Gauss' law (electrostatics)

In physics, more specifically in electrostatics, Gauss' law is a theorem concerning a surface integral of an electric field. In vacuum Gauss' law takes the form:

$\iint _{\mathrm {closed} \atop \mathrm {surface} }\mathbf {E} \cdot d\mathbf {S} =kQ_{\mathrm {tot} }\quad {\textrm {with}}\quad k={\begin{cases}{\frac {1}{\epsilon _{0}}}\quad &{\hbox{SI units}}\\4\pi &{\hbox{Gaussian units}}.\end{cases}}$

Here dS is an vector with length dS, the area of an infinitesimal surface element on the closed surface, and direction perpendicular to the surface element dS, pointing outward. The vector E is the electric field at the position dS, the dot indicates a dot product between the vectors E and dS. The double integral is over a closed surface that envelops a total electric charge Q_tot, which may be the sum over one or more point charges, or an integral over a charge distribution. If the closed surface does not envelop any charge then the integral is zero. The constant ε₀ is the electric constant.

The law is called after the German mathematician Carl Friedrich Gauss.

Application to spherical symmetric charge distribution

Gauss' law is a convenient way to computing electric fields in the case of spherical-symmetric charge distributions. For instance, a point charge is a spherical-symmetric charge distribution. Another example is a charged, conducting, spherical shell, the charge distribution being homogeneously distributed over the shell.

Take the origin of a spherical polar coordinate system in the center of symmetry of the charge distribution—the position of the point charge, or the center of the spherical shell, respectively. Because of symmetry, E has a radial component only (parallel to the unit vector e_r). Moreover, this component does not depend on the polar angles,

$\mathbf {E} ={\Big (}E_{r}(r),\quad E_{\theta }=0,\quad E_{\phi }=0{\Big )}.$

Take the surface of a sphere of radius r to integrate over (in the case that we are considering a spherical shell r must be larger than the radius of the spherical shell); the surface element is

$d\mathbf {S} =r^{2}\,\sin \theta \,d\theta \,d\phi \,\mathbf {e} _{r}\quad {\hbox{and}}\quad \mathbf {E} \cdot d\mathbf {S} =E_{r}(r)r^{2}\sin \theta d\theta d\phi$

Then,

${\begin{aligned}kQ_{\mathrm {tot} }&=E_{r}(r)\,r^{2}\iint _{\mathrm {surface} \atop \mathrm {of\,sphere} }\sin \theta \,d\theta \,d\phi =E_{r}(r)\,r^{2}\int _{0}^{\pi }\int _{0}^{2\pi }\sin \theta \,d\theta \,d\phi =E_{r}(r)\,r^{2}\,4\pi \\&\Longrightarrow \,E_{r}(r)={\frac {kQ_{\mathrm {tot} }}{4\pi r^{2}}},\end{aligned}}$

where r is the distance of the field-point to the origin. In the case of a point charge we have proved here Coulomb's law from Gauss' law. In the case of a charged spherical shell, we find that the electric field outside the shell is such that it seems that the total charge on the shell is concentrated in the center of the shell. Coulomb's law applies to the charge seemingly concentrated in the center of the shell.

Note

Gauss' law is an integral form of one of the four Maxwell's equations and as such is a common point of departure for development of electrostatic theory. From it Coulomb's law can be derived. Conversely, one can take Coulomb's law as point of departure and derive Gauss' law from it. If one does this, one finds that the inverse-square distance dependence arising in Coulomb's law exactly cancels the dependence on length squared of a surface. In other words, the magnitude of the enveloping surface is arbitrary. Given Q_tot, it is irrelevant for Gauss' law how tightly or loosely the surface envelops this charge.