Gaussian law

The Gauss's law , even Gauss , in describing electrostatics and electrodynamics the electrical flow through a closed surface. Since the law can be formulated in the same way for the classical theory of gravity , it describes the flow of the gravitational acceleration field through a closed surface. It is an application of Gauss-Ostrogradski's theorem . It is also known by this name.

Like Ampère's law , the analogue for magnetism , Gauss’s law is one of the four Maxwell’s equations (the first) and is therefore fundamental for classical electrodynamics . In the case of gravitation, an equation results which, apart from a few constants, is equivalent to the first Maxwell equation.

Integral form

The concept of flow is defined for each vector field . Imagine a body with the charge surrounded by an oriented , closed surface . The surface can be shaped in any way, it can be a ball or any bulged balloon. The field lines emanating from the charge according to the field concept now flow through this surface, just as water flows through the surface, if there were a source or a sink within the surface . ${\displaystyle Q}$ ${\displaystyle A}$

The flow of charge outside of flows in on one side but out on the other. The total flow therefore only depends on the enclosed charge . The gist of the law is that the total flow is actually the same . ${\displaystyle A}$ ${\displaystyle Q}$ ${\displaystyle Q}$

The surface A is subdivided into small vectorial surface elements , the amount of which is the area of ​​the element and the direction of which is perpendicular to the plane ( normal vector ). The flux through such an element is the component of the vector field in the direction of the element multiplied by its area; this is exactly what is expressed by the scalar product . The total flow through A is then the surface integral of this product over the entire surface. ${\displaystyle \mathrm {d} {\vec {A}}}$

${\displaystyle \Phi =\oint _{A}{\vec {E}}\;\cdot \mathrm {d} {\vec {A}}={\frac {Q}{\varepsilon _{0}}}}$

Here, the flow of the vector field through the surface A of the volume V , the charge Q contains (Caution: Do not confuse with the electrical flow , which as the integral of the electric flux density obtained). The electric field constant ensures the correct units. ${\displaystyle \Phi }$ ${\displaystyle {\vec {E}}({\vec {x}})}$ ${\displaystyle {\mathit {\Psi }}}$ ${\displaystyle {\vec {D}}}$ ${\displaystyle \varepsilon _{0}}$

Simple applications

In the case of some problems, such as the calculation of electrostatic fields in the vicinity of simple geometric bodies such as plates, line charges or spheres, one can also calculate without an integral by cleverly choosing the surface elements. For this purpose, a simple, closed enveloping surface consisting of a few surface elements for which the flow can easily be determined is placed around the given charge distribution . ${\displaystyle \Phi }$ ${\displaystyle \Phi }$

Point charge

Radial electric field of a positive sphere.

An electrically charged sphere ( electrical charge ) is surrounded by field strength vectors that run radially outwards. No direction is preferred. As a closed envelope surface in the sense of Gaussian law, a concentric sphere with the radius is placed , which is perpendicularly pierced by the field strength vectors . ${\displaystyle Q}$ ${\displaystyle {\vec {E}}}$ ${\displaystyle R}$

The envelope surface with the surface is thought to be composed of many small surface elements. Each has a surface normal with a value that is parallel to the vector of the field strength passing through. Then it follows from Gauss's law ${\displaystyle 4\pi R^{2}}$ ${\displaystyle \mathrm {d} A}$

${\displaystyle {\frac {Q}{\varepsilon _{0}}}=4\pi R^{2}\cdot E}$

with the result

${\displaystyle E={\frac {Q}{4\pi \varepsilon _{0}}}\cdot {\frac {1}{R^{2}}}}$.

If the distance is doubled, the field strength drops to a quarter.

Line charge

Cylindrical envelope around a charged wire

An electrically charged, infinitely long wire carries the charge per unit length . This corresponds to the charge density . ${\displaystyle L}$ ${\displaystyle Q}$ ${\displaystyle \lambda =Q/L}$

For reasons of symmetry, the vectors form right angles with the wire and perpendicularly pierce the cylinder wall marked in yellow. If you were to draw in these vectors, you would get the picture of a round brush . ${\displaystyle {\vec {E}}}$

A circular cylinder of the length with the wire as an axis is placed around a section of this wire as a closed envelope surface in the sense of Gaussian law . The envelope surface consists of three sub-areas: ${\displaystyle a}$

• Left and right cover with the areas ; Every surface normal is parallel to the wire and therefore forms right angles with the radial vectors of the field strength. This in turn ensures that the corresponding scalar products result in the value zero. The lids therefore make no contribution to the flux integral. ${\displaystyle \pi R^{2}}$
• Cylinder jacket with the surface area that is thought to be composed of many small surface elements. The surface normals have the amount and are parallel to the field strength vectors passing through. Therefore every scalar product has the value . The surface integral over the cylinder jacket results ${\displaystyle 2\pi Ra}$ ${\displaystyle \mathrm {d} A}$ ${\displaystyle \mathrm {d} AE}$

${\displaystyle {\frac {Q}{\varepsilon _{0}}}={\frac {\lambda \cdot a}{\varepsilon _{0}}}=2\pi RaE}$,

with the result

${\displaystyle E={\frac {\lambda }{2\pi \varepsilon _{0}}}\cdot {\frac {1}{R}}}$.

At twice the distance, the field strength drops by half. The wire doesn't actually have to be infinitely long. It is sufficient if the distance R at which the field strength is measured is much smaller than the length of the wire. Otherwise edge effects will occur and the proportion of base and cover must be taken into account.

Surface loading

Field lines of a positively charged, infinitely extended plane

A positively charged, infinitely large plane carries the charge per unit area . This corresponds to the charge density . ${\displaystyle Q}$ ${\displaystyle \sigma =Q/A}$

For reasons of symmetry, the vectors of the electric field strength are perpendicular to the plane. ${\displaystyle {\vec {E}}}$

As a closed envelope surface in the sense of Gaussian law, a cuboid of the height is placed around a partial surface , which is roughly halved by the charged plane. The vectors pierce both lids of the cuboid vertically. Its surface consists of three elements: ${\displaystyle 2H}$ ${\displaystyle {\vec {E}}}$

• Upper and lower cover with the areas ; every surface normal is perpendicular to the charged plane and is therefore parallel to . The flow goes out through each of the two covers . ${\displaystyle A}$ ${\displaystyle {\vec {E}}}$ ${\displaystyle \Phi _{\text{oben}}=\Phi _{\text{unten}}=AE}$
• The edge of the cuboid does not contribute to the flow because it forms right angles with the respective surface normals. This would not have changed if a prism with a different base or a cylinder had been chosen instead of the cuboid. The height is also irrelevant. ${\displaystyle \Phi }$ ${\displaystyle {\vec {E}}}$ ${\displaystyle H}$

So the total flow is

${\displaystyle \Phi _{\mathrm {gesamt} }=2AE}$.

Because of the charge contained in the cuboid, the following applies

${\displaystyle \Phi ={1 \over \varepsilon _{0}}Q_{\mathrm {eingeschlossen} }={1 \over \varepsilon _{0}}\sigma \cdot A}$.

A comparison of the right sides provides the result

${\displaystyle E={\frac {\sigma }{2\varepsilon _{0}}}}$.

The field strength is therefore independent of the distance to the (infinitely extended) charged plane. If the plane is limited, this result is only valid for sufficiently small distances. ${\displaystyle E}$ ${\displaystyle H}$

Two oppositely charged surfaces

Blue: negative plate E vectors.
Red: E-vectors of the positive plate

A positively charged, very large plane carries the charge per unit area . This corresponds to the charge density . At a distance there is a parallel plane of charge density . This arrangement is also known as a plate capacitor . To polarities differ, it was agreed that the field lines of the positive plate away to show (shown in red), and to the negative plate toward show (blue lines). ${\displaystyle Q}$ ${\displaystyle \sigma =Q/A}$ ${\displaystyle d}$ ${\displaystyle -\sigma }$

The arrows between the two plates are oriented in the same way, where the individual field strengths add up

${\displaystyle E={\frac {\sigma }{\varepsilon _{0}}}}$.

In the outer space, the arrows point in the opposite direction, there the field strengths compensate each other and it applies . To simplify matters, one says that the electric field is only present in the interior of a capacitor. ${\displaystyle E_{\text{gesamt}}=0}$

Differential form

Instead of the total macroscopic charge , the charge can also be expressed by the charge density at each point, where again the volume integral is over the entire volume enclosed by . One then obtains using the integral form and Gauss's theorem ${\displaystyle Q}$ ${\displaystyle \rho }$ ${\displaystyle Q}$ ${\displaystyle \rho }$ ${\displaystyle A}$ ${\displaystyle V}$

${\displaystyle \int \rho \mathrm {d} V=\varepsilon _{0}\oint {\vec {E}}\cdot \mathrm {d} {\vec {A}}=\varepsilon _{0}\int {\vec {\nabla }}\cdot {\vec {E}}\mathrm {d} V}$

where is the Nabla operator . Since the volume is arbitrary, the differential form follows the law ${\displaystyle \nabla }$

${\displaystyle \nabla \cdot {\vec {E}}=-{\frac {\rho }{\varepsilon _{0}}}}$

Time independence

The Gaussian law is often derived in the literature for the area of electrostatics . However, it also applies without restrictions to electrodynamics , although time-dependent processes must also be considered there.

In order to illustrate the importance of the time-independent Gaussian law and its strong connection with the conservation of charges, a thought experiment can be used: It is possible to generate charges and a certain positive or negative charge arises at a point in space at any point in time. What is required is the field of the electrical flux density , which penetrates a spherical surface with a radius that is concentric around the charge . Since Gauss's theorem does not contain any time dependency, this field would have to occur simultaneously with the generation of the charge through the envelope surface, even if it were, for example, = 1 light year away. However, this idea contradicts the theory of relativity , according to which information (the information about the existence of the charge) and energy (the field energy) can spread at most at the speed of light . Accordingly, in the example mentioned, the field would only arrive at the imaginary envelope surface one year later. ${\displaystyle {\vec {D}}}$ ${\displaystyle r}$ ${\displaystyle r}$

Since, according to all physical knowledge, both Einstein's theory of relativity and Gaussian law apply, it follows that a single charge can neither be generated nor destroyed. Only the "pairwise" simultaneous generation of positive and negative charges at the same place is possible (see pair formation (physics) ).

Application to gravity

Within the framework of Newton's theory of gravity, the principles presented above can also be applied to the gravitational field. The gravitational acceleration of a mass M is determined by the law of gravity to

${\displaystyle {\vec {g}}=-{\frac {GM}{r^{2}}}{\frac {\vec {r}}{r}}}$.

The flow through the surface of any volume is then

${\displaystyle \int \,\!\!\!\!\!\int _{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\bigcirc \,\,{\vec {g}}\;\cdot \mathrm {d} {\vec {A}}=-\int \,\!\!\!\!\!\int _{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\bigcirc \,\,{\frac {GM}{r^{2}}}{\frac {\vec {r}}{r}}\cdot {\vec {n}}\,\mathrm {d} A=-4\pi GM}$,

where is the normal vector . ${\displaystyle {\vec {n}}}$

Thus the gravitational acceleration field of a mass distribution can be determined with

${\displaystyle \int \,\!\!\!\!\!\int _{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\bigcirc \,\,{\vec {g}}\;\cdot \mathrm {d} {\vec {A}}=-4\pi GM}$

In differential form and for general mass distributions results

${\displaystyle \nabla \cdot {\vec {g}}=-4\pi G\rho ({\vec {r}})}$

which is the gravitational equivalent of the first Maxwell equation .

Individual evidence

1. ↑ Torsten Fließbach: Electrodynamics . 5th edition. Spectrum, Heidelberg 2008, p. 50 .
2. ↑ Torsten Fließbach: Electrodynamics . 5th edition. Spectrum, Heidelberg 2008, p. 52 .
3. ↑ Wolfgang Demtröder: Experimentalphysik 2: Electricity and optics . 3. Edition. Springer, Berlin Heidelberg 2004, p. 12 .
4. ↑ Wolfgang Demtröder: Experimentalphysik 2: Electricity and optics . 3. Edition. Springer, Berlin Heidelberg 2004, p. 20 .