September 14 2026

Today was the first day of my electricity grind. I went through around 3/4 of the first chapter of Purcell, learning about Coulomb’s law, the Electric field, Flux and Gauss’s law.

Like always, here’s a quick summary of what I have learned.

Electricity and Magnetism: Coulomb’s law, the Electric field, Flux and Gauss’s law

Charge

Charge is a fundamental property that particles have. Rather interestingly, by experiment we have observed that completely different particles like electrons or protons have the same magnitude of charge. Charge is the reason why we have electric forces between particles. We define electrons to have a charge of -e, and protons to have a charge of e, etc. e represents a fundamental unit of charge, being equal to $1.602 \times 10^{-19} C$, where “C” means Coulombs. Coulombs is the SI unit for charge.

Coulomb’s law

Caution

We (for some reason) use the convention that $\vec{r}_{21}$ represents a vector going to 2 from 1.

Coulomb’s law states the following:

For two charges $q_1$ and $q_2$, the following is true:

Note

Coulomb’s law

$$\vec{F}_2 = \frac{1}{4\pi \epsilon_0}\frac{q_1 q_2 \hat{r}_{21}}{r_{21}^2}$$

So like charges repel, and opposite charges attract. Wowie! A useful property of the electric force is that it adheres to superposition. This means that we can sum the vectors of electric forces across multiple particles to obtain a net force.

The Electric Field

Say if we focused on a single charged particle, at position $0$. Around it, we define an electric field that determines its force. The electric field is defined as:

Note

The Electric Field

$$\vec{E}(x, y) = \frac{1}{4\pi\epsilon_0} \sum^{n}_{i = 1} \frac{q_2 \hat{r_{0i}}}{r_{0i}^2}.$$

We can simply find the net electric force on that particle by using:

$$\vec{F}(x, y) = q\vec{E}(x,y)$$

Caution

In electrostatics, a field that depends only on the main charges’s position assumes that the other charges present are fixed. If they weren’t fixed, then they would move by the interaction with the main particle that we focus on, causing the Electric field to change. Alternatively, we can assume that the charge we have is infinitesimal.

We can also define the electric field for continuous charge distributions. Such distributions will be very important later on. Here, we can simply use an integral:

Note

The Electric Field (Version 2)

$$\vec{E}(x,y,z) = \frac{1}{4\pi\epsilon_0} \int \frac{\rho(x', y', z') \hat{r} dx' dy' dz'}{r^2}$$

Where $\hat{r}$ points from a infinitesimal charge towards $(x, y, z)$, or the location of the charge we consider.

Flux

Flux is sort of like water flowing through a pipe. If the pipe’s cross section was slanted against the direction of flow, then there would be less flow. Instead of something like the flow of water, we imagine the flow of charges. More precisely, we imagine how intense the electric field is across an area. Therefore, flux can be defined as:

Note

Flux

$$\Phi = \sum \vec{A_i} \cdot \vec{E_i}.$$

For continuous charge distributions:

$$\Phi = \int_{S} \vec{E} \cdot d\vec{A}.$$

Here, $\int_{S}$ means we integrate over infinitesimal, basically flat parts of the surface. It might seem very complicated to do this integral, but we actually have a very convenient and amazing law that allows us to skip all these calculations!

Note that the area vector is just the vector with the magnitude of the area pointing normal to it. Moreover, since Flux is defined with a dot product, it is a scalar quantity.

Gauss’s Law

Consider a sphere of uniform charge, and the field of a single isolated positive point charge $q$. The answer is really easy - $\vec{E}$ points along the same direction as $A$ for every small area element. As a result, the total flux is just:

$$\Phi = E \cdot A = \frac{q}{4\pi \epsilon_0 r^2} \cdot 4\pi r^2 = \frac{q}{\epsilon_0}$$

Now we can derive Gauss’s law. Imagine a closed surface that encloses a sphere. Now project each small section of the sphere that we considered onto the closed surface, and define $\theta$ to be the angle between the original surface and the new surface.

Note that the ratio of the areas are:

$$\frac{A_{new}}{A} = \left( \frac{r_{new}}{r} \right)^2 \frac{1}{\cos{\theta}}.$$ $$\frac{E_{new}}{E} = \left(\frac{r}{r_{new}}\right)^2.$$

The new flux through the projected area is:

$$\vec{A} \cdot \vec{E} = A_{new} E_{new} \cos{\theta} = AE\left( \frac{r_{new}}{r} \right)^2 \frac{1}{\cos{\theta}} \cdot \left(\frac{r}{r_{new}}\right)^2 \cos{\theta} = AE.$$

We see that the flux through the projected area is the same as the flux through the original area. Therfore, making correspondences for every little patch, we know the total flux through the arbitrary surface is the same as the total flux through the sphere, or $\frac{q}{\epsilon_0}$. This is really powerful - any arbitrary surface enclosing a charge q has the same flux of $\frac{q}{\epsilon_0}$! As a corollary, any charge q outside the surface creates a flux of 0.

In fact, we can generalize this to multiple charges. The total flux through an electric field is:

$$\Phi = \int_S \vec{E} \cdot d\vec{a} = \int_S (E_1 + E_2 + ... + E_N) \cdot d\vec{a}.$$

We just learned that every charge q inside the surface has a total flux of $\frac{q_i}{\epsilon_0}$, and every charge outside of it has a total flux of 0. We can just sum $\frac{q_i}{\epsilon}$ for every single charge inside the surface.

Therefore, the total flux through a surface can be represented very elegantly through Gauss’s law:

Note

Gauss’s law

$$\Phi_{total} = \int \vec{E} \cdot d\vec {A} = \frac{1}{\epsilon_0} \sum_i q_i = \frac{1}{\epsilon_0} \int \rho dv.$$