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Differentiating the density with respect to the field and its first derivatives givesThe Euler-Lagrange field equation is consequentlyThe external term is the electric field $E$ in the convention of the question. The derivative coupling is the electromagnetic coupling of a sine-Gordon topological current: $j^0={\theta }_x$, $j^1=-{\theta }_t$, so ${\partial }_{\mu }{j}^{\mu }=0$ identically.

When $E=0$ and $X=X_0+ut$, both $u$ and $\gamma =(1-u^2{)}^{-1/2}$ are constant. Put $y=\gamma (x-X)$ and $K(y)={\theta }_K(y)$. Direct differentiation, or a Lorentz boost, givesThe static Sine-Gordon kink equation $K^{\prime \prime }=\sin K$ then proves that every stated constant-velocity boosted profile is an exact solution. The range $|u|<1$ makes the boost real and subluminal.

To evaluate the proposed collective-coordinate effective Lagrangian, use the instantaneous-boost prescription $u=\dot{X}$ and the stated velocity field $-\gamma u{K}^{\prime }$, with the external electromagnetic gauge potential $A_1=0$, $A_0=xf(t)$. The kink hassoThe last identity uses the evenness of $K^{\prime }$. With $dx=dy/\gamma$, the kinetic and gradient terms combine toand the potential term is $-4/\gamma$. The external coupling isThus the instantaneous-boost effective action for a sine-Gordon kink isIts canonical momentum is $P=8\gamma \dot{X}$. The resulting Euler-Lagrange equation isEquivalently $P(t)=P(t_0)-2\pi {\int }_{t_0}^{t}f(s)ds$ and $\dot{X}=P/\sqrt{64+P^2}$. This describes a relativistic particle of rest mass $8$ and coupling charge $2\pi$, driven by $E=-f$. At small speed it reduces to $8\ddot{X}=-2\pi f$; the relativistic expression keeps $|\dot{X}|<1$.

There is also an exact force check at field level. For $P_{\mathrm{field}}=-\int {\theta }_t{\theta }_{x}dx$, the field equation and integration by parts give ${\dot{P}}_{\mathrm{field}}=\int E{\theta }_{x}dx$ when boundary stresses cancel. If $E=-f(t)$ and the field has kink boundary difference $2\pi$, this is $-2\pi f(t)$. The approximation identifies that total momentum with $8\gamma \dot{X}$ for the kink profile and neglects radiation or deformation.

The approximation's scope is important. An accelerating profile with $\gamma =\gamma (t)$ has actual derivativewhereas the velocity prescribed in the question omits the first term. Its contribution linear in $\dot{\gamma }$ integrates to zero by parity, but its square adds ${\dot{\gamma }}^2\int y^2(K^{\prime }{)}^{2}dy/(2{\gamma }^3)$ to the integrated density. Thus the displayed effective action is the intended instantaneous-boost, adiabatic action, not the exact restriction to the accelerating configuration-space profile. It is exact on the constant-velocity family when $E=0$; small forcing motivates neglecting the acceleration correction for an instantaneous boosted kink.

Take $z$ positive outward and define the separation as upper minus lower position. The Newtonian potential of a point mass gives the acceleration $a(z)=-{\partial }_z\varphi =-GM/z^2$. Expanding its difference across a separation $\zeta \ll z$ gives the tidal force equationThe upper particle falls slightly less rapidly, so the separation increases. If the upper particle carries the electric charge and the electric field points outward, the extra relative acceleration is $qE/m$. Reversing the field or placing the charge on the lower particle changes its sign. Write this signed acceleration as $a_e$ and put $k=2GM/z^3$. With initial relative velocity zero and nearly constant $k$ and $a_e$, the ordinary differential equation has solutionFor $k{t}^2\ll 1$, the leading changes due to the tidal force and electric field are respectivelywith the stated sign convention. Thus, for a nonzero fixed electric acceleration,A laboratory of size $L$ has $|{\zeta }_0|\le L$, so this ratio tends to zero as $L\to 0$. This is local tidal acceleration scaling: gravitational relative acceleration is proportional to separation, whereas the electric difference persists even at coincident positions.

The duration also matters. At fixed time, $\Delta {\zeta }_g\to 0$ in absolute size, but $\Delta {\zeta }_g/{\zeta }_0\simeq (GM/z^3)t^2$ need not tend to zero. To keep the electrically displaced particles in the shrinking laboratory, take $|a_e|t^2\lesssim L$; then $|\Delta {\zeta }_g|/L\lesssim GML/(z^3|a_e|)\to 0$. Alternatively a stationary terrestrial laboratory has a free-fall time of order $\sqrt{L/g}$, giving $|\Delta {\zeta }_g|/|{\zeta }_0|=O(L/z)$. Both make precise the small spacetime-region limit. An extended, long-running experiment can still measure arbitrarily small tidal forces.

The weak equivalence principle says that freely falling test particles with the same initial data have composition-independent motion. In particular, a freely falling frame removes the common gravitational acceleration of both particles. The electric accelerations depend on charge and cannot be removed for charged and uncharged particles simultaneously by the same change of frame.

The Einstein equivalence principle extends the freely falling laboratory statement to all local nongravitational experiments: their outcomes obey special relativity, with local Lorentz invariance and local position invariance. The strong equivalence principle also includes local gravitational experiments and bodies with significant gravitational binding energy. It asserts universality of their free fall and independence of local experimental outcomes from the laboratory's location and velocity. The two-particle experiment illustrates the local distinction between gravity and a nongravitational force; by itself it does not establish this stronger claim about self-gravitating bodies.

In a metric theory, the common freely falling trajectories are timelike geodesics of a Lorentzian metric. Normal coordinates make $g_{ab}={\eta }_{ab}$ and the Christoffel symbols zero at one point. Fermi normal coordinates do the same along a reference timelike geodesic. The residual effects are controlled by the Riemann curvature tensor through geodesic deviation, and enter the local metric at quadratic order in distance. Curvature is a physical tidal field, so no coordinate choice removes it over a finite region.

A relativistic theory should consequently couple matter universally to the metric tensor, recover nongravitational laws in local inertial frames and recover the Newtonian limit. General relativity implements this with the Einstein field equations; the contracted Bianchi identity is consistent with stress-energy conservation. The equivalence principle motivates this geometric description but does not alone uniquely determine the field equations. Extra gravitational fields may satisfy the Einstein equivalence principle for ordinary matter while failing the strong equivalence principle because local gravitational measurements or self-gravitating trajectories depend on those fields.

The three contributions to the Frank elastic energy penalize distinct slow distortions of the nematic director. Nematic splay measures local spreading or convergence and is governed by $K_1(\nabla \cdot \mathbf{n}{)}^2/2$. Nematic twist measures rotation around an axis perpendicular to the director, with cost $K_2(\mathbf{n}\cdot \nabla \times \mathbf{n}{)}^2/2$. Nematic bend measures curvature of director integral lines; for a unit vector, $(\mathbf{n}\cdot \nabla )\mathbf{n}=-\mathbf{n}\times (\nabla \times \mathbf{n})$, giving cost $K_3|\mathbf{n}\times \nabla \times \mathbf{n}{|}^2/2$. These terms respect the head-tail symmetry of a nematic liquid crystal. Positive elastic constants oppose these distortions.

For the one-dimensional reduction assume lateral uniformity, so $\theta$ and $\varphi$ depend only on $z$. A prime below means $d/dz$. Directly differentiating the nematic director givesThe squared curl is ${\sin }^2\theta {\theta }^{\prime 2}+{\cos }^2\theta {\varphi }^{\prime 2}$. Subtracting the squared twist invariant givesConsequently the one-dimensional twisted nematic energy has elastic coefficientsA normal electric field has $\mathbf{E}\cdot \mathbf{n}=E_0\sin \theta$. There is a genuine factor-of-two inconsistency in the source: its initial functional places the electric term inside the overall factor $1/2$, whereas its subsequent density and threshold use the full electric term. For the requested density and threshold, use the usual orientational coupling $-ϵE_0^2{\sin }^2\theta /(8\pi )$. Here $ϵ$ is the dielectric anisotropy of a nematic, in the printed Gaussian-unit convention. If the first functional is instead read literally, this coupling is halved and the resulting threshold voltage is multiplied by $\sqrt{2}$.

Put $c=ϵE_0^2/(8\pi )$ and $\mathcal{L}=f{\theta }^{\prime 2}/2+g{\varphi }^{\prime 2}/2-c{\sin }^2\theta$. Apply the Euler-Lagrange equations for two fields. Variation in $\varphi$, which does not appear explicitly, givesVariation in $\theta$ gives $(f{\theta }^{\prime }{)}^{\prime }-f_{\theta }{\theta }^{\prime 2}/2-g_{\theta }{\varphi }^{\prime 2}/2+2c\sin \theta \cos \theta =0$, orAt $\theta =0$, $f=K_1$, $g=K_2$, and $f_{\theta }=g_{\theta }=0$. With zero field, ${\theta }^{\prime }=0$ and ${\varphi }^{\prime }=\pi /(2L)$ satisfy both equations and the imposed azimuthal endpoints. Thus the quarter-turn planar texture is a stationary solution. Stationarity alone is not a claim of stability for arbitrary elastic constants.

To find the Fréedericksz transition, impose the additional strong planar anchoring $\theta (0)=\theta (L)=0$. Write $\alpha =\pi /(2L)$. The small-angle expansions areThe conserved $g{\varphi }^{\prime }$ and fixed total twist imply ${\varphi }^{\prime }=\alpha +O({\theta }^2)$ for a pure tilt perturbation. Linearizing the tilt Euler-Lagrange equation therefore givesSubstitute $\theta ={\theta }_0\sin (\pi z/L)$. A nonzero neutral amplitude requiresSince $V_c=E_{c}L$, the quarter-turn nematic instability threshold isThis is the first zero-endpoint mode: higher sine modes replace $K_1$ by $n^2{K}_1$ and cost more elastic energy. The quadratic energy change is $\frac{1}{2}\int [K_1{\theta }^{\prime 2}+\{(K_3-2K_2){\alpha }^2-ϵE_0^2/(4\pi )\}{\theta }^2]dz$, so the first mode has negative stiffness above this voltage. The result assumes $ϵ>0$ and $K_1+(K_3-2K_2)/4>0$. If the latter is negative the planar state is already unstable at zero field; for negative dielectric anisotropy of a nematic, a normal field opposes this tilt. The literal initial electric prefactor would give $V_{c,\mathrm{literal}}=\sqrt{2}{V}_c$.

For reflective operation of the twisted nematic field effect, orient each polarizer's transmission axis along the adjacent planar nematic director. Light entering through the top polarizer initially has $y$ linear polarization. At low voltage, adiabatic optical following in a twisted nematic rotates it through a quarter turn to $x$ as it travels downward, so it passes the bottom polarizer and reaches the mirror. On returning, it follows the reverse twist back to $y$ and exits the top polarizer: the pixel appears bright. At sufficiently high voltage, positive dielectric anisotropy of a nematic aligns most of the director with the normal field. The bulk no longer produces the quarter-turn optical rotation, so the crossed bottom polarizer blocks the incident light and the pixel appears dark. Elastic relaxation restores the bright twisted state after voltage is removed. The adiabatic requirement is more precisely $2\pi |\Delta n|/\lambda \gg |{\varphi }^{\prime }|$, involving birefringence, not thickness alone. Voltage switches director orientation and hence the reflected brightness; it does not require the liquid crystal to emit light.

In electrostatic equilibrium, the electric field inside a perfect conductor is zero. Apply the Faraday's law $\oint \mathbf{E}\cdot d\mathbf{l}=0$ to a thin rectangular loop crossing the surface, with its long sides tangential and its short sides shrinking to zero. It yields continuity of the tangential electric field. Since its interior value is zero, $E_x(0+)=E_y(0+)=0$. Earthing fixes the electric potential at the plane to zero. For completeness, a thin Gauss law pillbox gives $E_z(0+)=\sigma /{\epsilon }_0$; the normal component can be nonzero because of induced surface charge. These are the electrostatic boundary conditions at a conductor.

The vacuum Maxwell equations are $\nabla \cdot E=\nabla \cdot B=0$, $\nabla \times E=-B_t$ and $\nabla \times B=E_t/c^2$. Taking the curl of Faraday's equation and using $\nabla \times (\nabla \times E)=\nabla (\nabla \cdot E)-{\nabla }^{2}E$ gives $-{\nabla }^{2}E=-E_{tt}/c^2$. The same calculation for Ampère's equation gives ${\nabla }^{2}B=B_{tt}/c^2$. Hence each Cartesian component satisfies the electromagnetic wave equation.

Model the walls as a perfect conductor, with no oscillating fields in its interior. The tangential electric field is zero at the surface, otherwise it would drive unbounded conductor currents. The normal magnetic field is zero, since the divergence-free field has a continuous normal component and the interior oscillating field is zero. Thus $n\times E=0$ and $n\cdot B=0$. Normal $E$ and tangential $B$ may be nonzero because surface charge and surface current support their jumps.

The PDF writes the full magnetic field as $B=B_0(x,y)\hat{z}{e}^{i(kz-\omega t)}$. Taken literally, this cannot be a nonzero travelling wave: $\nabla \cdot B=ik{B}_z$, so for $k\ne 0$ it forces $B_z=0$. The subsequent requests require transverse components. The consistent intended reading is $B_z=B_0(x,y)e^{i(kz-\omega t)}$, with transverse fields determined by the Maxwell equations. Use that reading below.

All fields have the same $z,t$ factor. Put $\Delta =(\omega /c{)}^2-k^2$. The harmonic Maxwell equations are $\nabla \times E=i\omega B$ and $\nabla \times B=-i\omega E/c^2$. Their $x$ and $y$ components givetogether with ${\partial }_y{E}_z-ik{E}_y=i\omega B_x$ and $ik{E}_x-{\partial }_x{E}_z=i\omega B_y$. Substitution and solving, for $\Delta \ne 0$, givesThe corresponding electric components areFor a transverse electric mode of a rectangular waveguide, $E_z=0$. The electromagnetic wave equation becomesAt $x=0,a$, the tangential component $E_y=-i\omega {\partial }_x{B}_z/\Delta$ vanishes, so ${\partial }_x{B}_0=0$; at $y=0,b$, the corresponding condition $E_x=0$ gives ${\partial }_y{B}_0=0$. These are Neumann boundary conditions, and the normal-$B$ conditions give the same conclusions through the displayed transverse magnetic components. Separation giveswhere $m,n\ge 0$ are integers and $(m,n)\ne (0,0)$. The constant candidate is not a physical oscillating mode: its transverse derivatives vanish, so it has $E=0$ and Faraday's equation forces $B=0$ for $\omega \ne 0$. More generally a hollow single-conductor guide has no transverse-electromagnetic zero-cutoff mode: its transverse electrostatic potential is harmonic with the same constant boundary value, hence constant, leaving zero transverse field.

The dispersion relation is ${\omega }^2=c^2(k^2+{\kappa }_{mn}^2)$. The waveguide cutoff frequency for this mode is ${\omega }_{mn}=c{\kappa }_{mn}$; below it $k$ is imaginary and the field is evanescent. Since $a\ge b$, the lowest cutoff occurs at $(m,n)=(1,0)$:For a mode propagating in the positive $z$ direction ($k>0$), above its own cutoff ${\omega }_{mn}$, the phase velocity and group velocity areThese waveguide phase and group velocities have product $c^2$. The superluminal phase speed describes motion of a constant phase, while a narrow wave packet travels at the subluminal group velocity; it does not transmit information faster than light. At cutoff $k=0$ there is no travelling mode with nonzero longitudinal flux.

The uniform charge density is ${\rho }_e=3Q/(4\pi R^3)$. Spherical symmetry makes the electric field radial and constant in magnitude on each concentric sphere. A sphere of radius $r<R$ encloses charge $Q(r/R{)}^3$. Gauss's law therefore gives $4\pi r^2{E}_r=Q{r}^3/({\epsilon }_0{R}^3)$, soThis also defines the zero electric field at the centre by continuity.

Take the upper hemisphere to have $z>0$. Integrate the force $\mathbf{E}dq$ using the total interior electric field from part (a). This computes the net external force on the upper hemisphere, since the forces between pairs of its own charge elements cancel by antisymmetry of the Coulomb force. Horizontal components cancel by symmetry. The vertical component isA horizontal section at height $z$ has area $\pi (R^2-z^2)$, soThus the lower hemisphere repels the upper one vertically upward, withThe result is the same for either sign of $Q$, as expected for the force between hemispheres of a uniformly charged ball.

Take the inner conductor to be at potential $V$ above the outer. A cylindrical Gauss law surface of length $\ell$ encloses charge $\lambda \ell$ and gives $E_{r}2\pi r\ell =\lambda \ell /{ϵ}_0$. The potential difference isThus the coaxial cylindrical capacitor field and charge areThe perfectly conducting surfaces have no tangential electric field in this idealized steady model.

Let ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$ for a constant relative velocity $\mathbf{v}$. The Galilean transformation gives ${\nabla }^{\prime }=\nabla$ and ${\partial }_{t^{\prime }}={\partial }_t+\mathbf{v}\cdot \nabla$. In the magnetic Galilean limit of the pre-Maxwell equations, takeThe Ampère's law without displacement current and the solenoidal constraint are immediately unchanged. For Faraday's law, the constant-velocity vector identity is $\nabla \times (\mathbf{v}\times \mathbf{B})=\mathbf{v}\nabla \cdot \mathbf{B}-(\mathbf{v}\cdot \nabla )\mathbf{B}$. ThusThese are transformations of the magnetic nonrelativistic approximation. The exact charge-current transformation also contains the convective term $-{\rho }_e\mathbf{v}$ in ${\mathbf{J}}^{\prime }$; ignoring it requires the quasineutral magnetic limit. They are not exact transformations of the full Maxwell equations.

Since ${\mathbf{u}}^{\prime }=\mathbf{u}-\mathbf{v}$, the combination ${\mathbf{E}}^{\prime }+{\mathbf{u}}^{\prime }\times {\mathbf{B}}^{\prime }$ equals $\mathbf{E}+\mathbf{u}\times \mathbf{B}$. For an ideally conducting fluid the rest-frame electric field vanishes, so the ideal moving-conductor Ohm law is $\mathbf{E}+\mathbf{u}\times \mathbf{B}=0$. Substitution in Faraday's law givesThe ideal magnetohydrodynamic induction equation requires nonrelativistic fluid motion, negligible displacement current, and negligible nonideal terms in the electric-field relation. In particular, magnetic diffusion must be slow compared with advection: the magnetic Reynolds number $\mathrm{Rm}=UL/{\eta }_m\gg 1$, where ${\eta }_m=({\mu }_0\sigma {)}^{-1}$ is the magnetic diffusivity. The single-fluid moving-conductor Ohm law must apply on the scales considered, with corrections to that relation negligible; viscosity is not itself excluded by this induction equation.

For a smooth ideal evolution, magnetic flux freezing states that the magnetic flux through every material surface is conserved and that magnetic field lines are carried with the fluid. Their connectivity is preserved: ideal smooth advection does not reconnect them. These implications are stated here without proof, as requested.

Write $f=1-2M/r$ in the Schwarzschild metric. For the stationary Killing vector field $V={\partial }_t$, its dual is $V^{♭}=-fdt$ and the Papapetrou electromagnetic field isOutside the Schwarzschild event horizon, use the static orthonormal coframe $e^{\hat{0}}=\sqrt{f}dt$, $e^{\hat{r}}=dr/\sqrt{f}$, $e^{\hat{\theta }}=rd\theta$, $e^{\hat{\varphi }}=r\sin \theta d\varphi$. Then $F=(2M/r^2)e^{\hat{0}}\wedge e^{\hat{r}}$. With the electric field convention $E_{\hat{i}}=F_{\hat{i}\hat{0}}$ from the preceding part,Thus the stationary Killing vector field gives a Coulomb electric field, directed inward in this convention. If the charge is normalized by $E_{\hat{r}}=Q/r^2$, its charge is $Q=-2M$. A convention with $F\mapsto -F$ reverses that charge; the invariant content is a radial monopole electric field of magnitude $2M/r^2$.

For the axial Killing vector field $V={\partial }_{\varphi }$, $V^{♭}=r^2{\sin }^2\theta d\varphi$, givingIn the same static orthonormal coframe, this is a purely magnetic field withAt large $r$ these are the spherical components of $2{\mathbf{e}}_z$. This is therefore the regular magnetic test field asymptotic to a uniform axial field, distorted by the Schwarzschild black hole. Its net magnetic monopole flux is zero because the integral of $\cos \theta$ over a sphere vanishes. Both electromagnetic fields are regular at the future event horizon: in ingoing Eddington-Finkelstein coordinates, the electric field is $(2M/r^2)dv\wedge dr$, and the magnetic two-form already contains only regular spatial coordinates. Static observers themselves cease to exist on the event horizon.

Translation invariance in $y$, together with a $y$-independent incident field, permits a $y$-independent solution. In homogeneous space Maxwell's equations then separate into two scalar sectors.

For a TE wave, set $\mathbf{E}=\hat{\mathbf{y}}{E}_y(x,z)$. Its component satisfies $({\partial }_x^2+{\partial }_z^2+k^2)E_y=0$. The $y$ direction is tangent to the conducting surface, so vanishing tangential electric field imposes $E_y(x,h(x))=0$. Therefore TE scattering is exactly the two-dimensional Dirichlet scalar problem, with $E_y$ replacing the acoustic scalar. Its first-order boundary expansion and angular-spectrum formula are those above.

For a TM wave, set $\mathbf{H}=\hat{\mathbf{y}}{H}_y(x,z)$. Maxwell's equation $\nabla \times \mathbf{H}=-i\omega ϵ\mathbf{E}$ givesA surface tangent is $(1,0,h^{\prime })$, so vanishing tangential electric field givesThus TM also reduces to two dimensions, but to a Neumann problem, not the Dirichlet problem in part (a). This is scalar polarization reduction at a conducting corrugation.

For example the flat TM reflection has positive sign, $H^{[0]}=A_i{e}^{i{p}_{i}x}(e^{-i{q}_{i}z}+e^{i{q}_{i}z})$. Its linearized rough forcing isIn the transform domain this becomes $2A_i(k^2-p_i\xi )\hat{h}(\xi -p_i)$, and propagation divides by $i\beta (\xi )$ rather than prescribing the electric Dirichlet trace. The explicit slope term and different reflection sign show why simply reusing the TE boundary formula for TM would be incorrect.

Use units $c=1$, consistent with comparing $B$ directly to $E$, and metric signature $(+---)$. For an arbitrary worldline parameter $s$, an action for a charged relativistic particle isThe first term has variation $-m{\dot{x}}_{\mu }/\sqrt{{\dot{x}}^2}$ in its momentum, and the coupling gives $-q{A}_{\mu }$. The Euler-Lagrange equations therefore readWhen $s$ is proper time, $u^{\mu }={\dot{x}}^{\mu }$ has $u^{\mu }{u}_{\mu }=1$, giving the covariant Lorentz forceWith $F^{i0}=E_i$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, put $e=qE/m$, $b=qB/m$, $\omega =\sqrt{b^2-e^2}$ for nonzero charge. The velocity equations becomeThus ${\ddot{u}}^z=-{\omega }^2{u}^z$, with $u^z(0)=0$, ${\dot{u}}^z(0)=e$. Integrating with $u^0(0)=1$, $u^x(0)=0$ gives $u^z=(e/\omega )\sin \omega s$, $u^0=1+(e^2/{\omega }^2)(1-\cos \omega s)$ and $u^x=-(be/{\omega }^2)(1-\cos \omega s)$. Integrate once more from the origin:For zero charge the trajectory is simply rest, $x^0=s$, with all spatial coordinates zero. The charged trajectory oscillates in $z$ while drifting in the negative $x$ direction, the direction of $\mathbf{E}\times \mathbf{B}$. Its displacement-to-time ratio over a proper-time period is $-E/B$. A boost at that drift velocity removes the electric field and leaves circular magnetic motion; in the original frame the orbit has repeating cycloidal cusps where the particle is momentarily at rest. Energy oscillates rather than growing indefinitely because $B>E$. In SI units the corresponding magnetic-dominated condition is $cB>E$.

The canonical momentum is $p_i=\partial L/\partial {\dot{x}}_i=m{\dot{x}}_i+q{A}_i$. Its time derivative contains the total derivative along the trajectory: ${\dot{p}}_i=m{\ddot{x}}_i+q{\partial }_t{A}_i+q{\dot{x}}_j{\partial }_j{A}_i$. The Euler-Lagrange equation therefore givesThe magnetic field is $\mathbf{B}=\nabla \times \mathbf{A}$, so the last term is $(\dot{\mathbf{x}}\times \mathbf{B}{)}_i$. With the electric field $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, this is precisely the Lorentz force law:

The electric field of this point charge is a translated and scaled version of the radial field in part (ii). If $\mathbf{a}\notin V$ and $\mathbf{a}\notin \partial V$, its divergence vanishes throughout $V$, so the divergence theorem gives zero flux integral.

If $\mathbf{a}$ lies inside $V$, remove a small ball centered at $\mathbf{a}$. On the remaining volume the divergence is zero. Its boundary is $\partial V$ together with the small sphere, whose outward normal for the punctured volume points towards $\mathbf{a}$. That inner sphere contributes $-q/{ϵ}_0$, by the $4\pi$ sphere flux computed in part (ii). Thus the exterior flux must be $q/{ϵ}_0$:This is Gauss's law for a point charge. The assumption that the charge is not on the boundary is essential to these alternatives.

Let the electrolyte permittivity be $\epsilon$ and let ${\kappa }_D=1/{\ell }_D$ be the inverse Debye–Hückel screening length. We reserve $k$ for the imposed lateral wavevector. In the Debye–Hückel approximation, linearization about an electrically neutral bulk electrolyte givesaway from fixed charges. For ionic species of valence $z_i$ and bulk number density $n_i$, ${\kappa }_D^2={\sum }_i{n}_i{z}_i^2{e}^2/(\epsilon k_{B}T)$; the approximation requires $|z_{i}e\varphi |\ll k_{B}T$. Substitution of a lateral cosine into this screened Poisson equation leaves a vertical decay constantThe screened sinusoidal surface-charge mode thus decays faster than a laterally uniform charge mode. A nonzero $k$ retains a finite decay length even in the zero-salt limit.

The exterior media and their boundary conditions are not specified in the paper. Specifying the two surface charge densities alone does not fix the normal derivative on the inside unless the outside response is also fixed. We first take the confined-gap idealization, with the prescribed charge supplying all the displacement flux into the electrolyte between the sheets. The electrostatic interface boundary conditions are thenThis is, for example, the limit of negligible exterior displacement admittance. The electrostatic potential solving these conditions and the Debye–Hückel approximation equation isDifferentiation checks the two surface signs directly. This solution allows a sine as well as a cosine lateral component when the phase is nonzero.

In linear screened electrostatics, the quadratic charging free energy is $\frac{1}{2}\int \sigma \varphi dA$. It is equivalently $\frac{\epsilon }{2}\int [(\nabla \varphi {)}^2+{\kappa }_D^2{\varphi }^2]dV$ with the stated boundary conditions. The second term represents the linearized ionic response; integrating only the bare electric field energy would omit it. If $\mathcal{A}$ denotes membrane area, the wavelength-averaged energy density isUsing $⟨{\cos }^2(kx){⟩}_x=1/2$ and $⟨\cos (kx)\cos (kx+\theta ){⟩}_x=\cos \theta /2$ gives the phase registration of screened charged sheets energyThe positive cosine coefficient shows that, for nonzero charge amplitude and finite separation,At the minimum the electrostatic potential simplifies to $\varphi =\alpha \cos (kx)\sinh (pz)/[\epsilon p\cosh (pd/2)]$. Positive and negative charge patches face opposite signs across the gap. The two charge patterns therefore favor a half-wavelength lateral offset rather than like-charge alignment.

For comparison, if identical electrolyte fills all of space on both sides of infinitesimally thin charge sheets, the electrostatic interface boundary conditions are continuity of $\varphi$, decay at infinity, and the charge-induced derivative jumpsSuperposing the two screened sinusoidal surface-charge modes gives, in particular between the sheets,Its full-space continuation replaces each vertical distance by the corresponding absolute distance. Evaluating the electrostatic potential on both sheets now yieldsThe constant term is the two isolated-sheet self energies, and the cosine term is their screened interaction. The energy depends on the exterior convention, but both stated physical idealizations give the same phase registration of screened charged sheets: unlike charge patches oppose each other. In the full-space convention the optimum interaction energy is $-{\alpha }^2{e}^{-pd}/(4\epsilon p)$, giving an attractive normal force density $-{\alpha }^2{e}^{-pd}/(4\epsilon )$. Phase sensitivity becomes exponentially weak for $pd\gg 1$; if $\alpha =0$ or the separation tends to infinity there is no selected phase.

One can also display how an exterior dielectric response interpolates between these cases. Let $Y_o\ge 0$ be its normal displacement admittance for this lateral Fourier mode: $Y_o=0$ for the confined-gap idealization, $Y_o=\epsilon p$ for identical exterior electrolyte, and $Y_o={\epsilon }_o|k|$ for an ion-free exterior dielectric. DefineThe symmetric and antisymmetric surface-charge combinations giveSince $D_s<D_a$, its cosine coefficient is positive. This makes the phase-minimizing conclusion robust while exposing the boundary information needed for an absolute electrostatic energy.