# 0x111 Electrodynamics

## 1. Electrostatics

Electrostatics applies to the case when the source charges are stationary (though the test charge maybe moving).

The primary task here is to find the electric field of a given stationary charge distribution.

Law (Coulomb's law) The force on a test Charge $$Q$$ due to a single charge $$q$$ is given by

$\mathbf{F} = \frac{1}{4 \pi \epsilon_0} \frac{qQ}{r^2} \mathbf{\hat{r}}$

where the constant $$\epsilon_0$$ is called the permittivity of free space.

$\epsilon_0 =8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}$

Law (principle of superposition), the total force equals the sum of individual force from every charge

$F = \int_i F_i$

Everything in electrostatic can be explained by the Coulomb's law and principle of superposition, everything else is just to assemble tricks to deal with integral

### 1.1. Electric Field

Defintion (Electric Field) If there are several point charges $$q_1, q_2, ..., q_n$$ at $$r_1, ..., r_n$$, the total force on $$Q$$ is

$F=QE$

and $$E$$ is the elctric field of the source charge.

$E(r) = \frac{1}{4\pi \epsilon_0} \sum_{i=1}^{n} \frac{q_i}{r_i^2} \mathbf{\hat{r}_i}$

When the charge are distributed continuously, it can be written in the integral form by

$E(r) = \frac{1}{4\pi \epsilon_0} \int \frac{\rho(\mathbf{\hat{r}})}{r^2} \mathbf{\hat{r}_i} dq$

Law (Gauss's law) The flux through a enclosing surface is proportional to the charge inside, which is the integral form====================== of Gauss's law

$\oint \mathbf{E} \cdot d \mathbf{a} = \frac{Q}{\epsilon_0}$

By applying the divergence theorem to the left side:

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

and rewriting the right side with the charge density $$\rho$$

$Q = \int_\mathcal{V} \rho d\tau$

which leads to the differential form of Gauss's law:

$\nabla \cdot \mathbf{E} = \frac{1}{\epsilon_0}\rho$

### 1.2. Electric Potential

As $$\int_a^b \mathbf{E} \cdot d\mathbf{l}$$ only depends on $$a, b$$. the integral around a closed path is 0

$\oint \mathbf{E} \cdot d\mathbf{l} = 0$

With Stokes' theorem, we know

$\nabla \times \mathbf{E} = 0$

The potential formulation takes advantage of this constraint, it reduces the vector problem $$\mathbf{E}$$ into the scalar problem $$V$$

Definition (electric potential) We define the following function as the elctric potential, which only depends on the point $$\mathbf{r}$$

$V(\mathbf{r}) = - \int^{\mathbf{r}}_{\mathcal{O}} \mathbf{E} \cdot d\mathbf{l}$

By the fundamental theorem, it is easy to see

$\mathbf{E} = - \nabla V$

## 2. Reference

• [1] David J. Griffiths Introduction to Electrodnyamics