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The Bohr model assumes circular electron orbits with quantized orbital angular momentum $N\hslash$. It reproduces the hydrogen energy levels, but wavefunctions in quantum mechanics describe spatial probability distributions rather than definite trajectories. In particular, spherically symmetric Coulomb bound states have zero orbital angular momentum.

The puzzle is the coexistence of reproducible spectral lines and stable chemical atoms. A classical circular orbit in a Coulomb potential can have any negative energy: Coulomb's law alone selects no special radius or emitted frequency. Moreover, its electron has centripetal acceleration. The Larmor formula then predicts positive radiated power, so conservation of energy makes the orbit shrink. Simply drawing concentric circles explains neither stability nor a discrete spectrum.

Niels Bohr postulated nonradiating stationary states and emission with $h\nu =E_i-E_f$. His circular-orbit condition is $L=n\hslash$. These assumptions override classical radiation; the emitted frequency need not equal an orbital frequency.^[ref]

Consider a single electron of mass $m_e$ and a point atomic nucleus of mass $M$ and positive electric charge $Ze$, with $e>0$ the elementary charge. Use nonrelativistic classical mechanics, omit external fields and interactions with other electrons, and writeHere $\mu$ is the Coulomb two-body reduced mass, ${ϵ}_0$ the vacuum permittivity, and $h$ the Planck constant. The separation $r$ and relative speed $v$ describe the two-particle relative coordinate; neither body is silently pinned in place. The electron's own radius about the centre of mass is $Mr/(M+m_e)$.

For the assumed circular orbit, Coulomb's law provides the required centripetal acceleration, and the imposed orbital angular momentum condition isSubstitute $v=n\hslash /(\mu r)$ into the first equation and cancel $r^{-2}$ to obtainThe kinetic energy is $K=\mu v^2/2=\kappa /(2r)$ and the Coulomb potential energy is $U=-\kappa /r$, with zero energy at infinite separation. HenceThis derives the energy sequence from the stated circular model; the modern Schrödinger equation derivation is the distinct series-termination quantization of the Coulomb radial equation. The radius above scales as $n^2/Z$, whereas the binding energy scales as $\mu Z^2/n^2$. The Bohr radius article supplies the corresponding Coulomb potential length scale. The nonrelativistic assumption requires $v_n\ll c$, with $c$ the speed of light.

For an original numerical example choose $Z=1$, $M=1.67262192369\times 10^{-27}\mathrm{kg}$, $m_e=9.1093837015\times 10^{-31}\mathrm{kg}$, $e=1.602176634\times 10^{-19}\mathrm{C}$, $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, ${ϵ}_0=8.8541878128\times 10^{-12}\mathrm{F}{\mathrm{m}}^{-1}$, and $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. Treat these as stipulated constants for the calculation. They give $\mu /m_e=0.99945568$, $r_1=0.0529465\mathrm{nm}$, $E_1=-13.598287\mathrm{eV}$ and $v_1/c\simeq 0.00730$, where $\mathrm{eV}$ denotes an electronvolt. At $n=2$, $r_2=4r_1$ and $E_2=E_1/4$; the ground-state ionization energy is $-E_1$.

Taking $M\to \infty$ gives $\mu \to m_e$. A finite nuclear mass reduces every binding energy by $\mu /m_e$ and increases each separation by $m_e/\mu$. This follows from relative motion, not a fitted correction to individual spectral lines. Replacing the atomic nucleus by a heavier isotope changes the same factor, providing another check of the model.

The coulomb is the SI unit of electric charge. A constant electric current of $1\mathrm{A}$ transports $1\mathrm{C}$ in a time of $1\mathrm{s}$; a constant current of $2\mathrm{A}$ for $3\mathrm{s}$ transports $6\mathrm{C}$. The prefix in $1\mathrm{nC}=10^{-9}\mathrm{C}$ explains the much smaller charge scale used in the worked Coulomb's law example. ^[ref]

The Coulomb potential energy is the interaction potential energy of two point charges in vacuum. With separation $r>0$, electric charges $q_1,q_2$, $k=1/(4\pi {\epsilon }_0)$ and zero potential energy at infinite separation,This is mutual interaction energy; it does not include the divergent self-energies of ideal point charges. Unlike electric potential, which has units of volts, $U$ has SI units of joules. With charge 1 fixed and positive radial direction away from it, the conservative force in Coulomb's law givesThus the $1/r$ potential energy and the inverse-square law for force are two descriptions of the same interaction. ^[ref].

For $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$ and $r=0.20\mathrm{m}$, using the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$,At $r=0.40\mathrm{m}$, $U=-1.3485\times 10^{-7}\mathrm{J}$. Separating the pair from $0.20$ to $0.40\mathrm{m}$ therefore requires external work $\Delta U=+1.3485\times 10^{-7}\mathrm{J}$ if the initial and final kinetic energies agree; the electric force does $-1.3485\times 10^{-7}\mathrm{J}$ of work. Separating them all the way to infinity requires $+2.697\times 10^{-7}\mathrm{J}$. The negative initial potential energy means that the pair lies below the chosen separated reference; it does not mean a negative kinetic energy.

Like-signed electric charges have positive $U$, which decreases as they separate, so their electric force is repulsive. Opposite-signed electric charges have negative $U$, which decreases as they approach, so their electric force is attractive. This attractive interaction supplies the Coulomb potential used in the hydrogen atom; determining its stationary quantum states also requires quantum mechanics. ^[ref]. Dividing $U$ by a nonzero test electric charge $q_2$ gives the source's electric potential ${\Phi }_1=k{q}_1/r$. More generally, the electrostatic potential energy of point charges sums each unordered charge pair once. This energy description is useful because work between two separations follows from endpoint values, without integrating a complicated trajectory.

Coulomb's law gives the mutual electric force between stationary point charges in vacuum. Put electric charges $q_1$ and $q_2$ at ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, and define the displacement from charge 1 to charge 2 by $\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1$, with separation $r=|\mathbf{r}|>0$ and unit vector $\hat{\mathbf{r}}=\mathbf{r}/r$. In SI units, the force on charge 2 due to charge 1 isHere ${\epsilon }_0$ is the vacuum permittivity. The magnitude is $k|q_1{q}_2|/r^2$. If $q_1{q}_2>0$, the force on charge 2 points away from charge 1; if $q_1{q}_2<0$, it points toward charge 1. Interchanging the charges gives ${\mathbf{F}}_{1\leftarrow 2}=-{\mathbf{F}}_{2\leftarrow 1}$, consistent with Newton's third law in this electrostatic setting. ^[ref].

For a fully computed example, stipulate $q_1=+2\mathrm{nC}$, $q_2=-3\mathrm{nC}$, ${\mathbf{r}}_1=0$, ${\mathbf{r}}_2=0.20\mathrm{m}\hat{\mathbf{x}}$ and the approximate value $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$. ThenThe force is attractive, of magnitude $1.3485\times 10^{-6}\mathrm{N}$. The source electric field at charge 2 is ${\mathbf{E}}_1({\mathbf{r}}_2)=449.5N/C\hat{\mathbf{x}}$, pointing away from the positive source. Multiplication by the negative $q_2$ reverses the direction to give the attractive force. This field excludes charge 2's own self-field.

Holding the electric charges fixed while changing the separation from $R$ to $2R$ or $3R$ gives the normalized inverse-square law valuesFor the stipulated example, separations $0.20$, $0.40$ and $0.60\mathrm{m}$ give magnitudes $1.3485\times 10^{-6}$, $3.37125\times 10^{-7}$ and $1.498333\dots \times 10^{-7}\mathrm{N}$. The diagram computes these ratios and uses one common arrow-length scale for both sign choices.

The inverse-square law has a geometric explanation through Gauss's law: a source point charge gives fixed total electric flux through a concentric sphere, whose surface area is $4\pi r^2$. Dividing that flux by the area gives a radial electric field proportional to $1/r^2$. This connects a pairwise force calculation to a global conservation-like accounting of field sources. For several stationary point charges, the superposition principle adds the individual electric fields as vectors, and ${\mathbf{F}}_i=q_i{\mathbf{E}}_{\mathrm{other}}({\mathbf{r}}_i)$ uses all sources except charge $i$ itself. The Coulomb potential energy gives an alternative route when the desired quantity is work rather than instantaneous force.

Electric charge is an additive, signed property of matter that determines its coupling to the electromagnetic field. Write a particle's electric charge as $q$, or a collection's total electric charge as $Q$; the SI unit is the coulomb, $\mathrm{C}$. A stationary test particle in an electric field experiences electric force $\mathbf{F}=q\mathbf{E}$, while motion adds the magnetic part of the Lorentz force. Electric charge both produces an electric field and determines the response to one. For two stationary point particles in vacuum, Coulomb's law makes like signs repel and opposite signs attract. The choice of which sign to call positive is a convention, used consistently: protons carry $+e$ and electrons carry $-e$, where $e$ is the elementary charge. ^[ref].

Additivity means adding signed electric charges, rather than their magnitudes:for ordinary matter containing $N_p$ protons, $N_e$ electrons, and electrically neutral neutrons. A toy collection with seven protons and ten electrons has $Q=-3e$, however much neutral material accompanies it. Charge neutrality means $Q=0$; it does not mean an absence of charged constituents. Nor does it guarantee an absence of electric force: the spatial arrangement of positive and negative electric charges matters, as illustrated by an electric dipole. This makes a tiny imbalance in an otherwise nearly neutral object physically important. ^[ref].

Charge conservation states that a collection closed to transport of electric charge keeps its total signed $Q$. If $n$ electrons move from object A to object B without other exchange, then $\Delta Q_A=+ne$ and $\Delta Q_B=-ne$; the sum is unchanged. The amount of positive electric charge alone need not be conserved: creation of an electron and a Positron adds opposite electric charges with zero sum. These bookkeeping rules let us distinguish separation or transport of electric charge from creation of net electric charge. ^[ref].

A simple motivating experiment uses two charge-producing paddles, a conducting pail on an insulating support, a grounded outer shield and a high-input-impedance electrometer. Zero the pail, then isolate it. Rub the initially neutral paddles together, keeping their charged surfaces away from other objects. Lower each deeply into the pail without touching: the electrometer gives opposite signs. Inserting both together tests cancellation. For a separate contact test, insert one charged paddle without touching and withdraw it: the reading returns to zero. Repeat after zeroing, this time touching the inner wall before withdrawal: a residual reading shows transfer of electric charge. Zero the pail and reinsert the paddle to test how much electric charge remains on it. PASCO describes these procedures in Basic Electrostatics System manual, demonstrations 1A and 1B, printed pp. 13–16.

The crucial control is withdrawal without contact: a response during insertion can come from electrostatic induction, which redistributes the pail's existing electric charge. In the fixed-geometry approximation, the electrometer measures electric potential difference proportional to enclosed electric charge, with conversion set by effective capacitance; a voltage reading is not already a measurement in coulombs. Opposite readings and their cancellation test additivity and charge conservation within resolution. Leakage or accidental contact spoils that interpretation, and contact need not transfer all the paddle's electric charge. This macroscopic experiment cannot resolve elementary charge steps: that requires the much smaller particles of the oil drop experiment.

A finite electric dipole consists of equal and opposite electric charges separated by a nonzero displacement. It has charge neutrality but a nonzero electric dipole moment $\mathbf{p}=q\mathbf{d}$. An ideal point electric dipole is the mathematical limit $d=|\mathbf{d}|\to 0$ with $\mathbf{p}$ fixed, so $q$ must increase as $1/d$. Shrinking the separation while holding $q$ fixed instead makes the electric dipole moment vanish. The ideal electric dipole describes an exterior electric field; it does not reproduce the two near-charge singularities or the internal structure of a molecule. ^[ref].

Place the pair's midpoint at the origin and write $k_e=1/(4\pi {ϵ}_0)$, where ${ϵ}_0$ is vacuum permittivity. By superposition principle and Coulomb's law, its exact electric potential away from the two point charges isFor $r=|\mathbf{r}|\gg d$, retain the first nonzero term of a Taylor series. The electric potential and its negative gradient becomeHere $\hat{\mathbf{r}}=\mathbf{r}/r$ and the products in parentheses are dot products. These are exact exterior formulas for the ideal electric dipole and distant approximations for a finite pair. The electric field on the outer axis points along $\mathbf{p}$ with magnitude $2k_{e}p/r^3$; on the perpendicular bisector it points opposite $\mathbf{p}$ with magnitude $k_{e}p/r^3$. This directional structure is what a scalar total electric charge misses. ^[ref].

In a prescribed uniform external electric field ${\mathbf{E}}_{\mathrm{ext}}$, a rigid pair has opposite forces ${\mathbf{F}}_{\pm }=\pm q{\mathbf{E}}_{\mathrm{ext}}$. Its net force is zero, but summing their moments about the midpoint gives the torqueThe cross product makes the torque perpendicular to both the electric dipole moment and the external electric field. If $\theta$ is the angle from ${\mathbf{E}}_{\mathrm{ext}}$ to $\mathbf{p}$ in their plane, the signed turning torque is ${\tau }_{\theta }=-p{E}_{\mathrm{ext}}\sin \theta$. It tends to reduce a small positive $\theta$. ^[ref].

The external interaction potential energy, with zero chosen at $\theta =\pi /2$, isParallel alignment minimizes this potential energy; antiparallel alignment maximizes it even though the torque is zero in both positions. This omits the pair's constant internal binding potential energy and assumes a fixed permanent electric dipole moment. ^[ref].

A nonuniform external electric field can exert a net force on a neutral electric dipole. Before taking the small-pair approximation, the exact external force is $\mathbf{F}=q[{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}+\mathbf{d}/2)-{\mathbf{E}}_{\mathrm{ext}}(\mathbf{R}-\mathbf{d}/2)]$, where $\mathbf{R}$ is the midpoint. A first-order Taylor series gives, for a fixed electric dipole moment and an external electric field varying slowly across $d$,in electrostatics. The last equality uses the vanishing curl of the external electric field, with $\mathbf{p}$ held fixed while differentiating. Thus uniform external electric fields turn a misaligned electric dipole, whereas gradients can also move its midpoint. For example, the local source-free electric field ${\mathbf{E}}_{\mathrm{ext}}=Gx\hat{\mathbf{x}}+(E_0-Gy)\hat{\mathbf{y}}$ gives $\mathbf{F}=(G{p}_x,-G{p}_y,0)$, so reversing the electric dipole moment reverses the net force. This stipulated local model has zero divergence and zero curl.

A motivating experiment is Filsinger and colleagues' study of electrostatic deflection of polar molecules. Benzonitrile (${\mathrm{C}}_7{\mathrm{H}}_5\mathrm{N}$) in helium expanded from a pulsed valve into vacuum; two 1-mm skimmers narrowed the beam before a 15-cm deflector. A grounded trough and a high-voltage rod produced a vertical electric field gradient. Scanning a focused laser vertically downstream and recording benzonitrile ions from photoionization measured the beam's relative density. With voltage off, the profile spanned about 2 mm; applying voltage shifted and broadened it upwards. At 10 kV, many molecules lay outside the original profile. ^[ref].

Interpretation uses a quantum state's Stark effect energy $W(E)$: ${\mu }_{\mathrm{eff}}=-dW/dE$ gives $F_y={\mu }_{\mathrm{eff}}{\partial }_{y}E$. States whose energy falls with increasing electric field seek stronger fields. The voltage-off profile controls the mechanical aperture. Inferring state populations needs calculated trajectories, speed distributions and a rotational temperature model; profile displacement alone does not measure a permanent electric dipole moment uniquely. This is an electric Stark effect experiment, with photoionization occurring after deflection. ^[ref].

For the symmetric finite pair, the first omitted electric multipole expansion term after the electric dipole is third order, so typical relative errors scale as $(d/r{)}^2$. This stronger error estimate need not hold for an arbitrary neutral molecule with a nonzero electric quadrupole moment. A concrete check for a pair on the $z$ axis compares the exact and ideal electric fields at $r>d/2$ on the positive axis and at $r>0$ on the perpendicular bisector:Both perpendicular-bisector components are negative. At $r=5d$, the ratios are approximately $1.0203$ and $0.9852$: errors of $+2.03\%$ and $-1.48\%$ relative to the ideal electric field. The condition $r\gg d$ concerns observing the pair's own electric field; the separate condition for its external force approximation concerns variation of the applied electric field across the pair.

Electric flux measures how much electric field crosses a chosen oriented surface, with a sign distinguishing the two crossing directions. A strong electric field running along a surface contributes nothing to its electric flux: only the normal component counts. This lets Gauss's law turn a potentially complicated electric field into one number recording the net enclosed electric charge.

For a surface $S$ with chosen unit normal $\mathbf{n}$, the electric flux is the surface integralHere $dA$ is positive surface area, while $d\mathbf{S}$ is the vector area element. The dot product makes electric flux a signed scalar, rather than a vector like the electric field. Its SI unit is $\mathrm{N}{\mathrm{m}}^2/C$: electric field measured using the newton per coulomb, multiplied by surface area in square metres. Reversing the orientation of a surface replaces $\mathbf{n}$ by $-\mathbf{n}$ and reverses the electric flux. Calling it electric flux does not imply that electric charge is moving through the surface. ^[ref].

For an open patch in a plane, with surface area $A$ and a constant electric field, the unit normal is constant and the surface integral simplifies toThe angle $\theta$ is between the electric field and the chosen unit normal, not between the electric field and the plane. The cosine describes the signed projected surface area seen by the electric field. Thus electric flux is positive for $0\le \theta <90^{\circ }$, zero at $90^{\circ }$, and negative for $90^{\circ }<\theta \le 180^{\circ }$. ^[ref].

For example, take $E=100N/C$ and $A=0.02{\mathrm{m}}^2$, so $EA=2\mathrm{N}{\mathrm{m}}^2/C$. Three choices of angle giveKeeping the $60^{\circ }$ patch fixed and reversing its unit normal gives $120^{\circ }$ and changes $+1$ to $-1\mathrm{N}{\mathrm{m}}^2/C$. The electric field and physical patch have not changed. If the electric field is instead tangent to the patch, its dot product with the unit normal vanishes: electric flux is zero even though $E=100N/C$.

For a closed surface enclosing a volume $V$, choose the outward unit normal. Gauss's law givesHere ${\epsilon }_0$ is the vacuum permittivity, and $Q_{\mathrm{enclosed}}$ is the signed total electric charge inside. The electric field in the surface integral includes all sources, including external electric charges. Assume no ideal point charge lies on the integration surface. Zero enclosed electric charge implies zero total electric flux; it does not imply zero electric field everywhere on the closed surface or inside it. Positive and negative contributions to the surface integral can cancel. ^[ref].

To see the cancellation with sources inside, place point charges $+q$ at $(-a,0,0)$ and $-q$ at $(a,0,0)$, where $q>0$, $a=0.30\mathrm{m}$, and both lie inside the sphere of radius $R=1\mathrm{m}$ centered at the origin. This separated pair has the electric dipole geometry. Its enclosed electric charge is $q-q=0$, so its total outward electric flux is zero. At the point $P=(0,0,R)$ on the sphere, Coulomb's law and the superposition principle give the individual electric fieldsThe $z$ components cancel and the positive $x$ components add:For the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $q=1\mathrm{nC}$, this is about $4.740N/C$ along positive $x$. At $P$, the outward unit normal is $\hat{\mathbf{z}}$, so this nonzero electric field is tangent to the sphere. At other points of the sphere, its normal component can be positive or negative. The source-field formula applies away from the two point charges, where the electric field has zero divergence; it is singular at the sources. ^[ref].

The diagram shows streamlines of the three-dimensional electric field in the section $y=0$. The circle is a cross-section of the sphere, not the integration surface: Gauss's law uses the entire three-dimensional surface integral. A two-dimensional line integral of outward field around that circle would be a different quantity, with units $\mathrm{N}m/C$ rather than the electric flux units $\mathrm{N}{\mathrm{m}}^2/C$.

In electrostatics, the electric potential $\Phi$ tells us the interaction potential energy per unit test electric charge in the electric field of prescribed sources. Its most directly useful quantity is an electric potential difference: how much work per unit electric charge an external agent must supply to move that test electric charge slowly between two points. This replaces the bookkeeping of a vector force along a path with the difference of two scalar values. The test electric charge must be small enough not to rearrange the sources appreciably.

For motion from $A$ to $B$, write $\Delta \Phi =\Phi (B)-\Phi (A)$. With fixed sources and only the electric force doing work,Here $U$ is interaction potential energy, $K$ is kinetic energy, and the line integral uses the electric field produced by the other electric charges. For slow transport with unchanged kinetic energy, the external work is $W_{\mathrm{ext}}=\Delta U$, so $\Delta \Phi =W_{\mathrm{ext}}/q$ for nonzero $q$. This external work has the opposite sign to the work done by the electric field. The SI unit is the volt,A volt measures energy per unit electric charge, not energy itself. For example, crossing an electric potential difference of $+5\mathrm{V}$ changes the potential energy of $q=+2\mathrm{nC}$ by $+10\mathrm{nJ}$, but changes that of $q=-2\mathrm{nC}$ by $-10\mathrm{nJ}$. Both test electric charges encounter the same electric potential difference; their opposite signs reverse the energy change.^[ref]

The electrostatic potential is a scalar function of position. The electric field is a conservative vector field, and all three components follow from the negative gradient of the electric potential:The electric field points toward decreasing electric potential. A positive test electric charge feels electric force in that direction; a negative test electric charge feels electric force in the opposite direction, since $\mathbf{F}=q\mathbf{E}$. A freely accelerated negative test electric charge therefore moves toward higher electric potential while lowering its potential energy $U=q\Phi$.^[ref]

Only differences fix physical electrostatic potentials: adding the same constant $C_0$ everywhere, ${\Phi }^{\prime }=\Phi +C_0$, changes neither the gradient nor any electric potential difference. Choosing an electrical reference potential is this elementary gauge transformation. Setting zero at infinity is convenient for finitely many localized point charges; it is not available for an ideal uniform electric field extending to infinity. There we can choose a finite reference point. Assigning zero at ground is a convention, whereas physically grounding an electrical conductor also allows exchange of electric charge to impose a boundary value. Zero electric potential at a point never by itself implies zero electric field.^[ref]

For a fixed point charge $Q$ at the origin in vacuum, choose zero at infinity. Coulomb's law giveswhere ${\epsilon }_0$ is the vacuum permittivity. Use the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and $Q=+2\mathrm{nC}$. At $r=0.20\mathrm{m}$,A test electric charge $q=-3\mathrm{nC}$ has $U=-2.697\times 10^{-7}\mathrm{J}$ and radial electric force $F_r=-1.3485\times 10^{-6}\mathrm{N}$, directed inward. Moving it inward from $r_A=0.20\mathrm{m}$ to $r_B=0.10\mathrm{m}$ raises the electric potential from $89.9$ to $179.8\mathrm{V}$ but lowers its potential energy:This is the Coulomb potential energy calculation using electric potentials. The source's singular self-field at $r=0$ is excluded; when finding the force on a source point charge, use only the electric field of the other sources.^[ref]

For a uniform electric field $\mathbf{E}=300V/m\hat{\mathbf{x}}$, choose $\Phi (0)=12\mathrm{V}$. ThenMoving from $x_A=0$ to $x_B=0.04\mathrm{m}$ gives $\Delta \Phi =-12\mathrm{V}$. For $q=+2\mathrm{nC}$ the change in potential energy is $-24\mathrm{nJ}$ and the electric field does $+24\mathrm{nJ}$ of work. For $q=-2\mathrm{nC}$, the same prescribed displacement gives $\Delta U=+24\mathrm{nJ}$: the external agent must supply that work for slow transport. This uniform electric field approximates the interior of a parallel-plate capacitor sufficiently far from the edges; the fringing field of a capacitor limits the approximation.^[ref]

An equipotential is a level set $\Phi =\text{constant}$. On a smooth equipotential with nonzero gradient, every tangent displacement has $d\Phi =\nabla \Phi \cdot d\mathbf{l}=0$, so the electric field is normal to the equipotential. Thus moving a test electric charge along an equipotential requires no net electrical work. Around a point charge, the equipotentials are spheres; the circles in the diagram are their sections, not an assumption of two-dimensional Coulomb's law. A uniform electric field has parallel plane equipotentials. For contours separated by the same small electric potential difference, smaller normal separation means a stronger electric field, approximately $|\mathbf{E}|\approx |\Delta \Phi |/\Delta n$. At a point where $\nabla \Phi =0$, the electric field is zero and has no direction to call normal.^[ref]

The superposition principle adds electric potentials as signed scalars. For finitely many stationary point charges $q_j$ at ${\mathbf{r}}_j$ in vacuum,For example, the electric dipole with $+q$ at $(a,0,0)$ and $-q$ at $(-a,0,0)$ has zero electric potential everywhere in the midplane $x=0$, yet its electric field at the midpoint is $\mathbf{E}=-(2kq/a^2)\hat{\mathbf{x}}$. With $q=2\mathrm{nC}$ and $a=0.10\mathrm{m}$, this is $-3596N/C\hat{\mathbf{x}}$. At $(2a,0,0)$, the same model gives $\Phi =2kq/(3a)\approx 119.867\mathrm{V}$ and $E_x=8kq/(9a^2)\approx 1598.222N/C$. Cancellation of electric potentials is quite different from cancellation of vector electric fields.

A motivating measurement is conductive-paper electric potential mapping. A documented teaching experiment uses slightly conductive paper on an insulating board, two circular electrodes in highly conducting ink, a steady supply, and a digital voltmeter. Connect the supply to the two electrodes, keep the voltmeter common lead on the negative electrode, and use its other lead as a movable probe. Adjust the supply using the measured electric potential difference between electrodes; the published procedure uses $10\mathrm{V}$. The observable is the voltmeter reading $\Phi (P)-\Phi (\mathrm{COM})$ at the probe position $P$. Locate several points reading $2\mathrm{V}$, record their positions on a separate map, and join them into an equipotential; repeat at other readings. Lift the probe between contacts so it does not damage the paper. This procedure makes the electrical reference potential concrete: every reading compares two contacts.^[ref]

From neighboring readings infer an electric field component by a finite difference, $E_x\approx -[\Phi (x+h)-\Phi (x)]/h$. Reconstruct two in-plane components with displacements along two independent directions. Alternatively, draw the electric field normal to mapped equipotentials, toward lower electric potential. The second diagram illustrates the same measurement principle with bar electrodes spanning a uniform sheet, insulating top and bottom edges, and a synthetic model $\Phi (x)=(50V/m)x$. The model readings $4.0\mathrm{V}$ at $x=0.08\mathrm{m}$ and $5.0\mathrm{V}$ at $x=0.10\mathrm{m}$ imply $E_x=-50V/m$. These values are calculated illustrations, not reported experimental data. The documented circular-electrode experiment similarly estimates the electric field from readings at successive centimetre marks.^[ref]

Useful controls include checking the voltmeter zero, checking that the electric potential difference along each painted electrode is negligible, and repeating reference readings to detect supply drift. Reversing the supply should reverse the signs relative to the same reference contact. Contact resistance, leakage electric currents, and touching the sheet can distort the map. A voltmeter has finite input electrical resistance and draws a small electric current; voltmeter loading changes the quantity being measured if that electrical resistance is insufficiently large relative to the sheet's effective source electrical resistance.^[ref][ref]

The paper carries a steady electric current, so this is a planar analogue rather than a direct measurement of isolated point charges in three-dimensional vacuum. For constant electrical conductivity, Ohm's law and steady charge conservation give $\mathbf{j}=-\sigma \nabla \Phi$ and $\nabla \cdot \mathbf{j}=0$, hence the two-dimensional Laplace equation away from electrodes. Insulating sheet edges impose zero normal electric current, while finite electrodes fix boundary values. This derivation explains both the analogy and its limits: inhomogeneous paper, finite boundaries, and finite electrodes can change the contours. One should not fit the three-dimensional point charge formula $kQ/r$ to a planar map without testing that model. Small probe spacing resolves spatial variation better, but amplifies uncertainty in the inferred finite difference. The published laboratory instructions explicitly ask students to assess whether their proposed field fit is appropriate.^[ref][ref]

The scalar description has a crucial limit. Faraday's law of induction for a stationary closed path $\Gamma$ sayswhere ${\Phi }_B$ is magnetic flux. Nonzero changing magnetic flux gives nonzero circulation, incompatible with a single-valued electric potential alone. More generally, a scalar electric potential and a magnetic vector potential together giveConsequently, in the presence of induction, two voltmeter lead routes between the same contacts can produce different readings if the enclosed magnetic flux changes. This follows by subtracting the two path line integrals; it is a deduction from Faraday's law of induction, not a claimed outcome of the paper experiment. The steady setup and negligible induction justify interpreting its readings as electric potential differences. Merely introducing a scalar electric potential that depends on time cannot supply a nonzero curl of the electric field.^[ref]

A short mathematical justification connects the statements above. Along a smooth path, the chain rule gives $d\Phi =\nabla \Phi \cdot d\mathbf{l}$. Integrating and using $\mathbf{E}=-\nabla \Phi$ yields the endpoint formula by the fundamental theorem for line integrals, proving path independence. Conversely, vanishing circulation around every closed path lets us define $\Phi (P)=-{\int }_{P_0}^P\mathbf{E}\cdot d\mathbf{l}$ independently of the route, with $\Phi (P_0)=0$. Zero curl is a local test; for a continuously differentiable curl-free vector field, a simply connected open region suffices for a global potential. The explicit finite-source formula already gives a global electric potential away from its source points.^[ref]

Electromagnetism describes how electric charge and electric current interact through the electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$. The Maxwell equations describe the fields and their sources; the Lorentz force describes the response of a particle of electric charge $q$ and velocity $\mathbf{v}$:The electric field can change the particle's energy. The magnetic part of the Lorentz force is perpendicular to velocity, so magnetic forces do no work; they can bend a trajectory. These are local interactions: a particle responds to the electromagnetic field at its position. ^[ref].

The remarkable unification is that the same Maxwell equations connect Coulomb's law, Faraday's law of induction, and freely propagating electromagnetic waves. A changing electric field contributes to magnetic circulation through displacement current; together with Faraday's law, this permits waves in vacuum without conducting matter. The Poynting theorem then explains how the fields store and transport energy. This connects the force on a charge to the energy carried by light.

In his 1865 paper, James Clerk Maxwell derived transverse disturbances and compared the propagation speed inferred from electrical measurements with independently measured light speeds. Part VI, especially §§95–97, argues that their agreement supports identifying light as an electromagnetic wave. This was a comparison with finite experimental accuracy, rather than a numerical identity between the historical measurements. ^[ref]; ^[ref].

For two point electric charges in vacuum, Coulomb's law gives the interaction potential energywith zero at infinite separation. Here $r_{ij}$ is the distance between them and ${ϵ}_0$ is the vacuum permittivity. Opposite signs give negative potential energy and attraction; equal signs give positive potential energy and repulsion. For a finite collection of distinct point electric charges, sum over $i<j$ to count each pair once. A bulk ionic crystal requires the extended lattice sum, including both signs, rather than multiplying one attractive pair by the coordination number.^[ref]

At short distances, overlapping electron density makes the simple point-charge picture inadequate. Quantum mechanics, including the Pauli exclusion principle, matters to the steep repulsive contribution. Pure attraction proportional to $-1/r$ would collapse as $r\to 0$; it cannot set an equilibrium spacing on its own. A useful illustrative model isHere $r$ scales an assumed fixed arrangement, $A$ packages its attractive contribution, and $B/r^n$ represents short-range repulsion. The constants have units energy times length and energy times length to the power $n$, respectively. This is a model, not a fitted description of a named salt.^[ref]

Differentiating gives $r_0=(nB/A{)}^{1/(n-1)}$ and $U^{\prime \prime }(r_0)=(n-1)A/r_0^3>0$, so the stationary spacing is a stable minimum. For the diagram's stipulated $n=9$, put $x=r/r_0$ and $E_0=A(1-1/n)/r_0>0$. ThenThe dimensionless curves show how competing interactions select a spacing. Their energy scale and exponent are synthetic; neither is a measured lattice energy or a melting prediction.

Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every electron is motionless. ^[ref].

For two identical spheres sufficiently far apart, Coulomb's law and Newton's law of universal gravitation both give inverse-square forces. Equal-sign charges therefore balance attraction when the displayed charge-to-mass relation holds. For uniform mass density $\rho$, substituting $M=4\pi \rho a^3/3$ fixes the radius. Changing the separation does not restore balance if the charge-to-mass ratio is wrong. Spherically symmetric charge makes the monopole result exact outside each sphere; polarization can require a far-field approximation.

For a charge $q$ at $\mathbf{p}$, $R=|\mathbf{p}|>a$, outside a grounded sphere of radius $a$, the displayed image gives zero potential on the sphere and decay at infinity. On $|\mathbf{x}|=a$, one has $|\mathbf{x}-{\mathbf{p}}^{\prime }|=(a/R)|\mathbf{x}-\mathbf{p}|$, proving boundary cancellation. The induced force is inward, of magnitude $q^{2}aR/[4\pi {\epsilon }_0(R^2-a^2{)}^2]$, from Coulomb's law applied to the image field. The self-field is excluded.

For radii $a<b$ and a point charge $q$ at $r_0{\mathbf{e}}_z$, $a<r_0<b$, set $k=b/a$. The method of images gives the two families $(q{k}^n,r_0{k}^{2n}{\mathbf{e}}_z)$ and $(-qa{k}^n/r_0,a^2{k}^{2n}{\mathbf{e}}_z/r_0)$ for $n\in \mathbb{Z}$. The first family's $n=0$ term is the physical point charge. All other image charges lie outside the gap. Sum their Coulomb's law electric potentials with factor $1/(4\pi {ϵ}_0)$.

For a short proof, reflection of $(Q,z)$ in a grounded sphere of radius $R$ gives $(-RQ/z,R^2/z)$. On $r=a$ this pairs first-family index $n$ with second-family index $-n$; on $r=b$ it pairs $n$ with $1-n$. Each pair has zero boundary electric potential. The negative-index electric potential tails are $O(k^n)$ and the positive-index tails are $O(k^{-n})$, uniformly on the closed gap after excluding the physical singularity. These geometric series establish absolute convergence, boundary cancellation, and termwise differentiation away from the physical point charge. Uniqueness of Poisson equation identifies the solution. A spherical Gaussian surface just outside $a$ encloses the first family with $n\le -1$ and the second with $n\le 0$, so Gauss's law and the geometric series give $Q_a=-qa(b-r_0)/(r_0(b-a))$.

For one ionic species of number density $n_i$, the ion-sphere radius is $a_i=(3/(4\pi n_i){)}^{1/3}$. The ionic Coulomb coupling parameter compares the interaction energy from Coulomb's law at that separation with thermal energy $k_{B}T$. At fixed composition, $\Gamma \propto {\rho }^{1/3}/T$. Values near one mark significant corrections to the ideal ion gas; strong coupling can lead to an ordered solid. Screening, mixtures and quantum motion modify detailed transitions, so this estimate is not a universal crystallization boundary. It is separate from the electron degeneracy criterion based on the electron Fermi temperature.

An ionic crystal repeats an arrangement of ions in three dimensions. A unit cell describes that repetition, whereas a formula unit describes composition. The two accounting devices answer different questions. Coulomb's law is not restricted to an arbitrarily drawn line connecting one cation to one anion: every ion contributes to the electric field experienced by the others.

The Madelung constant, named for Erwin Madelung, is a dimensionless geometrical factor in the electrostatic binding of an ionic crystal. For a neutral lattice of electric charges $\pm e$ with nearest unlike-ion distance $r_0$, Coulomb's law gives the potential energy per ion pair as $-{\alpha }_M{e}^2/(4\pi {ϵ}_0{r}_0)$. At a positive reference site, ${\alpha }_M=-{\sum }_{j\ne 0}{z}_j{r}_0/r_j$, with neighboring electric charges $z_{j}e$. Both attraction and repulsion enter this lattice sum; summing with a neutral-bulk prescription is essential. Short-range overlap repulsion and other contributions must be added before inferring a stable spacing or a thermodynamic lattice energy.

The image charge for a grounded conducting sphere is obtained by the Kelvin transform:It lies inside the electrical conductor, outside the region where it represents the physical electric field. The electric potential isAt the physical point charge, the electric potential has its usual Coulomb's law singularity. There is no physical singularity at the image charge: the interior expression is identically zero.

For $|\mathbf{r}|=a$, direct expansion of squared distances givesConsequently the two terms cancel exactly, proving the zero Dirichlet boundary condition on the sphere. At infinity each term is $O(r^{-1})$, and more explicitlyThus the Dirichlet boundary condition at infinity also holds. The candidate solves the Poisson equation with the one physical point charge; uniqueness of Poisson equation, with these Dirichlet boundary conditions, identifies it as the required electric potential.

Put $k=b/a>1$ and ${\mathbf{e}}_z=(0,0,1)$. The complete construction of image charges between concentric grounded spheres consists of two families, indexed by every integer:The term $(q_0,{\mathbf{r}}_0)$ is the physical point charge; every other term is an image charge. In the first family $n\le -1$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. In the second family $n\le 0$ lies inside $r=a$ and $n\ge 1$ lies outside $r=b$. Thus no image charge lies in the physical gap.

To derive and verify these families, reflection of an axial point charge $(Q,z)$ in a grounded sphere of radius $R$ replaces it by the image charge $(-RQ/z,R^2/z)$. Reflection in $a$ sends $(q_n,{\mathbf{r}}_n)$ to $({\tilde{q}}_{-n},{\tilde{\mathbf{r}}}_{-n})$; reflection in $b$ sends it to $({\tilde{q}}_{1-n},{\tilde{\mathbf{r}}}_{1-n})$. Successive reflections therefore generate exactly the listed families. The method of images givesThe physical point charge itself is excluded when evaluating this formula. On $r=a$, pair the first-family term $n$ with the second-family term $-n$; their electric potentials cancel by the single-sphere distance identity. On $r=b$, pair $n$ with $1-n$ instead. These reindexings are justified by absolute convergence, as follows.

For $n\to -\infty$, both source positions tend to the origin while their electric charges have size $O(k^n)$. On the closed gap their distances are eventually at least $a/2$, so both electric potential terms are $O(k^n)$. For $n\to +\infty$, both distances are at least a positive constant times $k^{2n}$ and the electric charges are $O(k^n)$, giving $O(k^{-n})$ terms. Both tails are bounded by convergent geometric series. This proves absolute convergence and uniform convergence of the tails on the closed gap. On any compact set avoiding the physical point charge, the differentiated tails obey the same type of bounds: for example, the large positive tail of the electric field is $O(k^{-3n})$. Thus termwise differentiation is valid and the sum is a harmonic function except at the physical point charge, where it has the correct Coulomb's law singularity. The boundary cancellations and uniqueness of Poisson equation complete the construction.