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On the unit sphere with one point deleted, a distinct nonantipodal pair has a length minimizer exactly when its shorter great circle arc avoids the deleted point. Antipodal pairs admit an avoiding semicircle. If the unique shorter arc crosses the puncture, smooth detours approach its length without attaining it.

The drag coefficient measures a drag force against a specified scale of force. Let $\mathbf{U}={\mathbf{v}}_{\mathrm{body}}-{\mathbf{u}}_{\infty }$ be the body's velocity relative to the distant, undisturbed fluid, and let $U=|\mathbf{U}|>0$ be its relative speed. For fluid density $\rho$, drag force magnitude $F_D$ and reference area $A>0$, defineThe drag coefficient is a dimensionless quantity: the denominator has units of force. The reference area must be reported with it. For a sphere of radius $a$, the usual choice is the frontal area $A=\pi a^2$. Choosing its entire surface area $4\pi a^2$ instead divides the reported drag coefficient by four for the same physical drag force. Comparing coefficients without comparing their reference areas can therefore be misleading. ^[ref]

For a rigid sphere in steady flow through an unbounded Newtonian fluid with constant dynamic viscosity $\mu$, incompressible flow and the no-slip boundary condition, the creeping-flow limit requires the diameter-based particle Reynolds number ${\mathrm{Re}}_d=2a\rho U/\mu \ll 1$. The Stokes drag law gives $F_D=6\pi \mu aU$. Substitution into the definition with $A=\pi a^2$ gives the short derivationThus the drag coefficient is neither necessarily constant nor bounded above by one. The formula containing $U^2$ still describes linear drag here, because the drag coefficient varies as $1/U$. The relation $C_D=24/{\mathrm{Re}}_d$ is the leading small-Reynolds number approximation, not a measured curve valid at every Reynolds number. ^[ref]

For a worked example, stipulate dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$, density $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$, radius $a=10^{-4}\mathrm{m}$ and relative speed $U=10^{-3}\mathrm{m}{\mathrm{s}}^{-1}$. Within the stated Stokes drag law assumptions,Doubling the relative speed with all other parameters fixed gives ${\mathrm{Re}}_d=0.004$ and $C_D=6000$: the predicted drag force doubles to $3.76991118\times 10^{-7}\mathrm{N}$. Falsely holding the drag coefficient at $12000$ would instead quadruple the drag force. Both particle Reynolds numbers remain small, so the comparison is within the creeping-flow approximation.

The useful idea is to separate the force scale from the dependence on geometry and flow conditions. A drag coefficient can help compare experiments, but changing Reynolds number can change the coefficient itself; NASA Glenn explicitly stresses matching the relevant flow conditions. The elegant cancellation above also exposes a common mistake: a drag coefficient definition does not by itself impose a quadratic drag law.

For an off-axis nearly occluding sphere with radial offset $e<d$, the local minimum gap is $d(\varphi )=d+e\cos \varphi$ to leading order. Each circumferential strip has pressure-driven conductance proportional to $d(\varphi {)}^{5/2}$, and the strips act in parallel. Thus its speed at fixed load is multiplied relative to the concentric value byConvexity and the zero average of cosine make this greater than one for nonzero $ϵ$. It increases strictly with $ϵ$: pair the positive and negative cosine contributions to the derivative, obtaining factors $c[(1+ϵc{)}^{3/2}-(1-ϵc{)}^{3/2}]>0$ for $c>0$. Consequently an off-axis sphere falls faster in this narrow-gap, no-through-flow regime.

Place the same positive electric charge $Q$ uniformly on the sphere of radius $R$, with no external field. Every concentric interior Gaussian surface encloses zero charge; spherical symmetry then makes the interior electric field zero. Outside, it is again $Q/(4\pi {\epsilon }_0{r}^2)$. At $R$, the inner and outer limits differ by $Q/(4\pi {\epsilon }_0{R}^2)=\sigma /{\epsilon }_0$, with surface charge density $\sigma =Q/(4\pi R^2)$. The ideal surface charge makes the field discontinuous, so the plot displays one-sided limits rather than assigning a single field value on the charged sheet. An isolated charged conducting sphere in electrostatic equilibrium realizes this exterior/interior pattern.

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]

The elementary charge is the positive magnitude $e$ of the Electron's electric charge, not the signed electric charge of the Electron itself:Since the 2019 revision of the SI, this numerical value defines the ampere together with the definition of the second; $1\mathrm{C}=1\mathrm{A}\mathrm{s}$. Thus today's numerical exactness is a choice of units, whereas measured electric charges still have experimental uncertainty. An oil drop experiment can test charge quantization and the apparatus calibration; it does not supply a new uncertain defining value of $e$ in today's SI. ^[ref].

For ordinary isolated chemical atoms, ions and charged objects whose net electric charge is counted through protons and Electrons, charge quantization gives $Q=ne$ with signed integer $n$. A neutral chemical atom that loses two Electrons becomes an ion of electric charge $+2e$. A toy object with $10^9$ excess Electrons has $Q=-0.1602176634\mathrm{nC}$. Conversely, the scale $1\mathrm{nC}/e\simeq 6.24\times 10^9$ explains why laboratory electric charge often looks continuous: billions of microscopic increments fit into a small macroscopic amount. Averaged charge density is a continuum description, not a claim that each particle has an arbitrarily divisible electric charge.

The ordinary-matter qualifier matters. In the Standard Model, quarks have electric charges $+2e/3$ or $-e/3$, and antiquarks have the opposite signs. Confinement means these are constituents of hadrons, not isolated fractionally charged particles collected by an oil drop experiment. The usual valence quark sums give $2(2e/3)-e/3=e$ for a proton and $2e/3-2(e/3)=0$ for a neutron. Thus elementary charge should not be defined as the smallest nonzero electric charge of every constituent in particle theory. ^[ref].

The beautiful experimental idea behind the oil drop experiment is to determine a microscopic electric charge by watching a visible carrier move. In his 1913 paper, Millikan timed the descent and electrically driven ascent of individual oil drops, repeatedly changing a drop's electric charge through capture of air ions. His apparatus used parallel plates separated by $16\mathrm{mm}$, a roughly $5300\mathrm{V}$ source, and an optical scale to measure travel distances. He reported multiple relationships among a given drop's inferred electric charges. He also found that uncorrected Stokes's law gave a size-dependent inferred elementary unit, and introduced a pressure-dependent slip correction. These were measured relationships with calibration and model uncertainties, not exact readings of today's defined $e$. ^[ref].

A modern instructional implementation sprays low-volatility oil of known density through a plate aperture, illuminates the drops, and follows one through a calibrated optical scale. Measure the plate spacing $d$, electric potential difference $V$, air temperature and pressure. Time a fixed distance $L$ with the electric field off to get a terminal falling speed $v_f=L/t_f$. Then choose the field polarity to drive the same drop upward and measure $v_r=L/t_r$. Repeat timings at a fixed electric charge, and compare before and after an ion-capture event. The same drop is especially useful because its radius and mass need not be inferred anew for each charge state. ^[ref].

Here is the leading continuum model, retaining buoyancy. Assume a drop shaped as a sphere of radius $a$, oil and air densities ${\rho }_o$ and ${\rho }_a$, air dynamic viscosity $\eta$, and gravitational acceleration $g$. Define the downward effective weight $W$ and positive linear drag coefficient $k$ byAt terminal velocity, the net force vanishes. Falling with no electric field, the upward drag force balances weight minus buoyancy. Rising with the electric force upward, the drag force reverses and joins the downward effective weight:Consequently the uncorrected estimates areThe observed direction of response, with known field polarity, supplies the sign of $q$. A suspended drop gives the alternative balance $|q|E=W$, but it still requires a determination of $W$: balancing a drop alone does not determine its electric charge.

For an explicitly invented continuum example, stipulate $a=1\mu \mathrm{m}$, ${\rho }_o=900\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_a=1.2\mathrm{kg}{\mathrm{m}}^{-3}$, $\eta =1.8\times 10^{-5}\mathrm{Pa}\mathrm{s}$, $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $q=-5e$ and a downward $E=10^5\mathrm{V}{\mathrm{m}}^{-1}$. The balances predict $W\simeq 3.69\times 10^{-14}\mathrm{N}$, $v_f\simeq 0.109\mathrm{mm}{\mathrm{s}}^{-1}$ and $v_r\simeq 0.127\mathrm{mm}{\mathrm{s}}^{-1}$. Substituting those speeds recovers $|q|=5e$ within rounding. This checks the algebra; it is not a precision prediction for real micrometre drops, which need a slip correction.

The inference has two distinct parts. Compare inferred electric charges or charge changes with a common unit, seeking $q_i\simeq n_{i}e$ within uncertainty, rather than demanding exact ratios of rounded measurements. Then establish the absolute scale through electric field, density, dynamic viscosity and distance/time calibration. If the observed steps are all even multiples, their common spacing alone cannot exclude an unobserved smaller unit. Air motion, evaporation and mistimed crossings can alter inferred electric charge without an ion-capture event. Small Reynolds number supports Stokes's law, but does not by itself justify its no-slip continuum coefficient: a significant ratio of air mean free path to drop radius requires a slip correction. Millikan's comparison of different drop sizes and pressures addressed that systematic limitation; the modern PASCO manual also includes the correction. Agreement supports charge quantization for the observed carriers within experimental resolution, rather than proving a universal statement about every possible particle.

Multiplying $e$ by the Avogadro constant gives the Faraday constant, $F=N_{A}e$, the electric charge per mole of singly charged carriers. This connects counting individual electric charges with bulk electrochemistry. ^[ref].

For a real positive-definite matrix $A$ in finite dimension, write $A=Q\Lambda Q^T$ by the spectral theorem for real symmetric matrices. In the coordinates in an orthonormal basis $w=Q^{T}x$, the positive level set $q(x)=k$ becomesThis is an ellipsoid, with semi-major and semi-minor axes in two dimensions forming an ellipse; each semi-axis has length $r_j=\sqrt{k/{\lambda }_j}$ and direction given by the corresponding eigenvector. In the opening example, the shorter semi-axis is $\sqrt{k}/2$ along $e_{+}$, while the longer is $\sqrt{k/2}$ along $e_{-}$. A stiffer direction, meaning a larger eigenvalue, reaches the same energy at a smaller displacement. The ratio of longest to shortest semi-axis is $\sqrt{{\lambda }_{\max }/{\lambda }_{\min }}$, the square root of the spectral condition number of a positive-definite matrix.^[ref]

Strict positivity matters because ${\lambda }_{\min }>0$ givesHence a fixed positive level set cannot extend arbitrarily far. The Cholesky decomposition $A=L{L}^T$ gives another view: $q(x)=\Vert L^{T}x{\Vert }^2$, so the ellipsoid is the inverse image of a sphere under an invertible linear map. A singular positive semidefinite matrix has a nonzero kernel; adding a kernel vector preserves its quadratic form, so any nonempty positive level set is unbounded. These statements concern finite dimension; the positive definite symmetric operator entry explains why strict positivity alone need not give a uniform lower bound on an infinite-dimensional Hilbert space.

For a finite connected graph with at least one vertex, embedded as a plane graph, let $n$ count vertices, $m$ count edges, and $f$ count faces of a plane graph, including the outer face. ThenThe hypotheses refer to an actual embedding; regions in a crossing-filled graph drawing do not supply its $f$.^[ref]

In the plane graph $K_4$ pictured above, three bounded faces of a plane graph and the outer face give $4-6+4=2$. A square cycle graph gives $4-4+2=2$, while the graph path $A-B-C$ gives $3-2+1=2$. Changing a crossing into a new vertex changes the graph: the square-and-diagonals drawing then has five vertices, eight edges, and five faces of a plane graph, giving $5-8+5=2$.

A proof outline deletes edges on graph cycles. Each deletion preserves the connected graph and merges two faces of a plane graph, leaving $n-m+f$ unchanged. Eventually a tree remains, with $m=n-1$ and $f=1$. Adding the point at infinity identifies the plane picture with a sphere; the value two agrees with its Euler characteristic and the cell-count formula for Euler characteristic.

For a finite cell decomposition of the sphere into polygonal faces, the numbers of vertices, edges and faces satisfy the displayed formula. Triangulating faces preserves $V-E+F$ because each added diagonal adds one edge and one face. This is the spherical form of Euler formula for a connected planar graph, where the plane's exterior face is included.

A face of a plane graph is a connected component of the Euclidean plane after removing the drawn vertices and edges. Count the outer face too. This is the plane counterpart of a face of a planar map on the sphere.^[ref]

In vacuum, magnetization is zero and magnetic flux density obeys $\mathbf{B}={\mu }_0\mathbf{H}$. The applied quantities ${\mathbf{H}}_{\mathrm{app}}$ and ${\mathbf{B}}_0={\mu }_0{\mathbf{H}}_{\mathrm{app}}$ mean the field of the external sources at the sample position with the sample absent. The internal magnetic field intensity includes the sample's demagnetizing field; its internal magnetic flux density additionally includes ${\mu }_0\mathbf{M}$. A field quoted in the tesla is a $B$ quantity, even when an instrument informally calls it “$H$”.

For a uniformly magnetized ellipsoid, along a principal axis with demagnetizing factor $N_d$ in the International System of Units,For a sphere, $N_d=1/3$. The axial value approaches zero in the long slender ellipsoid limit and the normal value approaches one in the thin flat ellipsoid limit. Thus a stipulated intrinsic volume magnetic susceptibility ${\chi }_v=3$ would yield apparent values $1.5$ in a sphere and approximately $3$ in a long axial sample. These are calculated geometry effects, not measured material values. For fixed $N_d$, implicit differentiation gives the same conversion for local differential magnetic susceptibility, $dM/d{H}_{\mathrm{app}}={\chi }_{\mathrm{diff}}/(1+N_d{\chi }_{\mathrm{diff}})$.