Gauss's law

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In physics and mathematical analysis, Gauss's law gives the relation between the electric flux flowing out a closed surface and the electric charge enclosed in the surface.

1 Integral Form
2 {1 \over \epsilon_o} \int_V \rho\ dV

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Integral Form

In its integral form, the law states:

<math>\Phi = \oint_S \mathbf{E} \cdot d\mathbf{A}

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{1 \over \epsilon_o} \int_V \rho\ dV

\frac{Q_A}{\epsilon_o}</math>

where <math>\Phi</math> is the electric flux, <math>\mathbf{E}</math> is the electric field, <math>d\mathbf{A}</math> is the area of a differential square on the closed surface S with an outward facing surface normal defining its direction, <math>Q_\mathrm{A}</math> is the charge enclosed by the surface, <math>\rho</math> is the charge density at a point in <math>V</math>, <math>\epsilon_o</math> is the permittivity of free space and <math>\oint_S</math> is the integral over the surface S enclosing volume V.

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Differential Form

In differential form, the equation becomes:

<math>\nabla \cdot \mathbf{D} = \rho</math>

where <math>\nabla</math> is the del operator, representing divergence, D is the electric displacement field (in units of C/m²), and ρ is the free electric charge density (in units of C/m³), not including dipole charges bound in a material. The differential form derives in part from Gauss's divergence theorem.

And for linear materials, the equation becomes:

<math>\nabla \cdot \epsilon \mathbf{E} = \rho</math>

where <math>\epsilon</math> is the electrical permittivity.

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Coulomb's Law

In the special case of a spherical surface with a central charge, the electric field is perpendicular to the surface, with the same magnitude at all points of it, giving the simpler expression:

<math>E=\frac{Q}{4\pi\epsilon_0r^{2}}</math>

where E is the electric field strength at radius r, Q is the enclosed charge, and ε₀ is the permittivity of free space. Thus the familiar inverse-square law dependence of the electric field in Coulomb's law follows from Gauss's law.

Gauss's law can be used to demonstrate that there is no electric field inside a Faraday cage with no electric charges. Gauss's law is the electrostatic equivalent of Ampère's law, which deals with magnetism. Both equations were later integrated into Maxwell's equations.

It was formulated by Carl Friedrich Gauss in 1835, but was not published until 1867. Because of the mathematical similarity, Gauss's law has application for other physical quantities governed by an inverse-square law such as gravitation or the intensity of radiation. See also divergence theorem.

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Gravitational Analogue

Since both gravity and electromagnetism propagate relative to the squared distance between two objects, we can relate the two using Gauss's Law by examining their respective fields, G(r) and E(r).

<math>E(r) = \frac{1}{4 \pi \epsilon_{0}} \frac{Q}{r^2}</math>

In the same way that we evaluate the line integral for electromagnetism to get the result <math>\frac{Q}{\epsilon_{0}}</math>, we can choose a proper Gaussian Surface to quickly get an answer for gravitational flux. For a point mass, the most logical choice for our Gaussian surface is a sphere of radius <math>r</math> centered at the point.

We start with Gauss's Law:

<math>\Phi = \oint_S \mathbf{G}(r) \cdot d\mathbf{A}</math>

Each point mass will have a force acting radially, so we write our field equation as:

<math>\mathbf{G}(r) = G(r) \hat{r}</math>

For the infinitesimal, our area is just the area of the infinitesimal solid angle, which is defined as:

<math>d\mathbf{A} = r^{2} d\Omega \hat{r}</math>

Because of our wisely chosen Gaussian Surface, the area vector always has a direction pointing radially outward from the point charge or mass at the center of our sphere. Thus, it is just <math>\hat{r}</math>, and we can now write:

<math>\Phi = \oint_S G(r) \hat{r} \cdot \hat{r} r^{2} d\Omega</math>

We see the inner product of the two radial vectors is unity and that both the magnitude of our field, G(r), and the square of the distance between the surface and the point, <math>r^{2}</math>, remain constant. This gives us:

<math>\Phi = G(r) r^{2} \oint_S d\Omega</math>

The line integral is just the surface area of the unit sphere, <math>4 \pi</math>. If we combine this with our gravitational field equation from above, we have an expression for the gravitational flux of a point mass.

It is interesting to note that the gravitational flux, like its electromagnetic counterpart, is invariant of the radius of the sphere.

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