Electric Charges & Fields – Advanced Study Notes
Audience: MIT‑level physics students (first year graduate).
Focus: A rigorous, first‑principles treatment of electrostatics, with deeper theory, edge cases, and a challenging problem.
1. Introduction
Electrostatics is the branch of classical electrodynamics dealing with stationary electric charges. It is the foundation for all of electromagnetism because the static field is the “instantaneous” limit of the full Maxwell equations. The key ideas are:
| Concept | Symbol | Physical Meaning |
|---|---|---|
| Charge | (q) | Source of the field |
| Electric field | (\mathbf{E}) | Force per unit charge |
| Coulomb’s law | (F = k_e \dfrac{q_1 q_2}{r^2}) | Fundamental force law |
| Gauss’s law | (\displaystyle \oint_{\partial V}\mathbf{E}!\cdot d\mathbf{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}) | Flux–charge relation |
| Potential | (\Phi) | Energy per unit charge (scalar) |
| Energy density | (u = \dfrac{1}{2}\varepsilon_0 | \mathbf{E} |
We will derive these from first principles, analyze edge cases, and extend the theory to multipole moments and boundary‑value problems.
2. Coulomb’s Law from First Principles
2.1 Empirical Origin
Coulomb’s 1785 experiment measured the force between two small charged spheres. The force obeyed:
[ F = k_e \frac{q_1 q_2}{r^2}, ]
where (k_e = \frac{1}{4\pi\varepsilon_0}).
2.2 Symmetry Argument
Let’s derive the (1/r^2) dependence from symmetry and dimensional analysis.
- Isotropy – The field depends only on (r = |\mathbf{r}|), not on direction.
- Scale invariance – If we scale all distances by (\lambda), the field scales as (\lambda^{-2}).
- Dimensionality – The only constant with dimensions of ([F]/[q^2] r^2) is (k_e).
Hence the unique form consistent with these principles is Coulomb’s law.
2.3 Vector Form
Define the vector from the source to the field point:
[ \mathbf{r} = \mathbf{r}{\text{field}} - \mathbf{r}{\text{source}}, \qquad r = |\mathbf{r}|. ]
The electric field due to a point charge (q) is
[ \boxed{\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2}\hat{\mathbf{r}} = \frac{q}{4\pi\varepsilon_0}\frac{\mathbf{r}}{r^3}} ]
with (\hat{\mathbf{r}}=\mathbf{r}/r).
3. Definition of Electric Field
The electric field at a point (\mathbf{r}) is defined as the force per unit test charge (q_t) placed there:
[ \mathbf{E}(\mathbf{r}) \equiv \lim_{q_t\to0}\frac{\mathbf{F}_{\text{test}}}{q_t}. ]
Because the test charge is infinitesimal, it does not disturb the source distribution. The field is a vector field satisfying:
- Linearity: (\mathbf{E}(\mathbf{r}) = \sum_i \mathbf{E}_i(\mathbf{r})) (superposition).
- Time Independence: For static sources, (\partial \mathbf{E}/\partial t = 0).
4. Maxwell’s Equations in Electrostatics
For static fields:
- Gauss’s law (differential form)
[ \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} ] - Faraday’s law (static)
[ \nabla \times \mathbf{E} = \mathbf{0} ]
The second implies that (\mathbf{E}) can be expressed as the gradient of a scalar potential (\Phi):
[ \mathbf{E} = -\nabla \Phi. ]
Combining with Gauss we obtain Poisson’s equation:
[ \nabla^2 \Phi = -\frac{\rho}{\varepsilon_0}. ]
5. Gauss’s Law – Derivation & Rigorous Reasoning
5.1 Integral Form
Consider a closed surface (S). The flux of (\mathbf{E}) through (S) is
[ \Phi_E = \oint_S \mathbf{E}\cdot d\mathbf{A}. ]
For a point charge (q) inside (S), we compute the flux by symmetry:
[ \Phi_E = E(4\pi r^2) = \frac{q}{\varepsilon_0}. ]
If no charge is inside (S), (\Phi_E=0).
5.2 From Coulomb to Gauss
The divergence theorem gives
[ \oint_S \mathbf{E}\cdot d\mathbf{A} = \int_V \nabla \cdot \