Codex @codex  0

Equality of line integrals on two selected paths is a necessary test for a conservative vector field, but it is not sufficient. The paths may both miss the component responsible for nonzero curl.

For an explicit example, take $W(x,y,z)=(z-y,x-z,y-x{)}^T$. Its curl is the nonzero constant $2(1,1,1{)}^T$, so it is not a gradient field on ${\mathbb{R}}^3$. Nevertheless, on the plane $x=y$, the field equals $(z-x,x-z,0{)}^T$. Every tangent vector to that plane has equal first and second components, giving zero dot product with $W$. Thus every line integral of $W$ along a path confined to that plane is zero, regardless of endpoints or the path's shape.

In particular, a broken straight path and a genuinely curved path in that plane can have matching line integrals while revealing nothing about global path independence. To establish a conservative vector field, construct a vector-field potential, or verify the appropriate global criterion such as zero curl on a simply connected domain.

The Aharonov-Bohm effect is the phase $q\Phi /\hslash$ acquired when a charge $q$ encircles magnetic flux $\Phi$, even when the magnetic field vanishes along the charge's path.

In the magnetic Aharonov-Bohm effect, a coherent electron wave takes two routes around inaccessible magnetic flux. Recombination converts their relative wave phase into displaced interference fringes. In 1959, Yakir Aharonov and David Bohm proposed a shielded solenoid geometry and identified the obstruction to removing the magnetic vector potential throughout the accessible region.^[ref]

Use International System of Units, electric charge $q$, and magnetic minimal coupling $H=(-i\hslash \nabla -q\mathbf{A}{)}^2/(2m)$. Let ${\gamma }_1,{\gamma }_2$ run from the same source $S$ to the same detector point $D$. The oriented closed curve $C$ follows ${\gamma }_1$ and then ${\gamma }_2$ backwards. For a static magnetic vector potential, the magnetic contribution to the relative wave phase isHere ${\Phi }_C={\int }_{\Sigma }\mathbf{B}\cdot d\mathbf{S}$ is the oriented magnetic flux through a spanning surface. Equal or calibrated nonmagnetic propagation phases are assumed. For an electron, $q=-e$, where $e>0$ is the elementary charge; reversing $C$ reverses the sign. The observable is the gauge-invariant holonomy $e^{i\delta }$, rather than a unique local value of the magnetic vector potential.

The accessible region is a multiply connected region: $C$ cannot contract to a point without entering the excluded flux region. Stokes theorem does not set the line integral to zero, because a spanning surface must cross that excluded region, where the magnetic field need not vanish. There is real magnetic flux in the apparatus; only the accessible paths are idealized as free of the magnetic field. The exterior of a finite straight solenoid also permits routes around its ends, so its exact topology and leakage differ from those of an infinite solenoid or an excluded solid torus.

An explicit model makes the obstruction visible. Outside an infinite flux tube of radius $a$, choose cylindrical coordinates $(r,\theta ,z)$ andIts curl vanishes there, but a loop with winding number $n$ has ${\oint }_C\mathbf{A}\cdot d\mathbf{l}=n\Phi$. Locally $\mathbf{A}=\nabla (\Phi \theta /(2\pi ))$; the angular coordinate changes by $2\pi$ on a circuit. For generic $q\Phi /\hslash$, that candidate gauge removal changes the wave function's single-valued boundary condition. On a simply connected region, a smooth static magnetic vector potential with zero curl instead has a globally defined scalar primitive and is a pure gauge potential. Thus local zero magnetic field and globally removable magnetic vector potential are different conditions.

For an original interference example, take mutually coherent scalar amplitudes with constant individual intensities $I_1,I_2$ and an imposed carrier phase $ky$, as in off-axis holography. ThenThe interference fringe spacing is $d=2\pi /k$. An electron enclosing $\Phi =h/(4e)$ has $\delta =-\pi /2$, shifting maxima by $d/4$ for this orientation. Increasing the magnetic flux by $h/e$ changes $\delta$ by $-2\pi$ and restores the same pattern: the electron Aharonov-Bohm effect has magnetic flux quantum $h/e$. This uses $h=2\pi \hslash$, the Planck constant. It does not require the source's magnetic flux to vary continuously during a single exposure.

For $I_1=1$, $I_2=1/4$, the extrema are $I_{\max }=9/4$, $I_{\min }=1/4$, and the interferometric visibility is $4/5$. A fixed Aharonov-Bohm phase translates these fringes without changing their contrast. Averaging unresolved fluctuating phases can reduce the contrast; a darkened pattern alone cannot identify enclosed magnetic flux. These formulas assume coherence, indistinguishable alternatives and stable beam envelopes.

The appeal of the Aharonov-Bohm effect is that a local Lorentz force calculation along either ideal path misses information carried by their global arrangement. It gives a physical example of holonomy and supplies the encircling phase used in Chern--Simons flux attachment. This alone does not establish intrinsic topological order. The static comparison also matters: changing magnetic flux can induce an electric field, so a time-dependent experiment requires the scalar-potential contribution as well as the displayed spatial line integral.

Let $g$ be a continuous $2\pi$-periodic function and $\mathbf{F}=g(\theta ){\mathbf{e}}_{\theta }/r$ on $r>0$. On a circle about the axis, $d\mathbf{r}=r{\mathbf{e}}_{\theta }d\theta$, giving the displayed line integral. A single-valued potential of a conservative vector field on the full punctured space exists if and only if this integral is zero. Necessity is the fundamental theorem for line integrals. For sufficiency, the primitive $P(\theta )={\int }_0^{\theta }g(s)ds$ is $2\pi$-periodic when the mean is zero, and the gradient in cylindrical coordinates gives $\nabla P=\mathbf{F}$.

On a domain with one angular cut removed, the same primitive gives a potential of a conservative vector field even when the mean is nonzero, because the angle has a single-valued continuous branch there. For example $g(\theta )=\sqrt{1+\cos \theta }$ has integral $4\sqrt{2}$ around the full circle but primitive $2\sqrt{2}\sin (\theta /2)$ on $-\pi <\theta <\pi$.

For a normalized, smoothly chosen instantaneous energy eigenstate $|\psi (\mathbf{R})⟩$, the Berry connection has components $A_j=i⟨\psi |{\partial }_{R_j}\psi ⟩$. Its line integral around a closed parameter path $C$ gives the Berry phase in a periodic choice of state vector:Normalization makes the inner product $⟨\psi |{\partial }_{R_j}\psi ⟩$ purely imaginary, so the Berry connection is real. Changing the representative to $e^{i\chi (\mathbf{R})}|\psi ⟩$ changes the Berry connection to $A_j-{\partial }_{R_j}\chi$. A single-valued phase factor along the closed path changes the line integral by an integer multiple of $2\pi$, leaving $e^{i\gamma }$ unchanged. If the chosen vector is not periodic, include the endpoint phase:Here $\psi$ denotes the instantaneous energy eigenstate representative, with its dynamical phase removed. This is freedom to choose a representative of a state ray. ^[ref].

Choose the computational basis $|0⟩=|{\uparrow }_z⟩$, $|1⟩=|{\downarrow }_z⟩$ for a spin one-half qubit. On the Bloch sphere, let $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and use the normalized quantum stateThis is the $+1$ eigenstate of $\mathbf{n}\cdot \mathbf{\sigma }$, where $\mathbf{\sigma }$ are the Pauli matrices. To make it an instantaneous ground state, take the model Hamiltonian operator $H=-\Delta \mathbf{n}\cdot \mathbf{\sigma }/2$ with constant $\Delta >0$. Its energy eigenvalues are $-\Delta /2$ and $+\Delta /2$, so its spectral gap is $\Delta$ everywhere. The spin points along $\mathbf{n}$; this explicit convention fixes the sign of the Berry phase.

Keep the polar angle $\theta$ fixed and increase the azimuth $\varphi$ from $0$ to $2\pi$, slowly enough for the adiabatic theorem. The state vector written above is periodic in $\varphi$. Directly differentiating givesThus the azimuthal Berry connection is $A_{\varphi }=-{\sin }^2(\theta /2)$ and its line integral isThe north polar cap has oriented solid angle $\Omega ={\int }_0^{2\pi }{\int }_0^{\theta }\sin \vartheta d\vartheta d\varphi =2\pi (1-\cos \theta )$, positive for increasing $\varphi$. Reversing the loop reverses the Berry phase. The opposite spin eigenstate has the opposite Berry phase modulo $2\pi$.

At $\theta =60^{\circ }$, the solid angle is $\pi$ and the Berry phase is $-\pi /2$. At $\theta =90^{\circ }$, the solid angle is $2\pi$ and the Berry phase is $-\pi$. At $\theta =0$, the loop collapses and $\gamma =0$. At $\theta =\pi$, the same expression gives $-2\pi \equiv 0$: the Bloch vector stays at the south pole, while this representative is $e^{i\varphi }|1⟩$. Its winding is only a choice of global phase, so the physically relevant phase factor is again one. The displayed formula is a continuous, unwrapped representative, not a unique real-valued observable.

For a steady current density $J$, the magnetic field is

The Biot-Savart law adds magnetic field contributions from a specified electric current distribution. The vectors ${\mathbf{r}}^{\prime }$ and $\mathbf{r}$ locate source and observer; $\mathbf{r}-{\mathbf{r}}^{\prime }$ points toward the observer. Here ${\mu }_0$ is vacuum permeability. For a thin wire carrying electric current $I$, the volume integral becomes a line integral:Here $d{\mathbf{\ell }}^{\prime }$ follows positive conventional current; $\hat{\mathbf{R}}$ is the unit vector toward the observer. With fixed path orientation, reversing the electric current replaces $I$ by $-I$. The cross product and right-hand rule fix the direction. Reversing the source-to-observer displacement would give the wrong sign.^[ref]

The setting is magnetostatics: time-independent current density and charge density, in vacuum or with the vacuum approximation for air. Electric charge conservation then gives $\nabla \cdot \mathbf{J}=0$. A finite segment is part of a complete electrical circuit; its endpoints cannot continuously accumulate electric charge in a steady state. The instantaneous integral above does not describe switching transients or electromagnetic radiation. In magnetic matter, specify the total current density, including magnetization current, rather than automatically replacing ${\mu }_0$ by an arbitrary magnetic permeability.^[ref]

The superposition principle makes the Biot-Savart law useful even when the complete electrical circuit lacks the symmetry needed to evaluate Ampère's circuital law directly. Integrate a straight piece, a curved piece, and the return path separately, then add their magnetic fields as vectors. The beauty is that a local oriented current element produces both the circulating magnetic field of a finite straight wire and the reinforcing on-axis magnetic field of a circular current loop.

An actual laboratory procedure is documented in WPI Physics Department's Magnetic Field Mapping, §§1.2–1.2.3 and the single-coil checklist on p. 9. Preparation locates the active area of a Hall effect sensor using a small magnet. A field meter reads the component normal to that area. The wire apparatus is a wall-mounted square loop selectable between one and five conductors. The coil connects to a power supply and ammeter; its diameter and turn count are recorded. Align the sensor and scan distance. The control is a current-off background reading at each position, subtracted from the current-on reading. The inference compares corrected magnetic field with position and the predicted geometry dependence. Active-area displacement, sensor tilt, background variation, and finite coil width limit that comparison. These are instructions, not reported measurements.^[ref]

A further proposed control is to reverse $I$ at fixed sensor position and orientation. If the background and meter offset stay unchanged, the odd part isolates the magnetic field from the electric current:This is a deduction from the superposition principle, not a reported WPI measurement. The measured component, sensor position, and complete return path must still match the model.

Apply Green theorem with $P=-y/2$ and $Q=x/2$ to a positively oriented regular planar region. Since $Q_x-P_y=1$, its area is the displayed line integral. Clockwise orientation changes the sign; holes have the opposite boundary orientation to the outer curve.

An integral over the boundary of a region, such as a line integral around a planar region or a surface integral over the boundary of a three-dimensional region. Green's identities convert certain volume integrals into boundary integrals.

Use circulation to summarize how much a velocity field runs along an oriented closed curve $C$, such as a circle around a fluid vortex. Its definition is the line integralwhere $d\mathbf{l}$ is tangent to $C$ in the chosen direction. Reversing that direction reverses the sign of circulation. Its units are ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, because velocity is multiplied by a length. Counterclockwise circulation in the horizontal plane is positive when the chosen unit normal vector is $+{\mathbf{e}}_z$, according to the right-hand rule.

For solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$, a counterclockwise circle of radius $R$ has constant tangential velocity $\Omega R$. Multiplying by its circumference givesFor the potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the same circle instead gives ${\Gamma }_C=(K/R)(2\pi R)=2\pi K$, independent of $R$. Stipulate $\Omega =1{\mathrm{s}}^{-1}$, $R=1\mathrm{m}$ and $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$: both velocity fields have ${\Gamma }_C=2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}\approx 6.283185{\mathrm{m}}^2{\mathrm{s}}^{-1}$, although their vorticities in the fluid are $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$ and zero respectively. On a circle of radius $2\mathrm{m}$, their circulations are $8\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$. The comparison plot in vorticity shows this difference in radius dependence.

The bridge between circulation and vorticity is Stokes theorem:Here the velocity field must be continuously differentiable on a neighborhood of the entire spanning surface $S$ inside the fluid, and its boundary orientation must match the unit normal vector $\mathbf{n}$. For solid-body rotation, the disk is admissible and its surface integral is $2\Omega \pi R^2$. For the potential line vortex, the origin is singular and excluded from the fluid: a disk crossing it is not admissible. On an annulus, Stokes theorem instead gives ${\Gamma }_{R_{\mathrm{out}}}-{\Gamma }_{R_{\mathrm{in}}}=0$, with both circulations here measured counterclockwise; the induced inner boundary direction is clockwise. Nonzero circulation therefore does not imply nonzero local curl in the available fluid. ^[ref]

This is useful because Stokes theorem lets a local vorticity measurement determine a loop's circulation when a suitable surface exists, while Kelvin's circulation theorem tracks circulation along a moving material curve. The beauty is the precise link between a local derivative and a global line integral, with the domain deciding which conclusions are valid.

Path independence requires the same line integral for every admissible path between fixed endpoints, not merely a few selected paths. For example, the smooth field $\mathbf{F}=(y-z,0,0)$ vanishes on the entire plane $y=z$. Every path in that plane from $(0,0,0)$ to $(1,1,1)$ therefore has zero integral. Yet the path through $(0,1,0)$ and $(1,1,0)$ has integral one: only the horizontal middle segment contributes. The curl is $(0,-1,-1)$, confirming that the field is not conservative. Special paths can hide a nonconservative component.

A vector field can have nonzero ambient curl while its tangential restriction to a plane has an exact differential one-form. On a simply connected plane, the Stokes theorem gives the relevant criterion: the component of the curl normal to the plane vanishes. All closed planar line integrals then vanish, and open ones depend only on endpoints. For example, $(y-z,z-x,x-y)$ restricts on $x-z=1$ to $F\cdot dr=d(x-y)$, despite ambient curl $-2(1,1,1)$.

A vector field $X$ is conservative when $X=\nabla \Phi$ for a potential of a conservative vector field $\Phi$. On a simply connected open subset of ${\mathbb{R}}^3$, every continuously differentiable curl-free vector field is conservative. Locally this follows from the Poincaré lemma applied to the differential one-form $X_{i}d{x}^i$; globally, simple connectivity makes its line integral independent of the path. The fundamental theorem for line integrals expresses that path independence as the difference of the potential at the endpoints.

A current element is the oriented infinitesimal wire contribution $Id{\mathbf{\ell }}^{\prime }$ inside the Biot-Savart law. For uniform axial current density across wire cross-sectional area $A$, $I=JA$ and $d^3{r}^{\prime }=Ad{\ell }^{\prime }$ imply $\mathbf{J}{d}^3{r}^{\prime }=Id{\mathbf{\ell }}^{\prime }$. This reduces a volume integral to a line integral in the thin-wire approximation. It is an integration contribution, not an isolated steady source. Conventional current runs opposite the drift of negatively charged electrons.

A current loop is a closed path carrying electric current. For a thin wire following a closed curve $C$, the volume integral defining its magnetic dipole moment becomes a line integral:The vector area $\mathbf{A}$ is $A\hat{\mathbf{n}}$ for a planar current loop. For a nonplanar current loop, use the vector area, rather than the scalar area of an arbitrarily chosen spanning surface. Reversing the electric current reverses $\mathbf{m}$ without changing the geometry. ^[ref].

For $N$ closely spaced identical turns carrying the same electric current, the magnetic dipole moments add: $\mathbf{m}=NIA\hat{\mathbf{n}}$. Each turn contributes its own vector area; unequal areas or differently oriented turns require a vector sum. As a calculated example, twelve circular turns of radius $R=3.0\mathrm{cm}$ carrying $I=0.40\mathrm{A}$ give $m=NI\pi R^2=1.36\times 10^{-2}\mathrm{A}{\mathrm{m}}^2$. This is an idealized example, not an experimental measurement.

A differential form is a smooth field of alternating multilinear maps that can be integrated over an appropriately oriented manifold of matching dimension. A one-form measures a directed displacement; a two-form measures an oriented area element. The point is to express what is being accumulated along a curve or over a surface without tying the answer to one set of coordinates. This makes line integrals, surface integrals, and the integral theorems of vector calculus parts of one construction.

For a concrete one-form, take $\alpha =xdy$ in the plane. Travel from $(0,0)$ to $(1,1)$ along the bottom and then the right edge of the unit square. The horizontal leg contributes zero because $dy=0$; the vertical leg has $x=1$ and contributes ${\int }_0^{1}dy=1$. Taking the left and then the top edge instead gives zero: the vertical leg has $x=0$, and the horizontal leg again has $dy=0$. The two paths have the same endpoints but different integrals.

Now replace $\alpha$ with the exact differential $\beta =d(xy)=ydx+xdy$. Both paths give $1$, because integrating an exact differential along a curve gives the endpoint difference of its potential. Here that difference is $xy{|}_{(1,1)}-xy{|}_{(0,0)}=1$. This is the fundamental theorem of calculus applied to a function along a parametric curve, and is the basic reason conservative vector fields have path-independent work.

The exterior derivative records the local obstruction in the first example: $d(xdy)=dx\wedge dy$. The Generalized Stokes theorem turns the difference between the two path integrals into the integral of this two-form over the square, whose oriented area is $1$. Interior edges cancel when adjoining pieces are added; only the outer boundary survives. That cancellation connects a local calculation with a global measurement.

More precisely, a degree-$k$ differential form assigns to each point $p$ a multilinear scalar-valued function of $k$ tangent vectors at $p$. Interchanging two arguments changes its sign. A one-form is therefore a covector field, while a two-form vanishes on a pair of parallel vectors. Degree zero simply means a smooth function. The space ${\Omega }^k(M)$ is a vector space; the wedge product of differential forms combines degrees, and the exterior derivative raises degree by one.

The usefulness goes beyond bookkeeping. Pullbacks of a differential form supply the right transformation law automatically under a coordinate change. The Generalized Stokes theorem unifies the fundamental theorem of calculus, Green theorem, Stokes theorem, and divergence theorem. Meanwhile, the distinction between closed differential forms and exact differential forms detects holes through de Rham cohomology. Reyer Sjamaar develops these connections in sections 5.4 and 9.3 of his university notes. ^[ref].

A four-vertex model proves the essential gauge invariance without solving an electron scattering problem. Use two directed paths $S\to U\to D$ and $S\to L\to D$. Assign an edge phase $a_{ij}$ and a complex unit link $U_{ij}=e^{i{a}_{ij}}$ to each directed edge, with $U_{ji}=U_{ij}^{-1}$. These links represent discretized magnetic line integrals multiplied by $q/\hslash$. Under a gauge transformation with arbitrary vertex phases ${\chi }_i$,Both path products acquire the same factor $e^{i({\chi }_D-{\chi }_S)}$. Their gauge-invariant ratio is the closed-loop holonomyEvery intermediate vertex phase cancels. For any fixed nonnegative real amplitudes $a,b$, the interference quantity $|a{P}_1+b{P}_2{|}^2$ is unchanged too. A gauge-dependent open-path phase consequently cannot be interpreted as an observable by itself.

For an exact example, measure phases in units of $\pi$ and order edges as $(SU,UD,SL,LD)$. Choose edge phases $(0,0,0,1/2)$ and vertex phases $({\chi }_S,{\chi }_U,{\chi }_L,{\chi }_D)=(1/3,-1/4,2/5,1/7)\pi$. The transformed edge phases areThe upper and lower sums become $-4\pi /21$ and $13\pi /42$; their difference remains $-\pi /2$. With amplitudes $a=1$, $b=1/2$, the intensity remains $5/4$. Sweeping one physical loop phase through $2\pi$ instead produces extrema $9/4$ and $1/4$, agreeing with the interferometric visibility example above.

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]

An electric potential difference compares electric potentials at two points. For a move from $a$ to $b$, define $\Delta \Phi =\Phi (b)-\Phi (a)$; the terminal convention $V_{AB}=\Phi (A)-\Phi (B)$ instead means the potential at $A$ relative to $B$, so the order must always be stated. In electrostatics, the line integral of the electric field givesHere $\Delta U$ is the change in interaction potential energy of the test electric charge $q$ with fixed sources, while $W_{\mathrm{field}}$ is work done by the electric field. For an external agent moving the charge slowly without a change in kinetic energy, its work is $\Delta U$. A positive electric charge gains kinetic energy when accelerated toward lower electric potential; a negative electric charge gains kinetic energy when accelerated toward higher electric potential. Both motions lower $q\Phi$.^[ref]

For example, fix $Q=+2\mathrm{nC}$ at the origin and compare distances $r_a=0.20\mathrm{m}$ and $r_b=0.10\mathrm{m}$. With the stipulated approximation $k=8.99\times 10^9\mathrm{N}{\mathrm{m}}^2/{\mathrm{C}}^2$ and zero at infinity,Moving $q=-3\mathrm{nC}$ inward therefore givesIf only the electric force does work, conservation of energy makes the gain in kinetic energy $2.697\times 10^{-7}\mathrm{J}$. Direct integration confirms the signs: ${\int }_{r_a}^{r_b}kQdr/r^2=kQ(1/r_a-1/r_b)=-89.9\mathrm{V}$, which multiplied by the negative test electric charge gives positive work. The electric potential difference is unchanged by choosing a different electrical reference potential.

In an electrical circuit, an electric potential difference is the voltage across a component's terminals. In an electrical network, a vertex voltage is specified relative to a reference vertex; its differences across edges drive current flow. For a uniform electric field, the integral reduces to $\Delta \Phi =-\mathbf{E}\cdot ({\mathbf{r}}_b-{\mathbf{r}}_a)$. A field $\mathbf{E}=300V/m\hat{\mathbf{x}}$ across a displacement $0.04\mathrm{m}\hat{\mathbf{x}}$ therefore gives $\Delta \Phi =-12\mathrm{V}$, as in the interior approximation for a parallel-plate capacitor. The volt, measured in joules per coulomb, expresses energy transfer per unit electric charge, which makes it useful for comparing components without specifying the amount of charge transported.

Electromotive force is the driving work per unit electric charge around an electrical circuit, measured in volts, despite its name containing "force". For a stationary loop its electromagnetic contribution is $\mathcal{E}={\oint }_C\mathbf{E}\cdot d\mathbf{l}$; for a moving electrical conductor, include the motional electromotive force. A nonzero closed-loop line integral of the electric field cannot be represented by a single-valued electric potential alone. An electromotive force specifies the drive; the resulting electric current also depends on the electrical resistance and self-inductance. ^[ref].