Electrostatics: Coulomb's Law & Electric Field
Electrostatics: Coulomb's Law & Electric Field
Electrostatics studies the forces and fields produced by stationary electric charges. Coulomb's law quantifies the force between point charges and forms the foundation of classical electromagnetism.
Definition
The electric field \(\mathbf{E}\) at position \(\mathbf{r}\) due to a point charge \(q\) at the origin is \(\mathbf{E} = \frac{q}{4\pi\epsilon_0 r^2}\hat{r}\).
Key Result
Gauss's law states \(\oint \mathbf{E}\cdot d\mathbf{A} = Q_{enc}/\epsilon_0\), making symmetric charge distributions tractable.
Example 1
A sphere of radius \(R\) with uniform charge density \(\rho\): outside \(E = Q/(4\pi\epsilon_0 r^2)\); inside \(E = \rho r/(3\epsilon_0)\).
Example 2
An infinite line charge with linear density \(\lambda\) has field \(E = \lambda/(2\pi\epsilon_0 r)\) directed radially.
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Practice
- Use Gauss's law to find the field inside a uniformly charged sphere.
- Compute the potential due to an electric dipole at large distance.
- Why is the electric field zero inside a conductor?
- Find the capacitance of a parallel-plate capacitor with dielectric \(\epsilon_r\).
Show Answer Key
1. By Gauss's law with spherical symmetry: $\oint\mathbf{E}\cdot d\mathbf{A} = Q_{\text{enc}}/\epsilon_0$. For $r 2. Dipole: charges $+q$ and $-q$ separated by $d$, dipole moment $\mathbf{p}=q\mathbf{d}$. At large $r\gg d$: $V(r,\theta) = \frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$. Field components: $E_r = \frac{2p\cos\theta}{4\pi\epsilon_0 r^3}$, $E_\theta = \frac{p\sin\theta}{4\pi\epsilon_0 r^3}$. Falls off as $1/r^3$. 3. Inside a conductor in electrostatic equilibrium, $\mathbf{E}=0$. If $\mathbf{E}\neq 0$, free charges would move (current flows) until equilibrium is restored. All excess charge resides on the surface, and the interior is an equipotential. 4. Parallel plates area $A$, separation $d$, dielectric $\epsilon_r$: $C = \epsilon_r\epsilon_0 A/d$. The dielectric reduces the effective field by factor $\epsilon_r$ (polarization opposes the applied field), increasing capacitance by factor $\epsilon_r$ compared to vacuum.