See also: Charge density
Most calculations using Gaussian surfaces begin by implementing Gauss's law (for electricity):2
Thereby Qenc is the electrical charge enclosed by the Gaussian surface.
This is Gauss's law, combining both the divergence theorem and Coulomb's law.
A spherical Gaussian surface is used when finding the electric field or the flux produced by any of the following:3
The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S 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.
The flux out of the spherical surface S is:
The surface area of the sphere of radius r is ∬ S d A = 4 π r 2 {\displaystyle \iint _{S}dA=4\pi r^{2}} which implies Φ E = E 4 π r 2 {\displaystyle \Phi _{E}=E4\pi r^{2}}
By Gauss's law the flux is also Φ E = Q A ε 0 {\displaystyle \Phi _{E}={\frac {Q_{A}}{\varepsilon _{0}}}} finally equating the expression for ΦE gives the magnitude of the E-field at position r: E 4 π r 2 = Q A ε 0 ⇒ E = Q A 4 π ε 0 r 2 . {\displaystyle E4\pi r^{2}={\frac {Q_{A}}{\varepsilon _{0}}}\quad \Rightarrow \quad E={\frac {Q_{A}}{4\pi \varepsilon _{0}r^{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 do not count.
A cylindrical Gaussian surface is used when finding the electric field or the flux produced by any of the following:4
As 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. Imagine a closed surface in the form of cylinder whose axis of rotation is the line charge. If h is the length of the cylinder, then the charge enclosed in the cylinder is q = λ h , {\displaystyle q=\lambda h,} where q is the charge enclosed in the Gaussian surface. There are three surfaces a, b and c as shown in the figure. The differential vector area is dA, on each surface a, b and c.
The flux passing consists of the three contributions:
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.
Φ E = ∬ a E d A cos 90 ∘ + ∬ b E d A cos 90 ∘ + ∬ c E d A cos 0 ∘ = E ∬ c d A {\displaystyle {\begin{aligned}\Phi _{E}&=\iint _{a}EdA\cos 90^{\circ }+\iint _{b}EdA\cos 90^{\circ }+\iint _{c}EdA\cos 0^{\circ }\\&=E\iint _{c}dA\end{aligned}}}
The surface area of the cylinder is ∬ c d A = 2 π r h {\displaystyle \iint _{c}dA=2\pi rh} which implies Φ E = E 2 π r h . {\displaystyle \Phi _{E}=E2\pi rh.}
By Gauss's law Φ E = q ε 0 {\displaystyle \Phi _{E}={\frac {q}{\varepsilon _{0}}}} equating for ΦE yields E 2 π r h = λ h ε 0 ⇒ E = λ 2 π ε 0 r {\displaystyle E2\pi rh={\frac {\lambda h}{\varepsilon _{0}}}\quad \Rightarrow \quad E={\frac {\lambda }{2\pi \varepsilon _{0}r}}}
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.
Essential Principles of Physics, P.M. Whelan, M.J. Hodgeson, 2nd Edition, 1978, John Murray, ISBN 0-7195-3382-1 /wiki/ISBN_(identifier) ↩
Introduction to electrodynamics (4th Edition), D. J. Griffiths, 2012, ISBN 978-0-321-85656-2 /wiki/ISBN_(identifier) ↩
Physics for Scientists and Engineers - with Modern Physics (6th Edition), P. A. Tipler, G. Mosca, Freeman, 2008, ISBN 0-7167-8964-7 /wiki/ISBN_(identifier) ↩