Gauss's law

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Gauss's law in its integral form is particularly useful when, by symmetry reasons, a closed surface (GS) can be found along which the electric field is uniform. The electric flux is then a simple product of the surface area and the strength of the electric field, and is proportional to the total charge enclosed by the surface. Here, the electric field outside (r > R) and inside (r < R) of a charged sphere is being calculated (see Wikiversity).

In physics (specifically electromagnetism), Gauss's law, also known as Gauss's flux theorem, (or sometimes simply called Gauss's theorem) is one of Maxwell's equations. It relates the distribution of electric charge to the resulting electric field.

Definition

In its integral form, it states that the flux of the electric field out of an arbitrary closed surface is proportional to the electric charge enclosed by the surface, irrespective of how that charge is distributed. Even though the law alone is insufficient to determine the electric field across a surface enclosing any charge distribution, this may be possible in cases where symmetry mandates uniformity of the field. Where no such symmetry exists, Gauss's law can be used in its differential form, which states that the divergence of the electric field is proportional to the local density of charge.

The law was first<ref>Duhem, Pierre (1891). "4". Leçons sur l'électricité et le magnétisme [Lessons on electricity and magnetism] (in français). Vol. 1. Paris Gauthier-Villars. pp. 22–23. OCLC 1048238688. OL 23310906M. Shows that Lagrange has priority over Gauss. Others after Gauss discovered "Gauss's Law", too.</ref> formulated by Joseph-Louis Lagrange in 1773,<ref>Lagrange, Joseph-Louis (1869) [1776]. Serret, Joseph-Alfred; Darboux, Jean-Gaston (eds.). "Sur l'attraction des sphéroïdes elliptiques" [On the attraction of elliptical spheroids]. Œuvres de Lagrange: Mémoires extraits des recueils de l'Académie royale des sciences et belles-lettres de Berlin (in français). Gauthier-Villars: 619.</ref> followed by Carl Friedrich Gauss in 1835,<ref>Gauss, Carl Friedrich (1877). "Theoria attractionis corporum sphaeroidicorum ellipticorum homogeneorum methodo nova tractata" [The theory of the attraction of homogeneous spheroidal elliptic bodies treated by a new method]. In Schering, Ernst Christian Julius; Brendel, Martin (eds.). Carl Friedrich Gauss Werke [Works of Carl Friedrich Gauss] (in Latina and Deutsch). Vol. 5 (2nd ed.). Gedruckt in der Dieterichschen Universitätsdruckerei (W.F. Kaestner). pp. 2–22. Gauss mentions Newton's Principia proposition XCI regarding finding the force exerted by a sphere on a point anywhere along an axis passing through the sphere.</ref> both in the context of the attraction of ellipsoids. It is one of Maxwell's equations, which forms the basis of classical electrodynamics.<ref group="note">The other three of Maxwell's equations are: Gauss's law for magnetism, Faraday's law of induction, and Ampère's law with Maxwell's correction</ref> Gauss's law can be used to derive Coulomb's law,<ref>Halliday, David; Resnick, Robert (1970). Fundamentals of Physics. John Wiley & Sons. pp. 452–453.</ref> and vice versa.

Qualitative description

In words, Gauss's law states:

The net electric flux through any hypothetical closed surface is equal to 1/ε0 times the net electric charge enclosed within that closed surface. The closed surface is also referred to as Gaussian surface.<ref>Serway, Raymond A. (1996). Physics for Scientists and Engineers with Modern Physics (4th ed.). p. 687.</ref>

Gauss's law has a close mathematical similarity with a number of laws in other areas of physics, such as Gauss's law for magnetism and Gauss's law for gravity. In fact, any inverse-square law can be formulated in a way similar to Gauss's law: for example, Gauss's law itself is essentially equivalent to the Coulomb's law, and Gauss's law for gravity is essentially equivalent to the Newton's law of gravity, both of which are inverse-square laws.

The law can be expressed mathematically using vector calculus in integral form and differential form; both are equivalent since they are related by the divergence theorem, also called Gauss's theorem. Each of these forms in turn can also be expressed two ways: In terms of a relation between the electric field E and the total electric charge, or in terms of the electric displacement field D and the free electric charge.<ref name="GrantPhillips">Grant, I. S.; Phillips, W. R. (2008). Electromagnetism. Manchester Physics (2nd ed.). John Wiley & Sons. ISBN 978-0-471-92712-9.</ref>

Equation involving the E field

Gauss's law can be stated using either the electric field E or the electric displacement field D. This section shows some of the forms with E; the form with D is below, as are other forms with E.

Integral form

Electric flux through an arbitrary surface is proportional to the total charge enclosed by the surface.
No charge is enclosed by the sphere. Electric flux through its surface is zero.

Gauss's law may be expressed as:<ref name="GrantPhillips"/>

<math display="block">\Phi_E = \frac{Q}{\varepsilon_0}</math>

where ΦE is the electric flux through a closed surface S enclosing any volume V, Q is the total charge enclosed within V, and ε0 is the electric constant. The electric flux ΦE is defined as a surface integral of the electric field:

<math>\Phi_E = </math> \oiint<math>\scriptstyle _S</math> <math>\mathbf{E} \cdot \mathrm{d}\mathbf{A}</math>

where E is the electric field, dA is a vector representing an infinitesimal element of area of the surface,<ref group="note">More specifically, the infinitesimal area is thought of as planar and with area dN. The vector dR is normal to this area element and has magnitude dA.<ref>Matthews, Paul (1998). Vector Calculus. Springer. ISBN 3-540-76180-2.</ref></ref> and · represents the dot product of two vectors.

In a curved spacetime, the flux of an electromagnetic field through a closed surface is expressed as

<math>\Phi_E = c </math> \oiint<math> \scriptstyle _S</math> <math> F^{\kappa 0} \sqrt {-g} \, \mathrm{d} S_\kappa </math>

where <math>c</math> is the speed of light; <math>F^{\kappa 0}</math> denotes the time components of the electromagnetic tensor; <math>g</math> is the determinant of metric tensor; <math> \mathrm{d} S_\kappa = \mathrm{d} S^{ij} = \mathrm{d}x^i \mathrm{d}x^j </math> is an orthonormal element of the two-dimensional surface surrounding the charge <math>Q</math>; indices <math> i,j,\kappa = 1,2,3</math> and do not match each other.<ref>Fedosin, Sergey G. (2019). "On the Covariant Representation of Integral Equations of the Electromagnetic Field". Progress in Electromagnetics Research C. 96: 109–122. arXiv:1911.11138. Bibcode:2019arXiv191111138F. doi:10.2528/PIERC19062902. S2CID 208095922.</ref>

Since the flux is defined as an integral of the electric field, this expression of Gauss's law is called the integral form.

A tiny Gauss's box whose sides are perpendicular to a conductor's surface is used to find the local surface charge once the electric potential and the electric field are calculated by solving Laplace's equation. The electric field is perpendicular, locally, to the equipotential surface of the conductor, and zero inside; its flux πa2·E, by Gauss's law equals πa2·σ/ε0. Thus, σ = ε0E.

In problems involving conductors set at known potentials, the potential away from them is obtained by solving Laplace's equation, either analytically or numerically. The electric field is then calculated as the potential's negative gradient. Gauss's law makes it possible to find the distribution of electric charge: The charge in any given region of the conductor can be deduced by integrating the electric field to find the flux through a small box whose sides are perpendicular to the conductor's surface and by noting that the electric field is perpendicular to the surface, and zero inside the conductor.

The reverse problem, when the electric charge distribution is known and the electric field must be computed, is much more difficult. The total flux through a given surface gives little information about the electric field, and can go in and out of the surface in arbitrarily complicated patterns.

An exception is if there is some symmetry in the problem, which mandates that the electric field passes through the surface in a uniform way. Then, if the total flux is known, the field itself can be deduced at every point. Common examples of symmetries which lend themselves to Gauss's law include: cylindrical symmetry, planar symmetry, and spherical symmetry. See the article Gaussian surface for examples where these symmetries are exploited to compute electric fields.

Differential form

By the divergence theorem, Gauss's law can alternatively be written in the differential form: <math display="block">\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0 \varepsilon_r}</math>

where ∇ · E is the divergence of the electric field, ε0 is the vacuum permittivity, <math>\varepsilon_r</math> is the relative permittivity, and ρ is the volume charge density (charge per unit volume).

Equivalence of integral and differential forms

The integral and differential forms are mathematically equivalent, by the divergence theorem. Here is the argument more specifically.

Outline of proof

The integral form of Gauss' law is:

\oiint<math>{\scriptstyle _S}</math> <math>\mathbf{E} \cdot \mathrm{d}\mathbf{A}</math><math> = \frac{Q}{\varepsilon_0}</math>

for any closed surface S containing charge Q. By the divergence theorem, this equation is equivalent to:

<math display="block">\iiint_V \nabla \cdot \mathbf{E} \, \mathrm{d}V = \frac{Q}{\varepsilon_0}</math>

for any volume V containing charge Q. By the relation between charge and charge density, this equation is equivalent to: <math display="block">\iiint_V \nabla \cdot \mathbf{E} \, \mathrm{d}V = \iiint_V \frac{\rho}{\varepsilon_0} \, \mathrm{d}V</math> for any volume V. In order for this equation to be simultaneously true for every possible volume V, it is necessary (and sufficient) for the integrands to be equal everywhere. Therefore, this equation is equivalent to:

<math display="block">\nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0}.</math> Thus the integral and differential forms are equivalent.

Equation involving the D field

Free, bound, and total charge

The electric charge that arises in the simplest textbook situations would be classified as "free charge"—for example, the charge which is transferred in static electricity, or the charge on a capacitor plate. In contrast, "bound charge" arises only in the context of dielectric (polarizable) materials. (All materials are polarizable to some extent.) When such materials are placed in an external electric field, the electrons remain bound to their respective atoms, but shift a microscopic distance in response to the field, so that they're more on one side of the atom than the other. All these microscopic displacements add up to give a macroscopic net charge distribution, and this constitutes the "bound charge".

Although microscopically all charge is fundamentally the same, there are often practical reasons for wanting to treat bound charge differently from free charge. The result is that the more fundamental Gauss's law, in terms of E (above), is sometimes put into the equivalent form below, which is in terms of D and the free charge only.

Integral form

This formulation of Gauss's law states the total charge form:

<math display="block">\Phi_D = Q_\mathrm{free}</math>

where ΦD is the D-field flux through a surface S which encloses a volume V, and Qfree is the free charge contained in V. The flux ΦD is defined analogously to the flux ΦE of the electric field E through S:

<math>\Phi_D = </math> \oiint<math>{\scriptstyle _S}</math> <math>\mathbf{D} \cdot \mathrm{d}\mathbf{A} </math>

Differential form

The differential form of Gauss's law, involving free charge only, states: <math display="block">\nabla \cdot \mathbf{D} = \rho_\mathrm{free}</math>

where ∇ · D is the divergence of the electric displacement field, and ρfree is the free electric charge density.

Equivalence of total and free charge statements

Proof that the formulations of Gauss's law in terms of free charge are equivalent to the formulations involving total charge.

In this proof, we will show that the equation <math display="block">\nabla\cdot \mathbf{E} = \dfrac{\rho}{\varepsilon_0}</math> is equivalent to the equation <math display="block">\nabla\cdot\mathbf{D} = \rho_\mathrm{free}</math> Note that we are only dealing with the differential forms, not the integral forms, but that is sufficient since the differential and integral forms are equivalent in each case, by the divergence theorem.

We introduce the polarization density P, which has the following relation to E and D: <math display="block">\mathbf{D}=\varepsilon_0 \mathbf{E} + \mathbf{P}</math> and the following relation to the bound charge: <math display="block">\rho_\mathrm{bound} = -\nabla\cdot \mathbf{P}</math> Now, consider the three equations: <math display="block">\begin{align} \rho_\mathrm{bound} &= \nabla\cdot (-\mathbf{P}) \\ \rho_\mathrm{free} &= \nabla\cdot \mathbf{D} \\ \rho &= \nabla \cdot(\varepsilon_0\mathbf{E}) \end{align}</math> The key insight is that the sum of the first two equations is the third equation. This completes the proof: The first equation is true by definition, and therefore the second equation is true if and only if the third equation is true. So the second and third equations are equivalent, which is what we wanted to prove.

Equation for linear materials

In homogeneous, isotropic, nondispersive, linear materials, there is a simple relationship between E and D:

<math display="block">\mathbf{D} = \varepsilon \mathbf{E} </math>

where ε is the permittivity of the material. For the case of vacuum (aka free space), ε = ε0. Under these circumstances, Gauss's law modifies to

<math display="block">\Phi_E = \frac{Q_\mathrm{free}}{\varepsilon}</math>

for the integral form, and

<math display="block">\nabla \cdot \mathbf{E} = \frac{\rho_\mathrm{free}}{\varepsilon}</math>

for the differential form.

Relation to Coulomb's law

Deriving Gauss's law from Coulomb's law

Strictly speaking, Gauss's law cannot be derived from Coulomb's law alone, since Coulomb's law gives the electric field due to an individual, electrostatic point charge only. However, Gauss's law can be proven from Coulomb's law if it is assumed, in addition, that the electric field obeys the superposition principle. The superposition principle states that the resulting field is the vector sum of fields generated by each particle (or the integral, if the charges are distributed smoothly in space).

Outline of proof

Coulomb's law states that the electric field due to a stationary point charge is: <math display="block">\mathbf{E}(\mathbf{r}) = \frac{q}{4\pi \varepsilon_0} \frac{\mathbf{e}_r}{r^2}</math> where

Using the expression from Coulomb's law, we get the total field at r by using an integral to sum the field at r due to the infinitesimal charge at each other point s in space, to give <math display="block">\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \int \frac{\rho(\mathbf{s})(\mathbf{r}-\mathbf{s})}{

{|\mathbf{r}|^3}\right) = 4\pi \delta(\mathbf{r})</math>

where δ(r) is the Dirac delta function, the result is <math display="block">\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{1}{\varepsilon_0} \int \rho(\mathbf{s})\, \delta(\mathbf{r}-\mathbf{s})\, \mathrm{d}^3 \mathbf{s}</math>

Using the "sifting property" of the Dirac delta function, we arrive at <math display="block">\nabla\cdot\mathbf{E}(\mathbf{r}) = \frac{\rho(\mathbf{r})}{\varepsilon_0},</math> which is the differential form of Gauss' law, as desired. }}

Since Coulomb's law only applies to stationary charges, there is no reason to expect Gauss's law to hold for moving charges based on this derivation alone. In fact, Gauss's law does hold for moving charges, and, in this respect, Gauss's law is more general than Coulomb's law.

Proof (without Dirac Delta)

Let <math>\Omega \subseteq R^3 </math> be a bounded open set, and <math display="block">\mathbf E_0(\mathbf r) = \frac {1}{4 \pi \varepsilon_0} \int_{\Omega} \rho(\mathbf r')\frac {\mathbf r - \mathbf r'} {\left \

Deriving Coulomb's law from Gauss's law

Strictly speaking, Coulomb's law cannot be derived from Gauss's law alone, since Gauss's law does not give any information regarding the curl of E (see Helmholtz decomposition and Faraday's law). However, Coulomb's law can be proven from Gauss's law if it is assumed, in addition, that the electric field from a point charge is spherically symmetric (this assumption, like Coulomb's law itself, is exactly true if the charge is stationary, and approximately true if the charge is in motion).

Outline of proof

Taking S in the integral form of Gauss' law to be a spherical surface of radius r, centered at the point charge Q, we have

<math display="block">\oint_S\mathbf{E}\cdot d\mathbf{A} = \frac{Q}{\varepsilon_0} </math>

By the assumption of spherical symmetry, the integrand is a constant which can be taken out of the integral. The result is <math display="block">4\pi r^2\hat{\mathbf{r

\cdot\mathbf{E}(\mathbf{r}) = \frac{Q}{\varepsilon_0}</math>

where is a unit vector pointing radially away from the charge. Again by spherical symmetry, E points in the radial direction, and so we get <math display="block">\mathbf{E}(\mathbf{r}) = \frac{Q}{4\pi \varepsilon_0} \frac{\hat{\mathbf{r}}}{r^2}</math> which is essentially equivalent to Coulomb's law. Thus the inverse-square law dependence of the electric field in Coulomb's law follows from Gauss' law. }}

See also

Notes

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Citations

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References

  • Gauss, Carl Friedrich (1867). Werke Band 5. Digital version
  • Jackson, John David (1998). Classical Electrodynamics (3rd ed.). New York: Wiley. ISBN 0-471-30932-X. David J. Griffiths (6th ed.)

External links