P20-045 – Gauss’s Law

Gauss’s Law is one of the four fundamental equations that govern the behavior of electromagnetic fields. It is a very powerful way to calculate the electric field created by a distribution of charge, provided the distribution has sufficient symmetry to simplify calculations.

Gauss’s Law:

The electric flux through any closed surface is proportional to the net charge enclosed by that surface

\boxed{\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\frac{Q_{enc}}{{\varepsilon }_0}}

where $Q_{enc}$ denotes the charge enclosed by the closed surface which is often referred to as a Gaussian surface.

Important: in practice, the goal of Gauss’s Law is to derive the electric field $\overrightarrow{E}$ created by a charge distribution. Computing the electric flux is simply a means to an end and we will choose closed surfaces that are such that the electric field is constant in magnitude and direction over them, which will then allow us to simplify the integral. Indeed, if $\theta $ denotes the angle between $\overrightarrow{E}$ and $d\overrightarrow{A}$ then the electric flux can be written

\oiint_A{\overrightarrow{E}\cdot d\overrightarrow{A}}=\oiint_A{E{\mathrm{cos} \left(\theta \right)\ }dA}

Furthermore, if $E$ and $\theta $ are constant over the Gaussian surface, the electric flux simplifies to

\boxed{\oiint_A{E{\mathrm{cos} \left(\theta \right)\ }dA}=E{\mathrm{cos} \left(\theta \right)\ }\oiint_A{dA}=EA{\mathrm{cos} \left(\theta \right)\ }}

where $A$ denotes the surface area of the Gaussian surface chosen. This simplification then allows us to extract $E$ from the integral and to solve for it directly without having to carry out any calculus. Thus, it is important to choose a Gaussian surface wisely for this simplification to be valid.

Method: how to choose a Gaussian surface?

In order to allow the above simplification of the electric flux integral, we must carefully choose a Gaussian surface with the following properties:

• It encloses the charge or part of the charge that creates the electric field.
• It is such that electric field lines are parallel $\left(\theta =90{}^\circ \right)$ or perpendicular to it $\left(\theta =0{}^\circ \right)$. If this is not possible, then it is such that the angle $\theta $ between the electric field lines and $d\overrightarrow{A}$ is constant everywhere over the surface.
• The electric field is constant in magnitude everywhere over the surface.
• Its shape remains simple in order to make its surface area $A$ easy to evaluate.

Fundamental Theorem:

The electric flux through a closed surface can be zero without the electric field being zero. Having $Q_{enc}=0$ is sufficient.

If the electric field is zero on a closed surface, then the electric flux through the closed surface is zero and the enclosed charge is necessarily zero $\left(Q_{enc}=0\right)$.