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Gauss's Law

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'''Gauss's Law''' states that the electric flux through a closed surface is proporational to the [[electric charge|electrical charge ]] inside. This holds true regardless of the volume or shape of the closed surface. This is one of the most fundamental principles of electrodynamics, and is one of [[Maxwell's Equations]]. The law is named after [[Carl Friedrich Gauss]]. It is useful for calculating the [[electric field]] around distributions of charges with lots of symmetry.
In integral form, Gauss's Law is this:
<math>\Phi = \oint_S \vec{E} \cdot \mathrm{d}\vec{A}
= {1 \over \varepsilon_0} \int_V \rho\ \mathrm{d}V = \frac{Q_A}{\varepsilon_0}</math>
: where <math>\Phi </math> is the electric flux through the surface ''S'', <math>\vec{E}</math> is the [[electric field]], <math>\mathrm{d}\vec{A}</math> is a differential area 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>\varepsilon_0</math> is a constant for the permittivity of free space and the integral <math>\oint_S</math> is over the surface ''S'' enclosing volume ''V''. It can also be written in differential form as: <math> \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\varepsilon_0}</math> where <math>\rho</math> is the charge density. ==Proof of equivalency==It is easy to show that the differential and integral forms of Gauss's law are equivalent. This can be done by integrating the differential form over a volume: <math>\int_V \vec{\nabla} \cdot \vec{E} \mathrm{d}V = \frac{1}{\varepsilon_0} \int_V \rho \mathrm{d}V</math> As integral of charge density over volume is the charge contained within that volume, the integral can be replaced with Q<sub>A</sub>. Using the [[Divergence Theorem|Divergence theorem]], the left hand side can be changed from a volume integral of a [[divergence]] into a surface integral with no divergence across the surface of the volume, S. This produces the integral form: <math>\oint_S \vec{E} \cdot \mathrm{d}\vec{A}
= {1 \over \varepsilon_0} \int_V \rho\ \mathrm{d}V = \frac{Q_A}{\varepsilon_0}</math>
where ==Example of use==Gauss's law is useful when there is a lot of symmetry in a problem. An example of that is a [[sphere]] of uniform charge density, and radius R. We choose a surface over which we can easily evaluate: <math>\Phioint_S \vec{E} \cdot \mathrm{d}\vec{A}</math>  This surface is the electric flux through the known as a "Gaussian surface ''S'', ". We choose it so that <math>\veccos{E\theta}</math> is term in the [[dot product]] evaluates simply to either 1 or 0. In our example of a sphere, we choose a spherical Gaussian surface of radius r. This means that all the small vector elements dA point radially outwards and as the [[electric field]]due to the sphere will be radial, <math>\vec{E}</math> and <math>\mathrm{d}\vec{A}</math> is a differential area on will always be parallel. This means the closed surface ''S'' with an outward facing [[surface normal]] defining its direction, <math>Q_\cos{\theta}</math> terms will be: <math>\oint_S \vec{E} \cdot \mathrm{d}\vec{A}= \vec{E} \oint_S \mathrm{d}\vec{A} = \vec{E} 4 \pi r^2</math>  This is the left handside. The right hand side depends on the charge enclosed by the surfaceand therefore r. This means the electric field takes a different form inside and outside the sphere. Outside the sphere, the charge enclosed is <math>\frac{4}{3} \pi \rhoR^3</math> while inside it is <math>\frac{4}{3} \pi \rho r^3</math>. This means the charge density at a point in electric field inside and around the sphere is: <math>V\vec{E} =\begin{cases} \frac{1}{3} \rho r \hat{r} & 0 \leq r \leq R \\  \frac{\rho R^3}{3r^2} \hat{r} & r \geq R \end{cases} </math>,  where <math>\varepsilon_0hat{r}</math> is a constant for unit vector pointing radially outwards. Note that outside the sphere, the field drops off according to an [[permittivityinverse square law]] of free space and . This means that it is the integral <math>\oint_S</math> same as if the sphere were a point particle. This is over a much simpler calculation than integrating electric field component of each point particle in the surface ''S'' enclosing volume ''V''sphere==See also==*[[categoryAmpere's law]] [[Category:physicsElectromagnetism]][[Category:Electrical engineeringEngineering]]
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