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 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.

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[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 QA = 0 in Gauss's law, where QA 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 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 close 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.

Closed surface in the form of cylinder having line charge in the center and showing differential areas dAof all three surfaces.
Closed surface in the form of cylinder having line charge in the center and showing differential areas dAof all three surfaces.

Now the flux passing will be

\Phi_E = \oint\mathbf{E} \cdot d\mathbf{A} = \int_a\mathbf{E} \cdot d\mathbf{A} + \int_b\mathbf{E} \cdot d\mathbf{A} + \int_c\mathbf{E} \cdot d\mathbf{A}

For surfaces a and b, E and dA will be perpendicular. For surface c, E and dA will be parallel. As shown in the figure.

\Phi_E = \int_a\mathbf{E} \cdot d\mathbf{A}\cos 90^\circ + \int_b\mathbf{E} \cdot d\mathbf{A}\cos 90^\circ + \int_c\mathbf{E} \cdot d\mathbf{A}\cos 0^\circ
\Phi_E = 0 + 0 + E \int\ d \mathbf{A} = E \int\ d \mathbf{A}
\int\ d \mathbf{A} = \text{Surface area of cylinder} = 2 \pi\ r h

so,

\phi_E\ = E 2 \pi\ r h

Gauss's law is

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

As

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}

[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. 

[edit] External links

  • Fields - a chapter from an online textbook
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