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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


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