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Fig. 1 - The balloon-hair experiment is a great example of the interaction between the electric field of opposite charges.
Gauss's Law Description
In previous AP Physics courses, we have covered various aspects of electrostatics and electromagnetism, including how to calculate the total electric field from a group of charged particles. However, we never really explained where this formula comes from or how it was derived. In 1867, the German physicist and mathematician Carl Friedrich Gauss came up with a theory that allowed one to calculate the external electric field depending on the amount of charge enclosed in a surface. The theory he came up with is one of Maxwell's equations of electrodynamics.
This theory was coined as Gauss's law, it allows us to describe the electric flux through a Gaussian surface to the charge enclosed by a Gaussian surface. But what exactly is a Gaussian surface?
A gaussian surface is a three-dimensional closed surface where the field lines of an electric, magnetic, or gravitational field flow past.
It is important to note that through Gauss's law, the position and orientations of the charges enclosed by the gaussian surface are irrelevant. The only contributing factor to the surrounding electric field is the total amount of charge enclosed.
Gauss's Law Formula
Now we will introduce the formula for Gauss's law. It is given by
\[ \Phi_{\text{E}} = \frac{q_{\text{enclosed}}}{\epsilon_0} ,\]
where \( \Phi_{\text{E}}\) is the electric flux through the gaussian surface generated by the group of charges measured in \(\mathrm{\frac{V}{m}}\), \(q_{\text{enclosed}} \) is the total enclosed charge by the gaussian surface measured in coulombs \(\mathrm{C}\), and \(\epsilon_0\) is the vacuum permittivity given by a value of \( 8.85 \times 10^{-12} \, \mathrm{\frac{F}{m}} \). This is the simpler form of Gauss's law for a point charge, but we can also introduce the integral form as
\[ \oint_A \vec{E} \cdot \mathrm{d} \vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0}, \]
where \(\vec{E}\) is the electric field vector measured in units of \(\mathrm{\frac{V}{m}}\), \(\mathrm{d}\vec{A}\) is the infinitesimal area vector measured in \(\mathrm{m^2}\), and all the other quantities are the same as before. It is also important to note that the first equation involves the electric flux through the surface \(\Phi_{\text{E}}\), whilst the second equation involves the electric field vector \(\vec{E}\). Furthermore, the dot product between \(\vec{E}\) and \(\mathrm{d}\vec{A}\) shows that we are only considering the electric field components perpendicular to the gaussian surface.
Let's consider an example where we apply Gauss's law in its simpler form. Considering the image below, use Gauss's law to calculate the electric field.
Fig. 3 - The electric field from a group of charges can be calculated using Gauss's law.
Here we can see the group of charges has charge values of \(-2e\), \(1e\), \(3e\), and \(-4e\) respectively, giving a total enclosed charge of
\[ \begin{align} q_{\text{enclosed}} &= -2e + 1e + 3e + (-4e) \\ q_{\text{enclosed}} &= -2e .\end{align} \]
Finally, we can plug this into Gauss's law to find that
\[ \begin{align} \Phi_{\text{E}} &= \frac{q_{\text{enclosed}}}{\epsilon_0} \\ \Phi_{\text{E}} &= \frac{ -2 \times 1.6 \times 10^{-19} \,\mathrm{C} }{8.85 \times 10^{-12} \, \mathrm{\frac{F}{m}}} \\ \Phi_{\text{E}} &= -3.6 \times 10^{-8} \, \mathrm{V\,m} . \end{align} \]
It is important to note that the shape of the gaussian surface is irrelevant to the calculation. When working with the integral form, the shape becomes relevant, which we will see later on.
Gauss's Law Derivation
Previously, we have introduced two forms of Gauss's law so far, but we also have a third form, Gauss's law in differential form. In order to understand the derivation, we first need to quickly cover the divergence theorem. This theorem, also coined by Gauss, states that
\[ \int_V ( \nabla \cdot \vec{F} ) \, \mathrm{d} V = \oint_S \vec{F} \cdot \, \mathrm{d} \vec{S} ,\]
where \(\nabla \cdot \vec{F}\) is the divergence of the vector field \(\vec{F}\), \(\mathrm{d} V\) is the infinitesimal volume, and \(\mathrm{d} S\) is the infinitesimal surface. This shows how we can relate the volume integral of the divergence of a vector field, to a closed surface integral of the flux of the vector field.
Back to Gauss's law, let's start from the integral form, substituting the divergence theorem to give
\[ \begin{align} \oint_A \vec{E} \cdot \mathrm{d} \vec{A} &= \frac{q_{\text{enclosed}}}{\epsilon_0} \\ \int_V (\nabla \cdot \vec{E} ) \, \mathrm{d} V &= \frac{q_{\text{enclosed}}}{\epsilon_0} . \end{align} \]
Our next step is to introduce another integral form that relates the charge density and enclosed charge. By taking a volume integral of the charge density \(\rho\), we find that we can rewrite the total charge as
\[ Q = \int_V \rho \, \mathrm{d} V,\]
where \(Q\) is the total charge measured in coulombs \(\mathrm{C}\) and \(\rho\) is the charge density measured in \(\mathrm{\frac{C}{m^3}}\).
Despite the name charge density, \(\rho\) can be a function from one to three dimensions. Meaning that, depending on the specified charge density function, we can find the total charge by integrating over one, two, or three dimensions.
Thus, this allows us to rewrite Gauss's law as
\[ \int_V (\nabla \cdot \vec{E} ) \, \mathrm{d} V = \int_V \frac{\rho}{\epsilon_0} \, \mathrm{d} V .\]
Now we can see that we have two equivalent volume integrals on either side, allowing us to cancel them off. This leaves us with
\[ \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} .\]
This equation is what we refer to as Gauss's law in its differential form. It is important to note that the integral and differential forms of Gauss's law hold in all general cases, whereas the first equation we introduced is only applicable to point charges. By introducing these two calculus-based forms, we will be able to deal with much trickier problems involving the calculation of an electric field.
Gauss's Law Examples
Now let's consider an example where we use the integral form of Gauss's law.
Consider the figure below of a charged sphere with charge value \(q\), note that the sphere is in three-dimensional space. Using Gauss's law in integral form, calculate the electric field produced by the charged sphere.
Fig. 4 - We can calculate the electric field due to a charged sphere with Gauss's law.
From the diagram, we place the gaussian surface to be right on the edge of the sphere. To solve the integral, we could use cartesian coordinates, however, there is a trick we can use to make our calculation much easier, switching to spherical coordinates instead. You will not have come across spherical coordinates yet, but it will be something you cover in university physics.
\[ \mathrm{d} A = r^2 \sin(\theta) \mathrm{d} \theta \mathrm{d} \phi ,\]
where \(r\) is the radius of our sphere measured in \(\mathrm{m}\), \(\theta\) is the angle between the vertical and our unit vector measured in radians \(\mathrm{rads}\), and \(\phi\) is the angle around the equator measured in radians \(\mathrm{rads}\). Referring to the image below, we can see these quantities more clearly.
We can now write our integral as
\[ \oint_A \left(\vec{E} \cdot \hat{n} \right) r^2 \sin(\theta) \mathrm{d} \theta \mathrm{d} \phi = \frac{q}{\epsilon_0} .\]
Substituting in our integral limits for a sphere then gives
\[ \int^{\theta = \pi}_{\theta = 0} \int^{\phi = 2\pi}_{\phi = 0} \left(\vec{E} \cdot \hat{n} \right) r^2 \sin(\theta) \mathrm{d} \theta \mathrm{d} \phi = \frac{q}{\epsilon_0} .\]
The spherical symmetry of the problem implies that the electric field is only dependent on \(r\), therefore we can deduce that \( \vec{E} \cdot \hat{n} = E(r) \).
Thus we can solve our integral to give
\[ \begin{align} \int^{\theta = \pi}_{\theta = 0} \int^{\phi = 2\pi}_{\phi = 0} E(r) r^2 \sin(\theta) \mathrm{d} \theta \mathrm{d} \phi &= \frac{q}{\epsilon_0} \\ E(r) r^2 \int^{\theta = \pi}_{\theta = 0} \int^{\phi = 2\pi}_{\phi = 0} \sin(\theta) \mathrm{d} \theta \mathrm{d} \phi &= \frac{q}{\epsilon_0} \\ E(r) r^2 \int^{\theta = \pi}_{\theta = 0} 2\pi \sin(\theta) \mathrm{d} \theta &= \frac{q}{\epsilon_0} \\ E(r) r^2 2\pi [ -\cos(\theta)]^{\pi}_{0} &= \frac{q}{\epsilon_0} \\ E(r) r^2 2\pi \times 2 &= \frac{q}{\epsilon_0} \\ E(r) &= \frac{q}{4 \pi \epsilon_0 r^2} . \end{align} \]
This equation should be familiar to you as it is the electric field from a point charge! We have now proven this through Gauss's law
Gauss's Law of Magnetism
Gauss not only came up with this equation of electric fields but a similar equation for magnetic fields. This is also one of Maxwell's equations of electromagnetism.
Similar to Gauss's law for electric fields, Gauss's law for magnetic fields has a differential form and integral form. The differential form is given as
\[ \nabla \cdot \vec{B} = 0,\]
where \( \nabla \cdot\) indicates the divergence operator and \(\vec{B}\) is the magnetic field vector measured in units of \(\mathrm{\frac{A}{m}}\). On the other hand, its integral form is given as
\[ \oint_S \vec{B} \cdot \mathrm{d} S = 0,\]
where \(\mathrm{d} \vec{S}\) is the infinitesimal surface vector.
Comparing Gauss's law for magnetic fields versus electric fields, we can see that the divergence of a magnetic field vector is zero, whilst the divergence of an electric field vector is non-zero. The significance of these two equations tells us that magnetic monopoles do not exist. A magnetic dipole is an object that has both a north and a south pole, such as an atom with an orbiting electron, or a bar magnet. Unlike electric monopoles, such as charged particles, we do not have an equivalent when thinking about magnetic objects.
Gauss's Law - Key takeaways
- Gauss's law for electric fields is one of Maxwell's equations for electromagnetism.
- We use Gauss's law by enclosing a group of charges in a gaussian surface, which is a three-dimensional closed surface where the field lines of an electric, magnetic, or gravitational field flow past.
- Gauss's law for a point charge is given as \( \Phi_{\text{E}} = \frac{q_{\text{enclosed}}}{\epsilon_0} \).
- Gauss's law in integral form is given as \(\oint \vec{E} \cdot \mathrm{d} \vec{A} = \frac{q_{\text{enclosed}}}{\epsilon_0} \).
- Gauss's law in differential form is given as \( \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} \).
- Gauss's law for magnetic fields tells us that magnetic monopoles do not exist.
References
- Fig. 1 - Balloon hair, Wikimedia Commons (https://commons.wikimedia.org/wiki/File:Attractive-electric-force-between-hair-and-balloon.jpg) Licensed by CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0/)
- Fig. 2 - Gauss law formula diagram, StudySmarter Originals.
- Fig. 3 - Enclosed charges, StudySmarter Originals.
- Fig. 4 - Charged sphere, StudySmarter Originals.
- Fig. 5 - Spherical coordinates, StudySmarter Originals.
- Fig. 6 - Magnetic field lines, StudySmarter Originals.
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Frequently Asked Questions about Gauss's Law
What does Gauss's law explain?
How to calculate an electric field from a group of enclosed charges.
What is Gauss's law and how do we prove it?
Gauss's law is one of Maxwell's equations to calculate the electric field from enclosed charges. There is no proof for Gauss's law and it was confirmed with scientific experimentation.
What is the significance of Gauss's law and Coulomb's law in electricity?
Gauss's law focuses on the electric field generated from enclosed charges, on the other hand, Coulomb's law describes the force between electric charges.
What is the difference between Gauss's law and Gauss's theorem?
They are the same thing.
What are the applications of Gauss's law?
Calculating the electric field from a group of charges.
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