Electrostatics

Coulomb’s Law: The force between two point charges $q_1$ and $q_2$ separated by distance $r$:

$\mathbf{F} = \frac{1}{4\pi\varepsilon_0} \frac{q_1 q_2}{r^2} \hat{\mathbf{r}}$

The electric field due to a point charge $q$ at position $\mathbf{r}$:

$\mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{q}{|\mathbf{r}|^2} \hat{\mathbf{r}}$

Superposition Principle: The field due to a collection of charges is the vector sum of individual Fields.

Example: Infinite plane of charge with surface charge density $\sigma$.

Choose a Gaussian “pillbox” of area $A$ straddling the plane. By symmetry, $\mathbf{E}$ is Perpendicular to the plane. Gauss’s law:

$2EA = \frac{\sigma A}{\varepsilon_0} \implies E = \frac{\sigma}{2\varepsilon_0}$

The field is uniform and perpendicular to the plane, pointing away from positive charge.

Example: Uniformly charged sphere of radius $R$ with total charge $Q$.

For $r \gt R$: $\mathbf{E} = \frac{Q}{4\pi\varepsilon_0 r^2} \hat{\mathbf{r}}$ (identical to a point charge).

For $r \lt R$: $E = \frac{Qr}{4\pi\varepsilon_0 R^3}$ (linear in $r$).

The electric potential is defined by $\mathbf{E} = -\nabla V$ (for electrostatics, where $\nabla \times \mathbf{E} = \mathbf{0}$).

For a point charge: $V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$ (choosing $V(\infty) = 0$).

Theorem 2.1. $\nabla \times \mathbf{E} = \mathbf{0}$ in electrostatics implies $\mathbf{E}$ is Conservative, so the line integral $\int_A^B \mathbf{E} \cdot d\mathbf{l} = V(A) - V(B)$ is Path-independent.

Substituting $\mathbf{E} = -\nabla V$ into Gauss’s law:

$\nabla \cdot (-\nabla V) = -\nabla^2 V = \frac{\rho}{\varepsilon_0}$

This is Poisson’s equation:

$\nabla^2 V = -\frac{\rho}{\varepsilon_0}$

In regions with $\rho = 0$This reduces to Laplace’s equation:

$\nabla^2 V = 0$

Theorem 2.2 (Uniqueness --- statement). The solution to Laplace’s (or Poisson’s) equation in a Region is unique given either Dirichlet boundary conditions ($V$ specified on the boundary) or Neumann boundary conditions ($\partial V / \partial n$ specified on the boundary).

Problem. Two infinite conducting plates at $x = 0$ and $x = d$ are held at potentials $V = 0$ and $V = V_0$ respectively. Find the potential and field between them.

Solution. Between the plates, $\rho = 0$So $\nabla^2 V = 0$. By symmetry, $V$ depends only on $x$:

$\frac{d^2V}{dx^2} = 0 \implies V(x) = Ax + B$

Boundary conditions: $V(0) = 0 \implies B = 0$. $V(d) = V_0 \implies A = V_0/d$.

$V(x) = \frac{V_0}{d} x, \quad \mathbf{E} = -\frac{dV}{dx}\hat{\mathbf{x}} = -\frac{V_0}{d}\hat{\mathbf{x}}$

$\blacksquare$

Example: Infinite line charge with linear charge density $\lambda$.

By cylindrical symmetry, $\mathbf{E}$ points radially outward and depends only on $r$. Choose a Gaussian cylinder of radius $r$ and length $L$:

$\oint \mathbf{E} \cdot d\mathbf{A} = E \cdot 2\pi r L = \frac{\lambda L}{\varepsilon_0}$

$\mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{\mathbf{r}}$

Example: Coaxial cable. An inner conductor of radius $a$ carries linear charge density $+\lambda$And an outer conducting shell of radius $b$ carries $-\lambda$.

For $r \lt a$: $\mathbf{E} = \mathbf{0}$ (conductor interior).

For $a \lt r \lt b$: $\mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\,\hat{\mathbf{r}}$.

For $r \gt b$: $\mathbf{E} = \mathbf{0}$ (total enclosed charge is zero).

The potential difference between the conductors:

$V(a) - V(b) = -\int_a^b \mathbf{E} \cdot d\mathbf{l} = \frac{\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{b}{a}\right)$

Theorem 2.3 (Uniqueness for Dirichlet conditions). The solution to Poisson’s equation $\nabla^2 V = -\rho/\varepsilon_0$ in a volume $\mathcal{V}$ is unique if $V$ is specified on the Boundary $\mathcal{S}$.

Proof. Suppose $V_1$ and $V_2$ both satisfy Poisson’s equation with the same boundary Conditions. Define $U = V_1 - V_2$. Then $\nabla^2 U = 0$ in $\mathcal{V}$ and $U = 0$ on $\mathcal{S}$.

Apply Green’s first identity with $\phi = \psi = U$:

$\int_{\mathcal{V}} \left(U\,\nabla^2 U + \lvert\nabla U\rvert^2\right) dV = \oint_{\mathcal{S}} U\,\frac{\partial U}{\partial n}\, dA$

Since $\nabla^2 U = 0$ and $U = 0$ on $\mathcal{S}$:

$\int_{\mathcal{V}} \lvert\nabla U\rvert^2\, dV = 0$

Since the integrand is non-negative, $\nabla U = \mathbf{0}$ everywhere in $\mathcal{V}$So $U$ is Constant. With $U = 0$ on the boundary, $U = 0$ throughout $\mathcal{V}$. Hence $V_1 = V_2$. $\blacksquare$

Theorem 2.4 (Uniqueness for Neumann conditions). The solution is unique up to an additive Constant when $\partial V/\partial n$ is specified on $\mathcal{S}$.

Proof. The same argument applies, but now $\partial U/\partial n = 0$ on $\mathcal{S}$ and the Right-hand side of Green’s identity vanishes for a different reason. We again conclude $\nabla U = \mathbf{0}$So $U$ is constant. $\blacksquare$

The method of images replaces a problem with conductors by an equivalent problem with charges only, Exploiting the uniqueness theorem.

Point charge above a grounded plane. A charge $q$ is placed at distance $d$ above an Infinite grounded conducting plane ($V = 0$ at $z = 0$).

Replace the plane by an image charge $q' = -q$ at $z = -d$. The potential for $z \gt 0$ is:

$V(x,y,z) = \frac{1}{4\pi\varepsilon_0}\left[\frac{q}{\sqrt{x^2 + y^2 + (z-d)^2}} - \frac{q}{\sqrt{x^2 + y^2 + (z+d)^2}}\right]$

This satisfies $\nabla^2 V = 0$ for $z \gt 0$ (away from the charge), $V = 0$ at $z = 0$And $V \to 0$ as $r \to \infty$. By the uniqueness theorem, this is the correct solution.

The force on $q$ is the force due to the image charge:

$\mathbf{F} = -\frac{q^2}{4\pi\varepsilon_0 (2d)^2}\,\hat{\mathbf{z}}$

The induced surface charge density on the plane:

$\sigma(x,y) = -\varepsilon_0 \left.\frac{\partial V}{\partial z}\right|_{z=0} = -\frac{qd}{2\pi(x^2+y^2+d^2)^{3/2}}$

Example: Point charge inside a grounded sphere. A charge $q$ is at distance $a$ from the centre Of a grounded conducting sphere of radius $R$ ($a \lt R$).

The image charge is $q' = -qR/a$ located at distance $b = R^2/a$ from the centre, along the same Radial line.

For a localized charge distribution $\rho(\mathbf{r}')$The potential at large distance $r = \lvert\mathbf{r}\rvert \gg r' = \lvert\mathbf{r}'\rvert$ is expanded using $\frac{1}{\lvert\mathbf{r}-\mathbf{r}'\rvert} = \sum_{n=0}^{\infty} \frac{r'^n}{r^{n+1}} P_n(\cos\alpha)$ Where $\alpha$ is the angle between $\mathbf{r}$ and $\mathbf{r}'$:

$V(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_{n=0}^{\infty}\frac{1}{r^{n+1}}\int r'^n P_n(\cos\alpha)\,\rho(\mathbf{r}')\,d^3\mathbf{r}'$

Monopole term ($n = 0$):

$V_0 = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}, \quad Q = \int \rho(\mathbf{r}')\,d^3\mathbf{r}'$

This is the potential of a point charge at the origin.

Dipole term ($n = 1$):

$V_1 = \frac{1}{4\pi\varepsilon_0}\frac{\mathbf{p} \cdot \hat{\mathbf{r}}}{r^2}, \quad \mathbf{p} = \int \mathbf{r}'\,\rho(\mathbf{r}')\,d^3\mathbf{r}'$

Where $\mathbf{p}$ is the electric dipole moment.

Quadrupole term ($n = 2$): Depends on the quadrupole moment tensor:

$Q_{ij} = \int (3r_i' r_j' - r'^2 \delta_{ij})\,\rho(\mathbf{r}')\,d^3\mathbf{r}'$

$V_2 = \frac{1}{4\pi\varepsilon_0}\frac{1}{2r^3}\sum_{i,j} Q_{ij}\,\hat{r}_i\,\hat{r}_j$

For a neutral charge distribution ($Q = 0$), the dipole term dominates. If additionally $\mathbf{p} = \mathbf{0}$The quadrupole term dominates.

Polarization. When an external field $\mathbf{E}$ is applied to a dielectric, the material Develops a polarization $\mathbf{P}$The dipole moment per unit volume. This produces bound charges:

$\rho_b = -\nabla \cdot \mathbf{P}, \quad \sigma_b = \mathbf{P} \cdot \hat{\mathbf{n}}$

The displacement field $\mathbf{D}$ is defined as:

$\mathbf{D} = \varepsilon_0 \mathbf{E} + \mathbf{P}$

Gauss’s law in terms of $\mathbf{D}$:

$\nabla \cdot \mathbf{D} = \rho_f$

Where $\rho_f$ is the free charge density. This form is useful because $\mathbf{D}$ depends Only on free charges, not bound charges.

Linear dielectrics. For an isotropic linear dielectric:

$\mathbf{P} = \varepsilon_0 \chi_e \mathbf{E}, \quad \mathbf{D} = \varepsilon \mathbf{E}$

Where $\chi_e$ is the electric susceptibility and $\varepsilon = \varepsilon_0(1 + \chi_e)$ is the Permittivity. The relative permittivity (dielectric constant) is $\varepsilon_r = \varepsilon/\varepsilon_0 = 1 + \chi_e$.

Boundary conditions at dielectric interfaces (no free charges):

$D_{1n} = D_{2n} \implies \varepsilon_1 E_{1n} = \varepsilon_2 E_{2n}$

$E_{1t} = E_{2t}$

The tangential component of $\mathbf{E}$ is continuous, but the normal component changes. The angles of the field with respect to the normal satisfy $\varepsilon_1 \tan\theta_2 = \varepsilon_2 \tan\theta_1$.