Gauss' law

Gauss' law describes the flux of an electric field generated by a charge inside of a volume passing through the bounding surface of that volume. Given a system of charges or a charge distribution within the enclosing surface $S$ , the flux must be

\oint_{S}\mathbf{E}\cdot d\mathbf{S}=\frac{Q_{\text{enc}}}{\varepsilon_{0}}

where $Q_{enc}$ is the total charge enclosed within the surface and $d\mathbf{S}$ is the normal vector on the surface element $dS$ . This is true for any surface and and any charge distribution therein. It can be expressed in differential form through the divergence of the field:

\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_{0}}

where $\rho$ is the volume charge density. Alternatively, using $\mathbf{E}=-\nabla V$ :

\nabla ^{2}V=- \frac{\rho}{\epsilon_{0}}

which is Poisson's equation.

Gauss' law holds even for changing electric fields and moving charges, which may be subject to Electromagnetic induction.

Derivation#

Consider a point charge $q$ within a sphere of radius $R$ . The outgoing flux is going to be

\oint_{S}\mathbf{E}\cdot d\mathbf{S}=\int \frac{1}{4\pi\varepsilon_{0}}\left(\frac{q}{r^{2}}\hat{\mathbf{r}}\right)\cdot(r^{2}\sin\theta d\theta d\phi \hat{\mathbf{r}})=\frac{q}{\varepsilon_{0}}

The important thing to note here is that this occurs only because Coulomb's law goes like $\sim 1/r^{2}$ , as this cancels out perfectly with the area that goes like $\sim r^{2}$ . Any more or less and the flux would depend on the chosen area, but it does not. This can be easily extended to any number of particles by principle of superposition. Consider $N$ charges, all within the surface. The total field is

\mathbf{E}=\sum\limits_{i=1}^{N}\mathbf{E}_{i}

so the previous statement becomes

\oint_{S}\mathbf{E}\cdot d\mathbf{S}=\sum\limits_{i=1}^{N}\left(\oint\mathbf{E}_{i}\cdot d\mathbf{S}\right)=\sum\limits_{i=1}^{N} \frac{q_{i}}{\varepsilon_{0}}

calling the total enclosed charge $Q_{enc}=\sum\limits_{i=1}^{N}q_{i}$ we get Gauss' law

\oint_{S}\mathbf{E}\cdot d\mathbf{S}=\frac{Q_{enc}}{\varepsilon_{0}}

The differential form is derived by use of the divergence theorem:

\oint_{S}\mathbf{E}\cdot d\mathbf{S}=\int_{V}(\nabla\cdot\mathbf{E})\ d\tau

If we express the enclosed charge through a volume density by

Q_{enc}=\int_{V}\rho\ d\tau

we get

\int_{V}(\nabla\cdot\mathbf{E})\ d\tau=\int_{V} \frac{\rho}{\varepsilon_{0}}\ d\tau

The integrals are in the same variable and over the same domain, so to be equal, the integrands must be equal. Thus

\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_{0}}

Usage#

Gauss' law in integral form is by far the most convenient way of calculating the electric field of a system, provided symmetry works in one's favor.

Symmetries#

Gauss' law is most useful when in a system under specific kinds of symmetries. The three types of symmetries that well work are:

Spherical symmetry: make the Gaussian surface a concentric sphere.
Cylindrical symmetry: make the Gaussian surface a coaxial cylinder. Think of it like a electric cable coated in rubber.
Planar symmetry: make the Gaussian surface a "pillbox" that straddles the plane, half over and half under.

Points 2. and 3. technically require a infinite cylinder and an infinite plane to work, but can be used to good effect as approximations for very long cylinders and very large planes.

Spherical#

Consider a ball of radius $R$ and total charge $q$ . Imagine a sphere of radius $r>R$ (often called a Gaussian surface). In this case, $Q_{enc}=q$ . Symmetry allows us extract $\mathbf{E}$ out of the integral. In fact, $\mathbf{E}$ is always perpendicular to surface element (i.e. it's radial)¹, as is the surface normal, so the dot product between the two can be dropped

\int \mathbf{E}\cdot d\mathbf{a}=\int |\mathbf{E}|\ da

The magnitude of the electric fields depends on the distance, but the distance is constant over a sphere centered on the charge, so we can pull it out of the integral

\int|\mathbf{E}|\ da=|\mathbf{E}|\int da=|\mathbf{E}|4\pi R^{2}

Applying Gauss' law we get

|\mathbf{E}|4\pi R^{2}=\frac{q}{\varepsilon_{0}}

Since we know the direction of $\mathbf{E}$ , we can "reverse engineer" the $\mathbf{E}$ vector from just its magnitude: $|E|\hat{\mathbf{r}}=\mathbf{E}$ . Thus, we have

\mathbf{E}=\frac{1}{4\pi R^{2}} \frac{q}{\varepsilon_{0}}\hat{\mathbf{r}}

This way, we got the field generated by a ball (or even a sphere, really) without doing a single integral. The crucial part here is that the system is spherically symmetrical. If it weren't then $\mathbf{E}$ would not have been aligned with $d\mathbf{a}$ and we wouldn't have been able to pull $|\mathbf{E}|$ out of the integral.

Cylindrical#

Consider a long cylinder of charge density proportional to the distance from the axis $\rho=ks$ , where $s$ is the distance and $k$ is a constant. The cylinder has a length of $l$ and a radius of $R$ . The enclosed charge is

Q_{enc}=\int \rho d\tau=\int(ks)(s\ dsd\phi dz)=k\int_{0}^{R}s^{2}ds\int_{0}^{2\pi}d\phi\int_{0}^{l}dz=\frac{2}{3}\pi klR^{3}

by using Cylindrical coordinates. By symmetry, $\mathbf{E}$ points radially outwards, so

\int \mathbf{E}\cdot d\mathbf{a}=|\mathbf{E}|\int da=|\mathbf{E}|2\pi Rl

This is for the "long side" of the cylinder. The "top" and "bottom" of the cylinder contribute nothing because $\mathbf{E}$ is perpendicular to the surface there, so $\mathbf{E}\cdot d\mathbf{a}=0$ . By Gauss' law this is

|\mathbf{E}|2\pi Rl=\frac{1}{\varepsilon_{0}} \frac{2}{3}\pi klR^{3}

and thus

\mathbf{E}=\frac{1}{3\varepsilon_{0}}kR^{2}\hat{\mathbf{r}}

with $\hat{\mathbf{r}}$ the unit vector pointing radially outwards.

Planar#

Consider an infinite plane that carries a uniform surface charge $\sigma$ . We can use a "Gaussian pillbox" extending equal distances above and below the plane. Here we have

Q_{enc}=\sigma A

where $A$ is the area of the parts of the pillbox parallel to the plane (the "top" and "bottom"). By symmetry, $\mathbf{E}$ points away from the plane (upwards above the plane, downwards below the plane), so the integral yields

\int \mathbf{E}\cdot d\mathbf{a}=2A|\mathbf{E}|

and the sides contribute nothing because they are perpendicular to the field. Thus we have

2A|\mathbf{E}|=\frac{\sigma}{\varepsilon_{0}}A

and so

\mathbf{E}=\frac{\sigma}{2\varepsilon_{0}}\hat{\mathbf{n}}

where $\hat{\mathbf{n}}$ is the normal vector of the plane.

Symmetry by superposition#

While there are only three types of symmetries to work with, it's possible to compute the field of more complex systems if each of the components is individually symmetrical in one of these three ways. This is allowed by the superposition principle, as we can calculate the electric fields in isolation and then just sum them together.

Two parallel planes#

Consider two parallel infinite planes each carrying the same but opposite uniform surface charge density $\pm\sigma$ . By the result above, we know that each plane singularly produces a field of intensity

E=\pm\frac{\sigma}{2\varepsilon_{0}}

These fields are parallel to each other. The positive one points away from the plane, whereas the negative one points towards the plane. This means that the fields cancel each other out outside of the two planes and sum up in between to get

E_{out}=0, \quad E_{in}=\frac{\sigma}{\varepsilon_{0}}

where the inner field goes from the positive plane to the negative one.

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In dielectrics#

In a dielectric, the electric field is given by both the bound charges caused by dielectric polarization and by the free charges due to everything else. The total charge density is

\rho=\rho_{b}+\rho_{f}

and Gauss' law reads

\varepsilon_{0}\nabla\cdot\mathbf{E}=\rho=\rho_{b}+\rho_{f}=-\nabla\cdot\mathbf{P}+\rho_{f}

where $\mathbf{E}$ is the total field, due to both the polarization and external effects. Combining the divergence terms

\nabla\cdot(\varepsilon_{0}\mathbf{E}+\mathbf{P})=\rho_{f}

The expression in parentheses is called the electric displacement $\mathbf{D}$ . Thus, Gauss' law reads

\boxed{\nabla\cdot\mathbf{D}=\rho_{f}}

or in integral form

\boxed{\oint \mathbf{D}\cdot d\mathbf{a}=Q_{f,enc}}

where $Q_{f,enc}$ denotes the total free charge within the Gaussian surface. This form is nice because it only references free charges, which are generally artificial and therefore known, as opposed to bound charges. It's also possible to calculate displacement directly when the symmetry requirements above are met.

$\mathbf{E}$ can't not be radial. The ball is static and perfectly smooth, so there is not point of reference, like say, an axis of Rotation. This means any direction is equal, as any direction that's chosen could just as easily be chosen to be anything else. The only direction that has "character" to it, in the sense that it can be actually be defined, is the radial direction. ↩