\int |\nabla u{|}^2{d}^{3}x=0$, so $u$ is constant and its boundary value fixes it to zero. Conditions on a normal electric field instead determine a potential only up to an additive constant when the usual compatibility condition holds.

A cylindrical Gaussian surface around an infinite line charge gives $E_r=\lambda /(2\pi {ϵ}_{0}r)$. The electric potential difference is thereforeAn isolated infinite line charge has no finite potential zero at infinity. For the opposite pair, however, the logarithmic growth terms cancel and one can choose $V\to 0$ there. Define $r_b^2=(x-b{)}^2+y^2$ and $r_c^2=(x-c{)}^2+y^2$. superposition givesOn $x^2+y^2=bc$,Thus the circle, and its cylindrical extension, is an equipotential. This is an Apollonius circle: a constant ratio of distances gives a circle rather than a constant distance from either charge.

For the cylinder above the plane, call its centre height $H>a$ and set $s=\sqrt{H^2-a^2}$. The cylinder-plane image charges are $+\lambda$ at height $s$ and $-\lambda$ at height $-s$. Their distances are equal on the grounded plane, while their distance ratio is $(H+s)/a$ on the circular surface. The positive fictitious line lies inside the metal, so the exterior is charge-free and Uniqueness of the Dirichlet problem applies. A Gaussian surface enclosing the physical cylinder confirms that its charge per unit length is indeed $\lambda$. Its voltage isThe physical field energy per unit length at fixed cylinder charge is $U^{\prime }=\lambda V_c/2$. Differentiate with respect to height, keeping $a$ and $\lambda$ fixed:The electrostatic attraction per unit length of a cylinder above a plane is downward for either sign of charge. Placing the fictitious line at the cylinder centre would give the wrong answer at this separation. The long-cylinder approximation neglects the ends.

For inner radius $a$ and outer conductor's inner radius $b$, Gauss's law gives $E(s)=\lambda /(2\pi {ϵ}_0{ϵ}_{r}s)$. Integrating across the dielectric gives voltage $V=\lambda \ln (b/a)/(2\pi {ϵ}_0{ϵ}_r)$. Thus the capacitance per unit length isThe stated outer radius is interpreted as the radius of the dielectric-facing surface. Magnetic permeability affects inductance, but does not enter this electrostatic capacitance.

A transmission line stores electric and magnetic energy along its length, so voltage and current depend on both position and time. A distributed circuit uses inductance $L^{\prime }$ and capacitance $C^{\prime }$ per length, together with resistance $R^{\prime }$ and leakage conductance $G^{\prime }$. Applying voltage and charge balance to a short segment gives the telegrapher's equations,For a lossless line, both variables obey wave equations with speed $v=1/\sqrt{L^{\prime }{C}^{\prime }}$ and characteristic impedance $Z_0=\sqrt{L^{\prime }/C^{\prime }}$. Forward and backward waves obey $I^{+}=V^{+}/Z_0$ and $I^{-}=-V^{-}/Z_0$. Matching a termination $Z_L$ gives the voltage reflection coefficientA matched load absorbs without reflection; an open circuit has ${\Gamma }_L=1$, and a short has ${\Gamma }_L=-1$. Reflections can produce standing waves. A coaxial line confines the fields between conductors and, for a uniform dielectric TEM wave, has $v=1/\sqrt{\mu ϵ}$. Finite loss attenuates waves; frequency-dependent losses or material properties can also distort pulses. A long line is a wave-carrying distributed system, so it cannot generally be replaced by one lumped capacitor.

In microscopic SI form, Maxwell's equations couple charge density $\rho$ and current density $\mathbf{J}$ to electric and magnetic fields:Gauss's electric law expresses electric flux from charge and reduces to the inverse-square Coulomb field for a point charge. Gauss's magnetic law gives zero net magnetic flux through every closed surface; magnetic field lines do not begin or end on magnetic monopoles in this theory.

Faraday's induction law says changing magnetic flux produces a circulating electric field. For a fixed loop, its circulation is minus the rate of change of enclosed magnetic flux, explaining the sign of induction and Lenz's law. For example, changing a primary coil's electric current changes the flux through a stationary secondary coil and induces a voltage. A moving wire also has the magnetic part of the Lorentz force on its charges; the fixed-loop field equation and the full motional-emf calculation must be distinguished.

The Ampère-Maxwell law relates magnetic circulation to conduction current and displacement current. The latter is essential around a charging capacitor: a surface cutting the lead and another passing through the gap must predict the same circulation. Taking the divergence givesusing Gauss's electric law. Charge conservation therefore fixes the consistency of the time-dependent equations. In matter, the equivalent macroscopic equations use $\mathbf{D}$ and $\mathbf{H}$, free sources and constitutive relations. The Lorentz force density $\rho \mathbf{E}+\mathbf{J}\times \mathbf{B}$ describes exchange of momentum with charges.

In source-free vacuum, take the curl of Faraday's law and use $\nabla \times (\nabla \times \mathbf{E})=\nabla (\nabla \cdot \mathbf{E})-{\nabla }^2\mathbf{E}$. Both fields satisfyFor a plane wave, $\omega =ck$, $\mathbf{k}\cdot \mathbf{E}=\mathbf{k}\cdot \mathbf{B}=0$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$. The fields are transverse, mutually perpendicular and in phase. Their energy density and energy flux areThis is Poynting's theorem: field energy is transported and can do work on matter. Electromagnetic radiation is a propagating, energy-carrying solution of the coupled electric and magnetic field equations.

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.

An applied electric field displaces bound charges or reorients permanent dipoles, producing electric polarization $\mathbf{P}$, the dipole moment per unit volume. It gives bound charges ${\rho }_b=-\nabla \cdot \mathbf{P}$ and ${\sigma }_b=\mathbf{P}\cdot \mathbf{n}$. Definefor a linear isotropic dielectric. Then $\nabla \cdot \mathbf{D}={\rho }_f$ separates free charge from the bound-charge response. Electronic, ionic and orientational electric polarization respond on different timescales; real permittivity is frequency dependent, and its dissipative part describes dielectric loss.

A uniform dielectric filling a parallel-plate capacitor raises its capacitance to $ϵA/d$. At fixed charge this reduces field energy $Q^2/(2C)$; at fixed voltage the stored energy $C{V}^2/2$ increases because the voltage source supplies charge and energy. Insulation and energy storage are useful applications, limited by loss and dielectric breakdown. Static interface conditions are continuous tangential $\mathbf{E}$ and a jump in normal $\mathbf{D}$ equal to free surface charge.

In electrostatics, the Maxwell equations in matter areHere ${\rho }_f$ is free charge density and $\mathbf{P}$ is polarization density. For a homogeneous isotropic linear dielectric, the constitutive equation is $\mathbf{D}={ϵ}_0{ϵ}_r\mathbf{E}$. Integrating the first equation over a thin pillbox straddling an interface, with unit normal $\mathbf{n}$ from medium 1 to medium 2, gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Integrating the second around a vanishingly narrow rectangular loop gives $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$. These are the electrostatic interface boundary conditions. When there is no free surface electric charge, the normal electric displacement field and tangential electric field are continuous; the normal electric field generally is not.

Neglect end fringing field of a capacitor. In either axial segment of the coaxial cylindrical capacitor, a cylindrical Gaussian surface gives $D_r={\lambda }_f/(2\pi r)$, where ${\lambda }_f$ is free electric charge per length. Integrating the radial electric field between the conductors givesThe liquid and air segments share the same conductor electric potential difference, so their capacitances add in parallel. Thus the partially filled coaxial capacitor hasHere the question's $ϵ$ is relative permittivity, and air's relative permittivity is approximated by one.

Write $x$ for the rise above the external level. The wide reservoir keeps that level approximately fixed. At constant applied voltage, one must include work done by the source: the appropriate electrical contribution to the mechanical potential is $-C{V}^2/2$. The upward dielectric insertion force at fixed voltage is consequentlyThe excess hydrostatic pressure $\rho gx$ acts over the annular area $A=\pi (b^2-a^2)$. Alternatively its gravitational potential energy is $\rho gA{x}^2/2$, since liquid initially at the reservoir level must be raised through all heights from zero to $x$. Balancing the derivatives givesThis assumes $h+x<L$, a communicating liquid reservoir, negligible capillarity, and a sufficiently long capacitor that end effects are small. Differentiating gives the dielectric-rise voltmeter's sensitivity at a bias $V$:To obtain its numerical value in $\mu \mathrm{m}{\mathrm{V}}^{-1}$, multiply the right-hand side in SI units by $10^6$. A convenient parametric scale, with $\beta =b/a$, isFor a concrete estimate, choose $a=1\mathrm{mm}$, $b=2\mathrm{mm}$, mass density $\rho =10^3\mathrm{kg}{\mathrm{m}}^{-3}$, relative permittivity $ϵ=2$ and operating voltage $V=100\mathrm{V}$. These are illustrative choices, not supplied apparatus data. They giveThus this millimetre-scale example requires fine optical displacement measurement for volt-level changes. The parametric formula states exactly how the estimate changes with the chosen geometry, liquid and bias.

The displacement is quadratic in voltage, cannot distinguish its sign, and has zero first-order sensitivity at zero bias. Smaller gaps and larger relative permittivity increase sensitivity, but capillarity, leakage, dielectric breakdown, temperature-dependent liquid properties, and finite available rise then limit performance. A nonconducting liquid avoids a sustained conduction load, although charging still draws electric charge and leakage is possible. A high voltage can give a readable displacement; a low-voltage instrument would require fine displacement measurement and careful calibration.

The differential form of Faraday's law of induction isBy Stokes theorem, for a stationary oriented circuit it becomes $\mathcal{E}=\oint \mathbf{E}\cdot d\mathbf{\ell }=-d{\Phi }_B/dt$, where ${\Phi }_B=\int \mathbf{B}\cdot d\mathbf{S}$ is magnetic flux and $\mathcal{E}$ is electromotive force. Circuit orientation and surface normal obey the right-hand rule. Lenz's law is the physical meaning of the minus sign: the induced current's magnetic field opposes the change in linked magnetic flux. For a moving circuit, the force per unit electric charge is $\mathbf{E}+\mathbf{v}\times \mathbf{B}$, and its line integral is the corresponding electromotive force. Motion through a static magnetic field can therefore induce a voltage even when ${\partial }_t\mathbf{B}=0$.

Let $\theta$ be the wire's angle from the downward vertical, $r=L\sin \theta$, and choose $\mathbf{\omega }$ and $\mathbf{B}$ upward, as in the figure. The motional term along the wire has radial component $B\omega s$ at cylindrical radius $s$. In steady electrical equilibrium, integrating from the grounded axis to the bob givesThe Lorentz force on this positive electric charge is radially outward. Its ratio to the required centripetal term is $QB/(m\omega )\lesssim C{B}^2{L}^2/(2m)$, so the stated heavy-bob condition makes its effect on the zero-field conical pendulum geometry negligible. With wire tension $T_w$, $T_w\cos \theta =mg$ and $T_w\sin \theta =m{\omega }^{2}r$, giving $\cos \theta =g/({\omega }^{2}L)$. Hence the magnetically charged conical pendulum hasA nonzero-radius orbit requires ${\omega }^{2}L>g$. Reversing the sense of rotation or the magnetic field reverses the sign of $Q_0$.

The response to the added electric field requires an electrical convention. In the usual mechanical approximation, treat the pre-existing electric charge as $Q_0$, include its upward electric force $Q_{0}E$, and neglect all magnetic-force feedback. ThenThis predicts an increased radius for $Q_{0}E>0$. It is the result when the newly applied field is used only as a probe force on the original electric charge.

There is a first-order qualification if the bob remains electrically connected to the grounded pivot and the uniform added field's potential is included consistently. Put $s=\cos \theta$ and set the external electric potential to zero at the pivot. The external potential at the bob is then $ELs$, so the charging relation and full force balance becomeThe second equation follows by eliminating the wire tension between vertical balance and radial balance, including the outward magnetic Lorentz force. Expanding both equations to first order in $E$, and to leading order in $C{B}^2{L}^2/m$, with $s_0=g/({\omega }^{2}L)$, givesThe extra factor arises from the added field changing $Q$ by $-CEL{s}_0$ and therefore changing the radial magnetic force. That term is linear in $E$ too; the heavy-bob assumption alone does not justify retaining $Q_{0}E$ while dropping it. The two expressions refer to the two electrical approximations just specified. The expansion requires a small change relative to the initial radius and is nonuniform as that radius tends to zero.

For an electrical measurement, route the grounded connection through a sensitive electrometer operated near zero input voltage, and integrate the charging electric current when $B$ or $\omega$ is changed. Starting with the field off, the transferred electric charge is $Q_0$ in this ideal capacitance model. Reversing $B$ separates the induced signal from offsets. Alternatively, isolate the charged bob briefly and measure its electric potential with a high-input-impedance electrometer, accounting for the instrument's extra capacitance and for the fact that isolation changes the electrical boundary condition.

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.

The method of images replaces induced charges or currents on a boundary by fictitious sources outside the physical region, chosen to satisfy that boundary's conditions. For example, a point charge $q$ a height $d$ above a grounded conducting plane has an image $-q$ at the reflected point. The sum has zero potential on the plane, and the electrostatic uniqueness theorem identifies it with the physical solution in the upper half-space. Images do not describe a second physical charge below the metal.

For the superconducting plane, use the different boundary condition $B_z=0$ at its surface: magnetic flux cannot enter the ideal bulk. Choose the real ring moment $\mathbf{m}=m\hat{z}$, where $m=I\pi R^2$, at $z=h$. Its image has moment $-m\hat{z}$ at $z=-h$. This superconducting plane image dipole makes the normal fields cancel at $z=0$, while their tangential fields add. The image is a computational continuation; the physical field inside the plane is zero. The original sketch shows the dipole-scale field, appropriate to $h\gg R$; the fine field close to the ring wire is outside that approximation.

In the vacuum region the magnetic scalar potential isAt the surface $z=0$, it reduces to $-mh/[2\pi ({\rho }^2+h^2{)}^{3/2}]$. Its tangential derivative therefore givesBoth $B_z$ and $B_{\varphi }$ vanish there; the sign is inward radially for the chosen positive moment. The expression is the leading dipole result, not an exact finite-ring formula at arbitrary $R/h$.

For the force, calculate the external image field on the axis near the real dipole:Differentiate this field with the image location held fixed, then evaluate at $z=h$. The dipole levitation force above a superconducting plane isdirected upward. Differentiating a substituted separation $2h$ while moving both dipoles in the force derivative would incorrectly double the result.

To levitate ring mass $M$, balance this force against $Mg$:A light, thin wire and the smallest practical gap reduce current. At fixed gap and wire cross-section $s$, $M=2\pi Rs{\rho }_w$ gives $I_{\mathrm{lev}}\propto h^2\sqrt{s{\rho }_w}/R^{3/2}$, favouring a larger ring until the dipole-limit constraint on $R/h$ matters. There is no unique least current without geometric and material constraints. As a geometrically consistent order-of-magnitude example, take $R=1\mathrm{mm}$, $h=10\mathrm{mm}$, wire radius $50\mu \mathrm{m}$ and an assumed metallic-scale density ${\rho }_w=6.0\times 10^3\mathrm{kg}{\mathrm{m}}^{-3}$. Then $M=2.96\times 10^{-7}\mathrm{kg}$ and $I_{\mathrm{lev}}\simeq 2.8\times 10^2\mathrm{A}$. The large value reflects the weak $h^{-4}$ force in the stipulated far-field geometry. Uniformly shrinking this geometrical example reduces current as size to the power $3/2$, but increases required electric current density; arbitrarily thin wire is therefore not a free optimization.

This balance estimate is not a verified superconducting design. The actual wire must carry that current without quenching, its self-field counts toward the limit, the plane must remain in its assumed flux-excluding state, and leads or supports add load. Transport critical-current measurements specify temperature, applied field and the voltage criterion rather than a material-independent limiting current.^[ref]

For a linear nondispersive medium with real absolute permittivity ${ϵ}_a=ϵ{ϵ}_0$ and permeability ${\mu }_0$, instantaneous field energy density and energy flux arewith local balance ${\partial }_{t}u+\nabla \cdot \mathbf{N}=-\mathbf{J}\cdot \mathbf{E}$. energy stored locally and energy passing through a surface are distinct: integrating the balance relates their rates and the power given to charges.

Between the coaxial conductors, cylindrical Gauss and Ampère laws giveThe inner line charge is $\lambda =2\pi {ϵ}_{a}V/\ln (b/a)$, so the capacitance per length is $C=2\pi {ϵ}_a/\ln (b/a)$. Integrating magnetic energy over the annulus gives $\int B^2(2\pi rdr)/(2{\mu }_0)={\mu }_0{I}^2\ln (b/a)/(4\pi )=L{I}^2/2$, hence $L={\mu }_0\ln (b/a)/(2\pi )$. ThusThese are per-length parameters for ideal conductors, excluding internal magnetic penetration.

Now let the voltage and current amplitudes be $V_0,I_0$ and multiply both transverse fields by $e^{i(kz-\omega t)}$. Their divergences vanish in the annular dielectric. The two nontrivial curl equations reduce toConsequently both $k^2={\omega }^2{\mu }_0{ϵ}_a$ and $V_0=Z_0{I}_0$ are required for a forward wave. Arbitrary independent static $V$ and $I$ cannot acquire the same forward-wave factor and still satisfy Maxwell's equations. This TEM travelling-wave voltage-current relation and the dispersion constraint identify the transverse electromagnetic mode and its speed $v_p=1/\sqrt{LC}$.