Gaussian surface

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A cylindrical Gaussian surface is commonly used to calculate the electric charge of an infinitely long, straight, 'ideal' wire.

A Gaussian surface is a closed two-dimensional surface through which a flux or electric field is to be calculated. The surface is used in conjunction with Gauss's law (a consequence of the divergence theorem), allowing one to calculate the total enclosed electric charge by performing a surface integral.

Gaussian surfaces are usually carefully chosen to exploit symmetries of a situation to simplify the calculation of the surface integral. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constant can be pulled out of the integration sign.

1 Common Gaussian surfaces
2 References
3 External links

[edit] Common Gaussian surfaces

When performing the closed surface integral, one must choose a Gaussian surface that encompasses the required charge within the given situation. It is not necessary to choose a Gaussian surface that utilises the symmetry of a situation (as in the examples below) but, obviously the calculations are much less laborious if an appropriate surface is chosen. Most calculations using Gaussian surfaces begin by implementing Gauss' law:

$\Phi = \oint_S \mathbf{E} \cdot d\mathbf{A} = \frac{Q_A}{\varepsilon_o}.$

[edit] Spherical surface

A spherical Gaussian surface is used when finding the electric field or the flux produced by any of the following:

a point charge
a uniformly distributed spherical shell of charge
any other charge distribution with spherical symmetry

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.

As an example, consider a charged spherical shell of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting Q_A = 0 in Gauss's law, where Q_A is the charge enclosed by the Gaussian surface).

With the same example, using a larger Gaussian surface outside the shell where r > R, Gauss's law will produce a non-zero electric field. This is determined as follows:

$E \cdot\ 4\pi\ r^{2} = \frac{Q_A}{\varepsilon_0} \Rightarrow \; E=\frac{Q_A}{4\pi\varepsilon_0r^{2}}.$

This non-trivial result shows that a uniform spherical distribution of charge acts as a point charge when observed from the outside of the charge distribution; this is in fact a verification of Coulomb's law.

[edit] Cylindrical surface

A cylindrical Gaussian surface is commonly used to calculate the electric field produced by any of the following:

a long, straight wire with uniform charge distribution
any long, straight cylinder or cylindrical shell with uniform charge distribution

Note: Technically, these distributions must be infinitely long and straight over their entire length. Sufficiently long and straight charge distributions behave very similarly, so infinite length is often assumed when dealing with finite wires.

We begin by choosing the point P at which we wish to calculate the electric field. A closed, infinite cylinder whose axis of symmetry coincides with the wire, and whose surface contains point P is imagined. (Note that the surface never actually closes since it is infinite, but we can assume that it closes at an infinite distance away.) Since all of the charge of the wire is contained within the surface, we know that the total flux passing through the surface must be the enclosed charge divided by ε₀. In this case, the surface integral can again be reduced to a simple multiplication since the cylinder has rotational symmetry about the wire, and extends in both directions infinitely. EL(2πr) = q/ε. q can be expressed as a linear charge density d times the length of the wire, So the Ls cancel. So E = d/(2επr). Finally, we indicate the direction of the field, which will be directed away from the wire if the charge density is positive, and toward the wire if the charge density is negative. Suppose that the wire was not infinite? Then we could certainly use a Gaussian surface, but unfortunately there is no useful surface to pick to reduce the integral. Another method of attack is to use the differential form of the formula of the electric field for a point charge: dE = 1/(4πε)dq/(r²). We can then integrate from end to end of the wire adding up the infinitesimal dEs. It can be verified that by letting the lower and upper integrals go to negative infinity and infinity respectively, you will obtain the same formula found in the previous section.

[edit] Gaussian pillbox

This surface is most often used to determine the electric field due to an infinite sheet of charge with uniform charge density, or a slab of charge with some finite thickness.

[edit] References

Purcell, Edward M. (1985). Electricity and Magnetism. McGraw-Hill. ISBN 0-07-004908-4.
Jackson, John D. (1998). Classical Electrodynamics (3rd ed.). Wiley. ISBN 0-471-30932-X.
Tipler, Paul A. and Mosca, Gene (2004). Physics for Scientists and Engineers (Extended Version) (5th ed.). W.H. Freeman. ISBN 0-7167-4389-2.