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For two grounded conducting half-planes meeting at angle $\pi /3$, place a point charge $q$ at radius $d$ on the bisector $\theta =\pi /6$. The method of images closes after six copies: positive copies at angles $\pi /6+2k\pi /3$ and negative copies at $-\pi /6+2k\pi /3$, for $k=0,1,2$. Only the first positive copy is real. Opposite-sign reflected pairs make the electric potential vanish on both faces; the uniqueness theorem for the electrostatic potential proves the construction. Summing the five image charge electric fields gives the electrostatic force $\mathbf{F}=-q^2(5/4-1/\sqrt{3}){\hat{\mathbf{e}}}_r/(4\pi {ϵ}_0{d}^2)$, towards the intersection. The coefficient follows from distances $d$, $\sqrt{3}d$, and $2d$ to the adjacent negative, nonadjacent positive, and opposite negative copies.

In a grounded conducting enclosure, let the harmonic function $w_j$ equal one on boundary component $j$ and zero on the others. A point charge $q$ at ${\mathbf{r}}_0$ induces total electric charge $Q_j=-q{w}_j({\mathbf{r}}_0)$ on that component. The weighting harmonic function is its harmonic measure.

To prove the identity, let $\varphi$ be the zero-boundary electric potential, so ${\nabla }^2\varphi =-q{\delta }^{(3)}(\mathbf{r}-{\mathbf{r}}_0)/{ϵ}_0$. Green's second identity gives $-q{w}_j({\mathbf{r}}_0)/{ϵ}_0={\int }_{\partial D}{w}_j{\partial }_n\varphi dS$, where $\mathbf{n}$ points out of the vacuum domain into the metal. The electrostatic boundary conditions at a conductor give $\sigma ={ϵ}_0{\partial }_n\varphi$, proving the result. In an unbounded domain require the additional boundary term at infinity to vanish. Between concentric spheres, $w_a(r)=a(b-r)/(r(b-a))$; between parallel planes $z=0,d$, $w_0=1-z/d$ and $w_d=z/d$.

In 1939, Alan Hodgkin and Andrew Huxley reported intracellular action potential recordings from squid giant axons at Plymouth.^[ref]

Alan Hodgkin's later account describes advancing a capillary electrode from a cut end into intact axon, avoiding scraping the cell membrane. Its Figure 3 uses a seawater-filled capillary internally, with external seawater as zero: the readout is internal relative to external electric potential. This supports a voltage across the surface cell membrane. As a separate ionic control, the account reports that raising external potassium toward its internal concentration reversibly abolished resting voltage; this was later work, not a control asserted for the 1939 recording. Injury and electrode-solution junction potentials limit interpretation. Voltage recording alone does not identify every conducting ion. A voltage clamp additionally uses feedback to impose voltage and measures the required electric current.^[ref]

In the schematic, connect the intracellular electrode to the positive input and the external reference electrode to the negative input of a voltage recorder. After electrode-offset correction, a negative reading means ${\varphi }_{\mathrm{inside}}<{\varphi }_{\mathrm{outside}}$. A useful baseline check is to compare both electrodes in the same external chemical solution before insertion; this is a measurement check, not an additional historical observation claimed here. Swapping inputs reverses the readout sign without changing the biological cell's membrane potential.

An ion trap confines an ion using electromagnetic fields, allowing repeated interrogation of the same particle. The name also covers related traps for an electron. A Paul trap uses an oscillating electric quadrupole field; a Penning trap combines a static electric quadrupole field with a static magnetic field. Confinement creates time to measure an energy level, frequency or magnetic dipole moment; it does not itself cool the particle or identify its internal quantum state.

The obstacle is Earnshaw's theorem. In a charge-free neighbourhood the static electric potential $\Phi$ obeys Laplace's equation, ${\nabla }^2\Phi =0$. For a particle of fixed nonzero electric charge $Q$, the potential energy $Q\Phi$ is also a harmonic function. The maximum principle for harmonic functions excludes a strict interior minimum. For a quadratic example, the Hessian matrix has zero trace; it cannot have three positive eigenvalues. A static purely electric field therefore cannot supply three-dimensional stable confinement of a free point electric charge. Time variation in a Paul trap, and the velocity-dependent Lorentz force in a Penning trap, change the assumptions of this argument.

For a transverse monochromatic electromagnetic wave with angular frequency $\omega \ne 0$, ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$. The displayed electric potential and magnetic vector potential therefore give $-\nabla \Phi -{\partial }_t\mathbf{A}=\mathbf{E}$. Transversality gives $\nabla \cdot \mathbf{A}=0$, so the Lorenz gauge holds. For $\mathbf{E}=E_0(\cos \theta ,\sin \theta ,0)$ and $\theta =\omega t-Kz$, taking a curl yields $\mathbf{B}=(K/\omega )\hat{\mathbf{z}}\times \mathbf{E}$. The vacuum Maxwell equations require $K^2={\omega }^2/c^2$. The field has circular polarization, with $|\mathbf{B}|=|\mathbf{E}|/c$ and perpendicular electric and magnetic directions.

A conducting conical pendulum connected to a grounded pivot develops a motional electric potential difference $B\omega r^2/2$ between the rotation axis and a bob at radius $r$. If the bob dominates the capacitance, its electric charge is $CB\omega r^2/2$. An additional uniform vertical electric field changes both the force on that charge and, while the grounding connection remains, the background electric potential entering the charging relation. These effects must be treated consistently when calculating a first-order mechanical response.

The Field Mapper Kit (PK-9023), Product Manual 012-04346D by PASCO scientific (p. 3, Parallel plate capacitor) proposes a direct investigation of the plate-length-to-gap ratio using conductive-paper electric potential mapping. Its finite straight electrodes represent a cross-section of the plates. The apparatus is a gridded resistive sheet on a corkboard, silver-ink electrodes, pushpin electrical contacts, a $5$–$20\mathrm{V}$ direct supply capable of $25\mathrm{mA}$, and a voltmeter with input electrical resistance at least $10\mathrm{M\Omega }$. The manual specifies waiting 20 minutes for the ink to reach its maximum electrical conductivity, then checking each electrode: its internal voltage drop should be no more than $1\%$ of the applied electric potential difference.^[ref]

For the readout, keep one voltmeter lead on the reference electrode and scan the other across the sheet. Mark points having the same reading to trace an equipotential. A second method holds the two probe tips at a fixed separation and pivots them to find the largest electric potential difference; a suggested separation is one centimetre. Avoid printed grid marks when making contact.^[ref]

The inference follows from the measured electric potential, rather than assuming straight electric field lines. With probe separation $\Delta s$ and direction $\hat{\mathbf{t}}$, a local finite difference givesAn arrow toward rising electric potential follows $+\nabla \varphi$, so it must be reversed to show $\mathbf{E}$. To investigate the fringing field of a capacitor, compare contours well inside the gap with contours near and beyond the ends, then repeat at a changed gap while keeping the sheet boundary and plate length fixed. Curving contours and an outside gradient establish nonuniformity in this planar setup. Compare $-E_{y}d/V$ at recorded positions to distinguish simple voltage rescaling from geometric changes; no measured values are supplied here.

Additional useful controls are repeat reference readings to detect drift, reversal of supply polarity, and repeating selected readings with reduced probe spacing. Keep contact quality and sheet edges fixed during the aspect-ratio comparison. Finite probe separation averages the gradient; reducing it improves spatial resolution but magnifies the effect of voltage uncertainty. Voltmeter loading, imperfect ink electrodes, and inhomogeneous sheet electrical conductivity can mimic distortions. These proposed checks supplement the manufacturer's electrode check, rather than claiming a completed experiment.

The sheet carries a steady electric current. For uniform electrical conductivity, Ohm's law and electric charge conservation give the two-dimensional Laplace equation away from electrodes, as explained in conductive-paper electric potential mapping. Uncontacted sheet edges have zero normal electric current, whereas the diagram model fixes zero outer electric potential. These distinct boundary conditions must be matched before a quantitative comparison. The apparatus reveals a planar electric potential map; it does not directly measure the free-space three-dimensional capacitance of metal plates.

A membrane potential is an electric potential difference across a cell membrane. We use the conventionTaking the external electric potential as zero makes $V_m<0$ mean that the inside is at lower electric potential. “Negative inside” describes a difference, not an absolute electric potential, and does not imply that the entire cytosol has a large net negative electric charge.

Consider a cell membrane permeable only to potassium ions, with higher potassium concentration inside. Diffusion initially favors outward transport. If balancing negative ions cannot follow, that loss of positive electric charge produces a small inner excess of negative electric charge and an outer positive excess. The resulting electric field favors inward potassium transport. At the Nernst equilibrium potential, chemical and electrical contributions to electrochemical potential balance; ions still cross both ways, with zero net transport.^[ref]

In living biological cells, several ions can permeate and pumps maintain their concentration differences through active transport. Pumps transporting net electric charge also contribute electric current. Thus $V_m=E_{\mathrm{K}}$ is a special limit, not a universal rule; membrane potential need not always be negative.^[ref] The existing reversal potential example shows how opposing electric currents can balance without each ion reaching thermodynamic equilibrium. This makes selective permeability a way to convert chemical differences into electrical signals: changing the available ion channels changes which electrochemical gradients influence the voltage.

Membrane transport is the passage of substances across a cell membrane. For the small solutes considered here, distinguish the available route from its driving force: a lipid bilayer, ion channel, or carrier protein determines how passage can occur, while differences in chemical potential and membrane potential determine which direction is energetically downhill. A large driving force cannot make an unavailable route conduct. See Alberts et al.: principles of membrane transport.

For an ideal dilute chemical solution at uniform temperature and pressure, with the same reference molar chemical potential on both sides, the molar electrochemical potential change for transferring a solute from compartment A to B isHere $c_{\mathrm{A}},c_{\mathrm{B}}$ are concentrations, $T$ is absolute temperature, $R$ is the molar gas constant, $z$ is signed ion charge number, $F$ is the Faraday constant, and $V_{\mathrm{B}}-V_{\mathrm{A}}$ is the electric potential difference. The natural logarithm contains a dimensionless ratio. The electrical term vanishes for an uncharged solute. More generally, use thermodynamic activities in place of the ideal concentration ratio. This combines the ideal-solution chemical potential difference with electrical energy per mole; see IUPAC Gold Book: electrochemical potential. For a biological cell, taking B inside and A outside makes the voltage term use the existing membrane potential convention.

The electric potential difference across the inner mitochondrial membrane, contributing to the proton-motive force. A change in potential can alter both useful phosphorylation and leak behavior.

For an ideal MOSFET with a nondegenerate p-type semiconductor of acceptor density $N_A$, define ${\varphi }_F=(k_{B}T/e)\ln (N_A/n_i)>0$, where $n_i$ is the intrinsic carrier concentration. A positive surface electric potential ${\psi }_s$ lowers the local conduction band and valence band by $e{\psi }_s$. The surface electron density is $n_s=(n_i^2/N_A)\exp (e{\psi }_s/k_{B}T)$. Equality $n_s=N_A$ gives the conventional onset of strong inversion at ${\psi }_s=2{\varphi }_F$. The earlier crossover $n_s=p_s=n_i$ occurs at ${\psi }_s={\varphi }_F$. Neither condition requires the Fermi level to enter the conduction band.

Passive transport is net movement down the transported substance's electrochemical gradient without coupling that movement to another source of free energy. Facilitated diffusion is passive transport through an ion channel or a passive carrier protein. Providing such a route can speed passage, but does not turn an uphill transfer into a downhill one. For an uncharged solute under the ideal conditions above, net passive transport runs from higher to lower concentration; for an ion, the membrane potential also matters. See Cooper, The Cell: transport of small molecules.

In the figure, stipulate eight neutral solute tracers in A and two in B, with equal fixed volumes, identical surroundings, and no imposed electric potential difference. The concentration ratio is $c_{\mathrm{A}}/c_{\mathrm{B}}=8/2=4$, so A-to-B transfer has $\Delta \tilde{\mu }=-RT\ln 4\simeq -1.38629RT$, or $-3.458\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ at the assumed temperature $T=300\mathrm{K}$. If this solute can cross, thermodynamic equilibrium has equal expected concentrations: five tracers per side on average. Individual crossings occur in both directions; an observed ten-particle snapshot need not contain exactly five on each side. These stipulated counts illustrate a driving force, not a measured flow rate or a prediction of the time to equilibrate.

For a separate illustrative ion calculation, keep $c_{\mathrm{A}}/c_{\mathrm{B}}=4$, but assign $z=+1$ and $V_{\mathrm{A}}-V_{\mathrm{B}}=-60\mathrm{mV}$ at $300\mathrm{K}$. Using $R\simeq 8.31446\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$ and $F\simeq 96485.33\mathrm{C}\mathrm{mo}{\mathrm{l}}^{-1}$, B-to-A transfer gives $RT\ln 4-0.060F\simeq 3.458-5.789=-2.331\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Thus an appropriate open ion channel can permit passive transport toward the higher concentration: the electrical contribution outweighs the opposing chemical potential contribution. This stipulated membrane potential is not inferred from the tracer counts.

The static Lindhard dielectric function describes the real, zero-frequency linear response of a uniform free electron gas. Here ${ϵ}_1=\mathrm{Re}ϵ$, $\mathbf{r}$ is the position operator, and $\mathbf{q}$ is the perturbation wavevector, $q=|\mathbf{q}|$, $V$ the normalization volume, $e>0$ the elementary charge, and ${ϵ}_0$ the vacuum permittivity. The normalized one-electron states are ${\psi }_{\mathbf{k}}=V^{-1/2}{e}^{i\mathbf{k}\cdot \mathbf{r}}$ with $E_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$; $f(E)$ is their Fermi-Dirac distribution occupation. Both momentum sums exclude spin; the prefactor two counts the two equal spin populations. The density-operator matrix element is ${\delta }_{{\mathbf{k}}^{\prime },\mathbf{k}+\mathbf{q}}$. Coincident energy denominators are interpreted by the static limit or Cauchy principal value.

A weak total electric potential ${\varphi }_{\mathrm{tot}}$ perturbs an electron by $-e{\varphi }_{\mathrm{tot}}$. First-order state mixing produces $\delta n=e{\Pi }_0(q){\varphi }_{\mathrm{tot}}$, whereThe induced charge is $-e\delta n$. Self-consistent Poisson equation screening then gives ${\varphi }_{\mathrm{ext}}=ϵ{\varphi }_{\mathrm{tot}}$, with $ϵ=1+e^2{\Pi }_0/({ϵ}_0{q}^2)$. This is the random phase approximation: use the independent-electron response to the self-consistent total field.^[ref]

At zero temperature $f(E_{\mathbf{k}})=\Theta (k_F-k)$. Shifting the integration variable in the second occupation term givesThe two angular integrals are equal. Using the supplied integral, with $x=q/(2k_F)$, givesThe limits are $F(0)=1$, $F(1)=1/2$ and $F(x)\sim 1/(3x^2)$ for large $x$. The first recovers Thomas-Fermi screening; the logarithm produces the Kohn singularity of static Lindhard screening at $q=2k_F$.

Use the Fourier transform convention $\rho (\mathbf{q})=\int d^{3}r{e}^{-i\mathbf{q}\cdot \mathbf{r}}\rho (\mathbf{r})$. A point charge $Ze$ has total screened charge $Ze/ϵ(q)$ and induced charge $-Ze[1-ϵ(q{)}^{-1}]$. Angular inversion therefore gives the radial induced-charge screening kernelIn particular, ${\rho }_{\mathrm{ind}}(\mathbf{q}=0)=-Ze$: the total screening charge is negative although its local density can oscillate in sign.

Integrate the sine transform by parts twice; $g(0)=g(\infty )=0$ removes the boundary terms. For the Friedel density tail from a derivative singularity, write $Q=2k_F$ and $g^{\prime \prime }(q)\sim C/(q-Q)$. ThenIndeed, near $Q$ the principal value transform is ${\int }_{-\infty }^{\infty }\sin [(Q+s)r]ds/s=\pi \cos (Qr)$ for $r>0$. Smooth terms do not remove this leading oscillatory long-range contribution. Thus ${\rho }_{\mathrm{ind}}\propto \cos (2k_{F}r)/r^3$.

For a three-dimensional, single-valley gas with two spin states, the Friedel oscillation wavelength from carrier density isLow-temperature scanning tunneling spectroscopy near exposed dopants can map the oscillatory local density of states, with a Fourier feature near $2k_F$ at the Fermi energy. An energy-resolved tunnelling image measures local spectral weight, not directly the bulk energy-integrated charge: comparing occupied-bias maps and accounting for the surface geometry is necessary. In actual GaAs, the background relative permittivity and band effective mass change screening strengths; they do not change this wavelength at the stated density.

A plausible hypothesis is that broad specificity emerges from a flexible, chemically heterogeneous binding cavity containing overlapping recognition sites, coupled to an energy-driven cycle that exports many different bound molecules. The cavity recognizes combinations of physical properties rather than one unique chemical structure. Different substrates can use different subsets of contacts and different routes through the same multidrug efflux pump. This is a testable mechanism, not a claim that every pump has identical architecture.

The hypothesis has three parts. First, a large cavity contains patches providing hydrophobic contacts, aromatic contacts, hydrogen bonds and electrostatics. A chemically diverse set of substrates may share an aromatic region, a charged group or the ability to partition into the lipid bilayer. Each can make a favorable but different combination of contacts. Internal water can mediate alternative interactions rather than requiring a rigidly preorganized set of donors and acceptors. Second, local side-chain and backbone flexibility allows several cavity shapes. Third, the large-scale transport cycle changes accessibility and affinity so that occupancy of a permissive binding region can be converted into release on the far side.

These ideas have structural precedents. X-ray crystallography of AcrB identified an aromatic, multisite drug-binding region and different conformations in its three subunits, consistent with a functional rotation through binding and export states.^[ref] Further structures distinguished an access pocket and a deeper binding pocket separated by a flexible switch loop.^[ref] The observation supports overlapping recognition and staged movement; the proposed energetic and kinetic roles of each contact still require testing. AcrB is the proton-coupled inner-membrane component of a larger bacterial export assembly, rather than an ATP-powered pump by itself.

A conformational ensemble makes the hypothesis quantitative. Let the unbound pump occupy conformations $j$ with probabilities $p_j$, and let substrate $s$ have an association constant $K_{a,j,s}$ in conformation $j$. If these states equilibrate rapidly before significant transport, a simple equilibrium binding polynomial givesThe expression follows by adding the bound population contributed by each unbound state at low free substrate concentration. Distinct substrates can bind well overall even when their favorable conformations are different. For example, with $p_1=0.9$, $p_2=0.1$, $K_{a,1,s}=10^4{\mathrm{M}}^{-1}$ and $K_{a,2,s}=10^7{\mathrm{M}}^{-1}$, the apparent constant is $1.009\times 10^6{\mathrm{M}}^{-1}$. A minority state dominates the binding contribution. These are illustrative numbers, not measured pump parameters. A working pump is driven and may not equilibrate rapidly, so its transport kinetics cannot in general be reduced to this binding expression.

Permissive recognition is compatible with directional transport only when the cycle is coupled to a driving reaction. For an ATP-binding cassette transporter, ATP binding and ATP hydrolysis couple rearrangement of nucleotide-binding and transmembrane regions. For a proton-driven pump such as AcrB, proton movement down its electrochemical gradient couples to structural changes. Neither substrate affinity alone nor random thermal motion supplies sustained uphill transport. In a tightly coupled model exporting one drug molecule while importing $n$ protons, the molar cycle Gibbs free energy iswhereHere $a$ denotes dimensionless thermodynamic activity, $F$ the Faraday constant, $\psi$ electric potential, and $z_D$ the charge of the transported drug species. Forward operation requires $\Delta G_{\mathrm{cycle}}<0$; equilibrium stall has equality. This states an energetic bound under the stated coupling assumption, not a measured proton-to-drug ratio for AcrB. Mixed protonation states, coupling slippage and alternative pathways require a more detailed model.

Several predictions distinguish flexible multisite recognition from an indiscriminately sticky cavity. A mutation removing one aromatic contact should disproportionately affect a subset of substrates that use that region, while leaving others relatively intact. Altering a common gating element may impair many substrates despite preserving some equilibrium binding. Rigidifying a flexible recognition loop could narrow the range even when transport of a favored substrate survives. Different drugs may compete strongly, weakly or asymmetrically depending on overlap and pathway occupancy; failure of simple competitive kinetics would not alone prove distinct sites in a driven, multi-state system.

Test these predictions with a chemically varied panel, including transported compounds and matched nontransported controls. Measure direct transport in lipid vesicles or another defined system while controlling orientation, pump amount, membrane leak and the driving gradient. For a proton-coupled pump, vary pH and membrane potential separately where possible and include an energy-collapse control; establish that the collapse does not merely destroy membrane integrity. For an ATP-driven pump include no-ATP and hydrolysis-defective controls. Measure binding and driving-energy consumption independently, because drug-stimulated ATP hydrolysis or proton flow can be uncoupled from productive export. Growth-based resistance alone is insufficient: altered drug uptake, metabolism and expression also change survival.

Use cryo-electron microscopy or X-ray crystallography with several chemically different drugs to test different contact maps and cavity shapes, then validate those differences by targeted mutations. Confirm expression, assembly, membrane localization and protein folding for every altered pump. molecular dynamics can predict alternative entry routes and local reorganization, but force-field choices, drug protonation and limited sampling must be varied or checked; a docking score alone is not evidence of transport. single-molecule Förster resonance energy transfer or electron paramagnetic resonance can test whether different drugs stabilize different dynamic states and whether the driving reaction changes state occupancy.

Finally, membrane partitioning may explain part of the apparent range: a pump can collect chemically varied compounds that share access to the same membrane-facing route. Measure drug partitioning and free aqueous concentration, since equal added concentrations need not give equal local exposure at the pump. The strongest support would be substrate-specific contact and dynamics changes, preserved transport energetics, and direct export measurements showing that one adaptable recognition apparatus feeds a shared directional cycle.

Write $e=q>0$ for the elementary charge and $ϵ={ϵ}_0{ϵ}_r$ for the permittivity. Taking the bulk electric potential as zero, the Boltzmann distribution of point ions gives $c_i(x)=c_i^{\infty }\exp [-e{z}_i\varphi (x)/(k_{B}T)]$. Here $c_i$ is a number density, not a molar concentration. Combining this with Poisson's equation gives the planar Poisson-Boltzmann equationMultiplying by ${\varphi }^{\prime }$ and integrating from the neutral bulk, where $\varphi ={\varphi }^{\prime }=0$, yieldsFor the solution occupying $x>0$, with the field on the other side neglected, Gauss's law gives $\sigma =-ϵ{\varphi }^{\prime }(0)$. On the branch that decays to zero without crossing it,Source sign qualification: the unsigned square root printed in the paper gives $|\sigma |$. A negative wall electric potential, such as the next part's $-73\mathrm{mV}$, has ${\varphi }^{\prime }(0)>0$ and hence $\sigma <0$, providing a concrete counterexample to interpreting that printed expression as signed surface charge density. This is the planar Grahame equation, with an explicit sign convention.

Assume the structural surface charge density stays fixed while a divalent chloride is added, without changing $T$ or $ϵ$. Let $v=e{\varphi }_0/(k_{B}T)$ and $\delta c$ be the increment of the bulk divalent concentration. Electroneutral addition also increases bulk chloride by $2\delta c$. Varying the squared Grahame equation at fixed $\sigma$ givesAt a strongly negative wall, coions are small. Dropping their contribution to the denominator and $2e^v$ in the numerator gives the requested leading approximationThe $-3$ includes the change in bulk osmotic concentration of one added divalent counterion and its two monovalent coions. The numerator's increment is a bulk concentration; the denominator contains surface concentrations.^[ref] Extra counterions screen the wall and make its negative electric potential less negative. At fixed bulk calcium, $c_{\mathrm{Ca}}(0)=c_{\mathrm{Ca}}^{\infty }{e}^{-2{\varphi }_0/V_T}$ therefore decreases. Under the stipulated proportionality, synaptic activity decreases as well; this argument isolates electrostatic accumulation rather than assuming a separate specific magnesium effect on the ion channels.

For $\delta c=5\mathrm{mM}$, the printed linear approximation gives $\delta {\varphi }_0=9.53434\mathrm{mV}$. Linearizing the activity too gives $\delta A/A=-2\delta {\varphi }_0/V_T=-0.7422$, or approximately a $74\%$ reduction. Exponentiating this approximate voltage instead gives $A^{\prime }/A=0.4761$. Neither is a controlled finite-increment prediction: the added divalent concentration is 2.5 times its initial value.

A consistent finite-change calculation is possible within the same Poisson-Boltzmann equation. Define, using $\mathrm{mM}$ throughout,Fixed $\sigma$ requires $S_5(u^{\prime })=S_0(2.84129)$. The physical positive root is $u^{\prime }=2.58241$, givingThus the self-consistent finite-change estimate is about a 40.4% reduction. Both estimates rely on fixed surface charge, ideal ionic activities, and unchanged bulk calcium; the limitations of the large local concentrations remain.

Usable biological free energy has several interconvertible forms. At approximately fixed temperature and pressure, chemical reactions are described by chemical potentials, ${\mu }_i={\mu }_i^{\circ }+RT\log a_i$, and ${\Delta }_{r}G={\sum }_i{\nu }_i{\mu }_i$. ATP hydrolysis, for example, has ${\Delta }_{r}G={\Delta }_r{G}^{\circ \prime }+RT\log (a_{\mathrm{ADP}}{a}_{\mathrm{Pi}}/a_{\mathrm{ATP}})$ under the specified biochemical standard-state convention. The activities and proton/ion convention matter. Breaking a bond alone costs energy; the net free-energy change includes the products, solvent and all bond changes, not a store released simply by snapping one bond.

Redox reactions provide free energy when electrons move between donors and acceptors with a favourable potential difference: $\Delta G=-nF\Delta E$. Electron-transfer chains can couple that drop to ion transport, building an electrochemical gradient. For transfer from an outer to an inner reservoir,Both concentration difference and electrical potential contribute. ATP synthase couples downhill ion flow to nucleotide free energy; driven pumps can perform the reverse conversion and maintain gradients used by transporters and excitability.

Mechanical work appears as $Fdx$ or torque times angle, and can be stored elastically, for example $\kappa x^2/2$. Molecular motors convert chemical free energy into movement or force; mechanically driven reactions can conversely reshape chemical transitions. Electrical energy can also be stored in membrane capacitance, while bulk ionic concentration gradients store additional chemical free energy. Light supplies photon energy $h\nu$, which photosynthetic machinery can convert into redox and chemical potential differences.

Coupling is essential: an overall favourable reaction can drive an unfavourable one only when their processes are linked, rather than merely present in the same cell. At fixed conditions the extracted work is bounded by the consumed free-energy decrease. Dissipation produces heat and entropy, so energy conservation does not mean all energy remains available for work. Thermal fluctuations can drive random motion, but a persistent directional cycle requires a nonequilibrium free-energy source; an equilibrium asymmetric structure alone is not a free motor.

In a contractible region, the Maxwell equations $\nabla \cdot \mathbf{B}=0$ and the Poincaré lemma imply the existence of a magnetic vector potential $\mathbf{A}$ such that $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law of induction then givesThe expression in parentheses is a gradient there, so a scalar electric potential $\varphi$ can be chosen withThese statements are local, or global when the domain has the required topology and the electromagnetic fields are regular. A nonzero flux through a topologically nontrivial closed surface can obstruct a single globally defined vector potential, even though local expressions remain valid. The electric potential and magnetic vector potential are not unique: $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ and $\varphi \mapsto \varphi -{\partial }_t\chi$ leave both electromagnetic fields unchanged.

For the Lorenz gauge condition $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$, Gauss's law becomesThe Ampere-Maxwell law givesCancel the two gradient terms using the Lorenz gauge condition. HenceBoth potential equations have the same d'Alembert operator, with their respective charge density and current density sources. In a source-free region these are homogeneous wave equations with propagation speed $c$.

Take $d$ as the centre-to-centre separation and $ϵ={ϵ}_0{ϵ}_r$. Use the ideal lossless two-conductor transmission line model, with nonmagnetic surroundings and negligible internal-wire inductance. Opposite line charges $+{\lambda }_q$ and $-{\lambda }_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their electric fields, or adding their logarithmic electric potentials, gives the potential differenceThis is the two-wire capacitance at large separation. For opposite electric currents $+I$ and $-I$, the magnetic fields between the wires add. The linked magnetic flux per unit length is approximatelyThus the external inductance of a thin two-wire line is $L^{\prime }\simeq {\mu }_0\ln (d/a)/\pi$. The lossless telegrapher's equations give a forward wave with $V/I=\sqrt{L^{\prime }/C^{\prime }}$, so its characteristic impedance isThe product $L^{\prime }{C}^{\prime }={\mu }_0ϵ$ also gives the propagation speed $c/\sqrt{{ϵ}_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln (d/a)$ by $\mathrm{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite electrical conductivity and internal-wire magnetic energy require a more detailed line model.

Let $\mathbf{n}=\mathbf{r}/a$ be the unit normal from the electrical conductor into the vacuum. The electrostatic boundary conditions at a conductor, together with the vanishing interior electric field, giveWrite $u=\cos \theta$ in spherical coordinates, $D=|\mathbf{r}-{\mathbf{r}}_0|$ on $r=a$, and $z^{\star }=a^2/r_0$. Differentiating the electric potential at fixed angles givesHere the distance identity in the preceding answer was used before simplifying. Hence the induced surface charge density isFor positive $q$ this surface charge density is negative everywhere, with largest magnitude nearest the point charge.