Gaussian surface

A Gaussian surface is a closed surface in three dimensional space through which the flux of an electromagnetic field is calculated. It is an arbitrary closed surface $S=\partial V$ used in conjunction with Gauss's law in order to calculate the total enclosed electric charge by performing a surface integral, or alternatively, to calculate the electric field of a particular charge distribution.

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 integral.

1 Common Gaussian surfaces
- 1.1 Spherical surface
- 1.2 Cylindrical surface
2 Sample question
- 2.1 Gaussian pillbox
3 See also
4 References
5 External links

Common Gaussian surfaces

When performing the closed surface integral, the Gaussian surface (commonly abbreviated G.S. or g.s.) does not necessarily encompass all the charge; i.e., there can be arbitrary charges outside the volume: as mentioned, $\,Q(V)$ 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:

$\Phi_E = \iint_{\partial V}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\;\;\;\subset\!\supset \mathbf E\;\cdot\mathrm{d}\mathbf A = \frac{Q(V)}{\varepsilon_o}.$

Thereby $\,Q(V)$ 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.

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 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.

Cylindrical surface

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

an infinitely long line of uniform charge
an infinite plane of uniform charge

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

$q = \lambda\ h$

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

$\Phi_E = \iint_a\mathbf{E} \cdot d\mathbf{A} %2B \iint_b\mathbf{E} \cdot d\mathbf{A} %2B \iint_c\mathbf{E} \cdot d\mathbf{A}$

For surfaces a and b, $\,\mathbf E$ and $\,d\mathbf A$ will be perpendicular. For surface c, $\,\mathbf E$ and $\,d\mathbf A$ will be parallel, as shown in the figure.

$\Phi_E = \iint_a\mathbf{E} \cdot d\mathbf{A}\cos 90^\circ %2B \iint_b\mathbf{E} \cdot d\mathbf{A}\cos 90^\circ %2B \iint_c\mathbf{E} \cdot d\mathbf{A}\cos 0^\circ$

$\Phi_E = 0 %2B 0 %2B E\cdot \int d {A} = E\cdot A$

$\int\ d {A}\hat = \text{Surface area of cylinder} = 2 \pi\ r h$

so,

$\phi_E\ = E \cdot 2 \pi\ r h$

Gauss's law is

$\phi_E\ = \frac{q}{\varepsilon_0}$

$q = \lambda\ h$

So,

$\phi_E\ = \frac{\lambda\ h}{\varepsilon_0}$

Comparing both makes the following equation

$E 2 \pi\ r h = \frac{\lambda\ h}{\varepsilon_0}$

$E = \frac{\lambda}{2 \pi\varepsilon_0\ r}$ '

Sample question

Here is a critical thinking question about which the above concepts can be applied.

A slab of insulating material has thickness 2w and is oriented so that its faces are parallel to the yz-plane and given by the planes x=w and x= - w. The y- and z-dimensions 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 y-z plane, so the electric field can only point in the x-direction. But at the origin in the x-direction, 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 cylindrically-symmetric inner charge distribution acts as a line charge centered at the symmetry-axis of the cylinder.

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. 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.

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.

^ Introduction to electrodynamics By: Griffiths D.J

External links

Fields - a chapter from an online textbook