📂Electrodynamics

Coulomb's Law and Electric Fields

Coulomb’s Law¹

The force exerted on a test charge $Q$ placed at a distance $\cR$ away from a fixed point charge $q$ is known as the Coulomb force, and its equation is as follows.

$$ \mathbf{F} = \dfrac{1}{4\pi \epsilon_{0}} \dfrac{qQ}{\cR ^2} \crH $$

This is referred to as Coulomb’s Law.

Description

Coulomb’s Law is an empirical law derived from repeated experiments. Therefore, it cannot be mathematically proven. It is easier to understand if one thinks of it as analogous to axioms in mathematics. $\epsilon_{0}$ represents the permittivity of free space, and its value is $8.85 \times 10^{-12} \dfrac{\mathrm C^2}{\mathrm N \cdot \mathrm m^2}$. On the other hand, the formula at the top of the text is expressed in International System of Units. If it is represented in Gaussian system, it looks like this:

$$ \mathbf{F} = \dfrac{qQ}{\cR ^2} \crH $$

This replaces the constant of proportionality in the International System of Units with $1$. That is to say $\dfrac{1}{4\pi\epsilon_{0}} \equiv 1$. In other words, the easy way to convert SI units to Gaussian units is to replace $\epsilon_{0}$ with $\dfrac{1}{4\pi}$.

Electric Field

Point Charge Distribution

Let’s now imagine that there are several point charges around the test charge $Q$. Then, the force exerted on $Q$ can simply be added linearly from each of the point charges. This means that the interaction between $Q$ and $q_{1}$ is not affected by $q_{2}, q_{3}, \dots$. This is called the Superposition Principle.

$$ \begin{align*} \mathbf{F} &= F_{1}+F_{2}+\cdots + F_{n} \\ &= \dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}Q}{{\cR_{1}}^2}\crH_{1} +\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{2}Q}{{\cR_{2}}^2}\crH_{2}+\cdots +\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{n}Q}{{\cR_{n}}^2}\crH_{n} \\ &= Q\left( \dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{1}}{{\cR_{1}}^2}\crH_{1} +\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{2}}{{\cR_{2}}^2}\crH_{2}+\cdots +\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q_{n}}{{\cR_{n}}^2}\crH_{n} \right) \\ &= Q\mathbf{E} \end{align*} $$

Here, the portion within the parentheses is defined as the electric field created by the source charges $q_{1},\ q_{2},\ \cdots ,\ q_{n}$, and it is denoted as $\mathbf{E}$.

$$ \mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \sum \limits_{i=1}^n \dfrac{q_{i}}{{\cR_{i}}^2}\crH_{i} $$

Continuous Charge Distribution

When the charge is continuously distributed, it is expressed with an integral instead of a sum.

$$ \sum \rightarrow \int \\ \mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \int \dfrac{1}{\cR^2}\crH dq $$

In the case of line charge, $dq=\lambda dl^{\prime}$. Here, $\lambda$ is the linear charge density. The electric field created by a line charge is as follows.

$$ \mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \int _\mathcal{P} \dfrac{\lambda (\mathbf{r}^{\prime})}{\cR^2} \crH dl^{\prime} $$

In the case of surface charge, $dq=\sigma da^{\prime}$. Here, $\sigma$ is the surface charge density. The electric field created by a surface charge is as follows.

$$ \mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \int _\mathcal{S} \dfrac{\sigma (\mathbf{r}^{\prime})}{\cR^2} \crH da^{\prime} $$

In the case of volume charge, $dq=\rho d\tau^{\prime}$. Here, $\rho$ is the volume charge density. The electric field created by a volume charge is as follows.

$$ \mathbf{E}(\mathbf {r}) =\dfrac{1}{4\pi \epsilon_{0}} \int _\mathcal{V} \dfrac{\rho (\mathbf{r}^{\prime})}{\cR^2} \crH d\tau^{\prime} $$