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Neglect the entrance and exit capacitor fringing fields and use the coaxial electric field between the cylindrical plates. Let $U=V(R_1)-V(R_2)$, with $R_2>R_1$. The radial electric field isFor an electron, of electric charge $-e$, to follow the circular path $R=(R_1+R_2)/2$, its inward centripetal force must be $m_e{v}_0^2/R$. ThereforeThe inner plate must be at higher electric potential: the electric field points outward and the electron's electric force inward. The electric force is perpendicular to the velocity, so no work is done and the speed remains $v_0$. The required voltage is independent of the opening angle $\varphi$, which only sets the transit time $R\varphi /v_0$. For a narrow gap, $\ln (R_2/R_1)\simeq (R_2-R_1)/R$. This is a nonrelativistic, negligible-capacitor fringing field answer; finite open cylindrical sections do not have an exactly radial electric field at their ends.

In vacuum, Gauss's law states that the outward electric flux through a closed surface is the enclosed electric charge divided by the vacuum permittivity:Take opposite electric charges $\pm Q$ on the two electrical conductors, with surface charge densities $\pm \sigma =\pm Q/A$. A Gaussian pillbox crossing one isolated large sheet gives $2ES=\sigma S/{ϵ}_0$, so each sheet contributes $\sigma /(2{ϵ}_0)$ on either side. By the superposition principle, the electric fields add between the sheets and cancel outside. Ignoring the fringing field of a capacitor, the gap electric field is $E=\sigma /{ϵ}_0$. The electric potential difference and capacitance of this parallel-plate capacitor are consequentlyThe approximation requires $d$ much smaller than both transverse sheet dimensions. A uniform filling with permittivity $ϵ$ replaces ${ϵ}_0$ by $ϵ$.

Put electric charge $+Q$ on the inner electrical conductor and $-Q$ on the outer one. In the gap, spherical symmetry makes the electric field radial and constant on a spherical Gaussian surface. Gauss's law givesThe inner-to-outer electric potential difference is thereforeThus the capacitance of the spherical capacitor isAs $b\to \infty$ this gives the capacitance of an isolated conducting sphere, $4\pi {ϵ}_{0}a$; for $b-a\ll a$ it approaches the parallel-plate capacitor result with area $4\pi a^2$.

Use opposite electric charges $\pm Q$ over a common length $L$, and write the inner linear charge density as $\lambda =Q/L$. Away from the ends, axisymmetry and Gauss's law on a coaxial cylindrical Gaussian surface of length $\ell$ giveThe electric potential difference and capacitance of the coaxial cylindrical capacitor areThis neglects the fringing field of a capacitor at the ends, as justified when $L\gg b$.

Let $U>0$ be the drop in electric potential along the positive electric charge's path. By conservation of energy, $qU=m{v}^2/2$, so the nonrelativistic exit speed is $v=\sqrt{2qU/m}$. Apply a uniform magnetic field perpendicular to the incident velocity and to the desired orbital plane. The magnetic Lorentz force is perpendicular to the velocity, does no work, and supplies the centripetal force:For example, $\mathbf{v}=v\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{z}}$ give initial deflection along $-y$ for positive $q$. Reversing the magnetic field reverses the bending. A uniform electric field alone would give constant acceleration and a parabolic path, not uniform circular motion. The displayed result assumes $qU\ll m{c}^2$; the exact relativistic momentum relation would give $R=\sqrt{q^2{U}^2/c^2+2mqU}/(qB)$.

Let the original cylinder axis be $\mathbf{r}=0$, and let the hole axis be at transverse vector $\mathbf{a}$, with $|\mathbf{a}|=a$. The charge density is reproduced by a full cylinder of radius $R_1$ and charge density $+\rho$, superposed with a cylinder of radius $R_2$ centred at $\mathbf{a}$ and charge density $-\rho$. This uses the superposition principle for fixed insulating electric charges; the hole is not a hole in an electrical conductor.

For a complete uniformly charged infinite cylinder of radius $R$, a coaxial Gaussian surface of length $\ell$ gives, by Gauss's law,Consequently defineThe complete electric field, including the exterior if desired, isIn the insulating material, $|\mathbf{r}|<R_1$ and $s=|\mathbf{r}-\mathbf{a}|>R_2$, soInside the hole both contributions use their interior formulas, and all dependence on $\mathbf{r}$ cancels:The electric field in the hole is uniform, points from the original axis towards the hole axis for $\rho >0$, and vanishes for a concentric hole. The two expressions agree at the hole wall, as expected with no imposed surface charge density there. Its uniformity is the useful uniform electric field in an off-axis cylindrical cavity result.

Choose zero electrostatic potential at infinity. By Coulomb's law and the superposition principle, the two point charges giveHere ${ϵ}_0$ is the vacuum permittivity. For a nondegenerate zero-equipotential curve, the electric charges must have opposite signs. Let $d_1,d_2$ be the two distances. The equation $\Phi =0$ becomes $q_1{d}_2=-q_2{d}_1$, and henceFor a circle $r=\rho >0$ centred at the origin, its $\cos \theta$ coefficient must vanish: $q_1^2{r}_2=q_2^2{r}_1$. If $r_1,r_2>0$ and $r_1\ne r_2$, this givesSubstitute $q_2^2/q_1^2=r_2/r_1$ in the constant term:ThereforeThis is an Apollonius circle: the distance ratio $d_2/d_1$ is constant. The distinct-position assumption matters. For example, $r_1=r_2=1$, $q_2=-q_1$ gives zero electrostatic potential everywhere away from the coincident point charges, so there is no distinguished zero circle. If only one position is at the origin and the other is not, no nontrivial positive-radius circle centred at the origin is possible.

Apply the superposition principle to the result of part (a). If $F(a,h)=-qh/[2\pi {ϵ}_0(a^2+h^2{)}^{3/2}]$, then the exact normal electric field from the two real point charges and their image charges isFor $d\ll h$, Taylor's theorem gives $E_z=-d{\partial }_{h}F+O(d^2)$. Sincethe leading electric dipole result isThe electric dipole moment is $\mathbf{p}=-qd\hat{\mathbf{z}}$. The circle $a=\sqrt{2}h$ is the zero of the leading normal electric field and hence of the induced surface charge density. Inside it, induced surface charge density is negative; outside it, positive. The tangential electric field is zero everywhere on the grounded electrical conductor, so the leading full electric field vanishes on that circle. The corresponding dipole-induced charge on a grounded plane integrates to zero over the whole plane: the negative central region is balanced by the positive exterior.

For finite $d$, $a=\sqrt{2}h$ is only the leading-order zero. With $u=h^{2/3}$ and $v=(h+d{)}^{2/3}$, the exact zero satisfies $a^2=uv(u+v)$, giving $a=\sqrt{2}(h+d/2)+O(d^2/h)$. For example, inserting any $d>0$ in the exact expression at $a=\sqrt{2}h$ does not give zero. This is a controlled electric dipole approximation, not an exact finite-separation identity.

The electric field within a conductor in electrostatic equilibrium vanishes. The normal electric field boundary condition, with outward normal vector $+\hat{\mathbf{z}}$, therefore gives $\sigma ={ϵ}_0{E}_z(0^{+})$. Directly below the monopole, $a=0$:For the electric dipole,The exact finite-separation answer iswhose Taylor series agrees with the boxed leading term. Although the real point charges have zero total electric charge, their unequal distances leave negative induced surface charge density immediately beneath them.

Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

The original diagram is a series RLC circuit, so the same electric current $i=\dot{q}$ passes through all three elements. Choose capacitor electric charge $q$ positive on the plate into which positive $i$ enters. Kirchhoff's voltage law givesDividing by inductance $L$ givesThe correspondence to the forced harmonic oscillator is $m\leftrightarrow L$, $b\leftrightarrow R$, $k\leftrightarrow 1/C$, and $x\leftrightarrow q$. The stored energy is $L{i}^2/2+q^2/(2C)$; its balance is $\dot{E}=Vi-R{i}^2$, confirming the damping sign.

Doping changes the mobile carrier population by introducing controlled impurity levels. In an intrinsic semiconductor, the filled valence band is separated from the nearly empty conduction band by a band gap $E_g$. Thermal excitation produces an electron-hole pair, so the electron and hole number densities obey $n=p=n_i$. A electron hole represents a missing valence-band electron and behaves as a mobile positive charge carrier. For nondegenerate bands in thermal equilibrium,Here $N_C,N_V$ are the effective density of states, $E_C,E_V$ the band edges, and $E_F$ denotes the chemical potential, conventionally called the Fermi level. The semiconductor mass-action law holds in this nondegenerate equilibrium regime.

In a tetrahedrally bonded elemental semiconductor such as silicon, a substitutional group-V atom has an extra outer electron relative to the host. It is a donor impurity: a shallow level just below $E_C$ can ionise, releasing an electron to the conduction band and leaving a fixed positively charged donor. The resulting n-type semiconductor has electrons as majority carriers and holes as minority carriers. Conversely a group-III substitution supplies one fewer bonding electron. An acceptor impurity can take an electron from the valence band, leaving a mobile electron hole and a fixed negative acceptor; the resulting p-type semiconductor has holes as majority carriers.

The impurity state is shallow because the electron or hole responds to a screened Coulomb potential and an effective mass, rather than the bare atomic potential. A simple hydrogenic donor estimate iswhere ${ϵ}_r$ is the relative permittivity. This is a useful scaling model, with real band anisotropy and central-cell effects omitted. Dopants are fixed in the lattice; their ionisation changes mobile carrier numbers without making a uniformly doped bulk sample carry a large net electric charge.

The charge neutrality condition isWhen shallow dopants are almost fully ionised and $N_D-N_A\gg n_i$, an n-type region has $n\simeq N_D-N_A$ and $p\simeq n_i^2/n$. A p-type region similarly has $p\simeq N_A-N_D$ and $n\simeq n_i^2/p$. Thus compensation by the opposite impurity type reduces the majority-carrier density; the total dopant concentration is not generally the mobile carrier density. At sufficiently high impurity density the degenerate semiconductor regime requires the full Fermi-Dirac distribution and these exponential approximations fail.

There are three useful temperature regimes. At low temperature, dopant freeze-out leaves many shallow levels un-ionised. In the extrinsic regime, the dopants are largely ionised and set the majority-carrier density. At high enough temperature, intrinsic pair creation dominates and $n,p$ approach $n_i$. The Fermi level moves towards the conduction band with donor doping and towards the valence band with acceptor doping. The electrical conductivity isDoping therefore controls conductivity and its carrier type, but ionised impurities also scatter carriers and can reduce charge carrier mobility. Carrier density and mobility must both be considered.

A p-n junction forms when p-type and n-type regions are joined within a semiconductor. Initially the concentration gradients drive diffusion current: electrons diffuse from the n region to the p region and holes in the opposite direction. Recombination removes mobile majority carriers near the join, exposing fixed positive donors on the n side and fixed negative acceptors on the p side. This forms a depletion region and an electric field directed from n to p. Its drift current opposes diffusion. At equilibrium the net current vanishes and the Fermi level is uniform, although the band edges vary spatially through band bending.

For an abrupt nondegenerate junction with fully ionised dopants, its built-in potential isUnder the depletion approximation, the space-charge densities are $-e{N}_A$ on a p-side width $w_p$ and $+e{N}_D$ on an n-side width $w_n$. Integrating Poisson equation and imposing zero field outside gives $N_A{w}_p=N_D{w}_n$ andMost of the depletion width therefore extends into the more lightly doped side. The schematic shows the impurity levels before contact and the resulting space charge and potential after contact; there is no externally extractable equilibrium battery voltage from the built-in field alone.

A forward bias, with p positive relative to n, reduces the barrier from $V_{\mathrm{bi}}$ towards $V_{\mathrm{bi}}-V$ and injects minority carriers across the junction. In the ideal low-injection, diffusion-controlled regime this produces the Shockley diode equation, $I=I_s[e^{eV/(k_{B}T)}-1]$. Reverse bias increases the barrier and the depletion region; only a small thermally generated minority-carrier current flows before breakdown. Reverse-biased depletion regions also give a voltage-dependent junction capacitance. Real devices have recombination, leakage, series resistance and breakdown mechanisms beyond the ideal law. Controlled doping creates majority and minority carriers, and their junction produces a depletion barrier that makes rectification possible.

The Hall effect is a transverse electric response to a magnetic force on moving charge carriers. Take conventional current density $\mathbf{J}=J_x\hat{\mathbf{x}}$ and magnetic field $\mathbf{B}=B_z\hat{\mathbf{z}}$ in a conducting slab. For one carrier type of number density $n$ and signed electric charge $q$, the drift velocity is ${\mathbf{v}}_d=\mathbf{J}/(nq)$. Carriers initially deflect sideways until a Hall electric field balances the magnetic Lorentz force:The Hall coefficient is thereforeFor electrons it is $-1/(ne)$, whereas a simple hole-dominated conductor gives $+1/(pe)$. The sign reveals the effective carrier sign; the magnitude estimates their density in the one-carrier approximation.

For width $w$ along $y$ and thickness $t$ along $z$, $I=J_{x}wt$. Define the signed Hall voltage by $V_H=E_{y}w$, namely the electric potential at the lower-$y$ edge minus that at the higher-$y$ edge. ThenThis explicit contact convention fixes the voltage sign. A Hall bar with longitudinal and transverse contacts measures both electrical resistivity and the Hall response. Reverse the field and take the part of the transverse voltage odd in $B_z$ to suppress longitudinal-voltage pickup from misaligned contacts; reversing current provides another useful check.

A Hall measurement is useful for carrier identification, semiconductor doping measurements and magnetic-field sensing. Combining a one-carrier $R_H$ with electrical conductivity gives a mobility magnitude $\mu =\sigma |R_H|$. However, mixed electron and hole conduction is not described by a single density: in weak fields,The response weights mobilities differently from ordinary conductivity. Complicated energy bands, anisotropic Fermi surfaces and multiple bands can likewise change the simple interpretation in metals; a positive Hall coefficient does not mean that the material contains positively charged fundamental electrons.

Twice the quadratic coefficient from the previous part is the band curvature $E_{\pm }^{\prime \prime }(G_1/2)$. Thus the band effective masses areHere $U=|V(G_1)|$ and $E_B={\hslash }^2{G}_1^2/(8m)$. In the weak-potential regime $U\ll E_B$,Both have units of mass; their magnitudes are much smaller than $m$ in this asymptotic regime. The upper energy band has a local minimum and positive effective mass, while the lower energy band has a local maximum and negative effective mass. The exact lower-branch sign changes at $U=2E_B$ in the two-wave formula; the intended weak-coupling limit lies below that threshold.

The physical meaning follows from the semiclassical equations for an electron wave packet in one energy band:An upper-band electron near its minimum accelerates in the direction of the applied force, whereas a lower-band electron near its maximum accelerates opposite to it. At the extremum itself the group velocity is zero. A nearly filled lower energy band is conveniently described by electron holes with positive electric charge $+e$ and positive effective mass $m_h^{\ast }=-m_{-}^{\ast }$. Negative curvature is a response of a crystal energy eigenstate, not a negative bare electron mass. These statements assume slow forcing that does not induce transitions across the band gap.

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.

Choose the guide axis as $z$, with $0<x<a$ and $0<y<b$. The electric field is along $y$, parallel to the walls' $b$ dimension. Take a homogeneous lossless nonmagnetic dielectric with relative permittivity $ϵ$, ideal conducting walls, and time dependence $e^{-i\omega t}$. The lowest transverse electric mode of a rectangular waveguide isAt $x=0,a$, this tangential electric field vanishes. At $y=0,b$ the electric field is normal to the wall; a nonzero value is supported by surface electric charge and violates no conducting boundary condition. The field has zero divergence. The associated normal magnetic field also vanishes at each wall, as verified below. The fundamental sine across the wider dimension minimizes the nonzero transverse wave number; ${\mathrm{TE}}_{00}$ is not a physical propagating mode of a hollow simply connected guide.

Below the waveguide cutoff frequency $f_c=c/(2a\sqrt{ϵ})$, put $\beta =i\kappa$ with $\kappa >0$ to select decay for $z>0$. The evanescent transverse electric mode in a rectangular waveguide then hasFor the fixed frequency in the question,The decay length increases monotonically and diverges as $ϵ\to 4^{-}$; its formal limit at $ϵ\to 0^{+}$ is $a/\pi$. At $ϵ=4$ the mode is at cutoff, with zero axial wave number. For $ϵ>4$, $\beta =(\pi /a)\sqrt{ϵ/4-1}$ is real and the mode propagates without axial attenuation in this ideal lossless model. There is no real finite evanescent decay length to continue beyond cutoff. The paper's graph notation $l$ denotes the same $\ell$ as its preceding formula.

From Faraday's law of induction, $\nabla \times \tilde{\mathbf{E}}=i\omega \tilde{\mathbf{B}}$. Writing $q=\pi /a$ givesThus $B_x$ vanishes on the $x$ walls and $B_y$ vanishes on the $y$ walls, meeting the normal magnetic boundary conditions. The fields also satisfy $\nabla \cdot \mathbf{B}=0$. For real $E_0$ the physical fields are $E_y=E_0\sin (qx)e^{-\kappa z}\cos \omega t$, $B_x=-(\kappa /\omega )E_0\sin (qx)e^{-\kappa z}\sin \omega t$, and $B_z=-(q/\omega )E_0\cos (qx)e^{-\kappa z}\sin \omega t$.

The time-averaged Poynting vector isAll products are imaginary because the electric field and magnetic field are in quadrature. A single decaying evanescent wave stores reactive energy without transmitting average power into a semi-infinite lossless guide. A finite below-cutoff section can transmit power when both independent evanescent solutions are present; that is a different boundary-value problem.

The skin effect also gives exponential field decay, but in a good metal $k$ has both real and imaginary parts and there is a real inward mean Poynting vector. That flux supplies Joule heating $\sigma |\tilde{E}{|}^2/2$. Here attenuation comes from geometric cutoff in a lossless region, with no conductivity and no dissipative power sink; exponential decay alone does not establish absorption.

An electric dipole consists of equal and opposite electric charges separated by a displacement $\mathbf{d}$. Its electric dipole moment is $\mathbf{p}=q\mathbf{d}$, directed from negative to positive electric charge. At distances large compared with the separation, its leading electric potential and electric field areThere is no monopole term because the net electric charge is zero. The axial electric field points along $\mathbf{p}$ with magnitude $2p/(4\pi {ϵ}_0{r}^3)$, while the equatorial electric field points oppositely with half that magnitude.

A uniform applied electric field exerts zero net force on an ideal fixed electric dipole, but it exerts torque $\mathbf{\tau }=\mathbf{p}\times \mathbf{E}$. Its orientation-dependent potential energy is $U=-\mathbf{p}\cdot \mathbf{E}$, so alignment with the electric field is stable. A nonuniform applied electric field can also exert a net force, $\mathbf{F}=\nabla (\mathbf{p}\cdot \mathbf{E})$ for a fixed permanent electric dipole moment.

A permanent electric dipole moment exists without an applied electric field. An induced electric dipole moment instead develops in response to the local electric field; in an isotropic linear model $\mathbf{p}=\alpha \mathbf{E}$. Creating that induced moment requires accounting for the work of polarization, so its mechanical potential energy in a specified applied electric field is $-\alpha E^2/2$, rather than $-\alpha E^2$. In matter, the dipole moment per unit volume is the polarization density $\mathbf{P}$.

Free charge is charge assigned to externally supplied or mobile carriers in a macroscopic description. Bound charge describes the net charge associated with a material's polarization density $\mathbf{P}$. Its bulk and surface densities arewhere the normal points out of the polarized material. Uniform bulk polarization density has no bulk bound charge, but it can still have surface bound charge.

The electric field is sourced by the total electric charge. Introducing the electric displacement field separates the free source in the macroscopic field equation:This does not mean that bound charges are physically absent from $\mathbf{D}$: they are represented through $\mathbf{P}$. In a linear isotropic dielectric, $\mathbf{P}={ϵ}_0{\chi }_e\mathbf{E}$ and $\mathbf{D}=ϵ\mathbf{E}$, with $ϵ={ϵ}_0(1+{\chi }_e)$.