A Gaussian surface is a closed threedimensional surface through which a flux of an electric field is to be calculated. It is thus not an arbitrary closed surface . Rather the surface is used not only in conjunction with Gauss's integration law (a consequence of the divergence theorem), but also the Gauss relation of Maxwell electrodynamics, involving the Coulomb form of the electrical field, is considered, 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.
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When performing the closed surface integral, the Gaussian surface does not necessarily encompass all the charge; i.e., there can be arbitrary charges outside the volume: as mentioned, only counts the interior contribution. Furthermore: 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:. May be used for items such as charged disks, electric dipoles, or triangles with a charge at each.
Thereby is the electrical charge contained in the interior, V, of the closed surface.
This is Gauss's law, combining both the divergence theorem and Coulomb's statement.
A spherical Gaussian surface is used when finding the electric field or the flux produced by any of the following:
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 nonzero electric field. This is determined as follows:
This nontrivial result shows that any 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. And, as mentioned, any exterior charges don't count.
A cylindrical Gaussian surface is used when finding the electric field or the flux produced by any of the following:
As an example "field near infinite line charge" is given below;
Consider a point P at a distance r from an infinite line charge having charge density (charge per unit length) λ (lambda). Imagine a closed surface in the form of cylinder around line charge in its wall.
If h is the length of cylinder, then charge enclosed in cylinder is
where, q is the charge enclosed in Gaussian surface. There are three surfaces a, b and c as shown in figure. Now Take differential area "dA" with vector area dA on each surface i.e a, b and c.
Now the flux passing consists of the three contributions
For surfaces a and b, and will be perpendicular. For surface c, and will be parallel, as shown in the figure.
so,
Gauss's law is
As
So,
Comparing both makes the following equation
Here is a critical thinking question about which the above concepts can be applied.
A slab of insulating material has thickness 4w and is oriented so that its faces are parallel to the yzplane and given by the planes x=w and x=  w. The y and zdimensions of the slab are very large compared to d and may be treated as essentially infinite. The slab has a uniform positive charge density p.
Explain why the electric field due to the slab is zero at the center of the slab (x=0)? The electric field of the slab must be zero by symmetry. There is no preferred direction in the yz plane, so the electric field can only point in the xdirection. But at the origin in the xdirection, neither the positive nor negative directions should be singled out as special, and so the field must be zero.
Again, only the inner distribution counts, and again, any cylindricallysymmetric inner charge distribution acts as a line charge centered at the symmetryaxis of the cylinder.
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. The pillbox has a cylindrical shape, and can be thought of as consisting of three components: the disk at one end of the cylinder with area πR², the disk at the other end with equal area, and the side of the cylinder. The sum of the electric flux through each component of the surface is proportional to the enclosed charge of the pillbox, as dictated by Gauss's Law. Because the field close to the sheet can be approximated as constant, the pillbox is oriented in a way so that the field lines penetrate the disks at the ends of the field at a perpendicular angle and the side of the cylinder are parallel to the field lines. by: pratibha
