Electrostatics | Physics - Wyatt's Notes

2.1 Coulomb’s Law and the Electric Field

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.

2.2 Gauss’s Law Applications

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 $r$).

2.3 Electric Potential

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.

2.4 Poisson’s and Laplace’s Equations

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).

2.5 Worked Example

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

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

$\blacksquare$

2.6 Gauss’s Law: Cylindrical Symmetry

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:

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

2.7 The Uniqueness Theorem

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. $\blacksquare$

2.8 Method of Images

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.

2.9 Multipole Expansion

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.

2.10 Dielectrics

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 $\varepsilon_1 \tan\theta_2 = \varepsilon_2 \tan\theta_1$.

flowchart TD
    A[2_Electrostaticsx] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Intuition

Electrostatics studies how stationary charges create electric fields and potentials. Gauss’s law is a bookkeeping tool: the total flux through any closed surface equals the enclosed charge divided by the permittivity, regardless of surface shape. This makes symmetric problems trivial to solve. The potential is like a topographic map of electrical height, where charges sit on hills. The uniqueness theorem guarantees that any solution satisfying the boundary conditions is the only solution, so clever guesses like the method of images are valid. Dielectrics respond to external fields by polarising, creating bound charges that partially cancel the applied field.

Common Mistakes

Mistake 1: Using Gauss’s law when the symmetry is insufficient Gauss’s law is always true, but it is only useful for computing $\mathbf{E}$ when the charge distribution has sufficient symmetry (spherical, cylindrical, or planar) to pull $E$ out of the integral. Applying it to asymmetric distributions without the symmetry argument gives an integral equation that cannot be solved analytically.

Mistake 2: Confusing the potential of a point charge with the potential energy The electric potential $V = q/(4\pi\varepsilon_0 r)$ is the potential energy per unit charge, not the total energy. To find the energy of a system of charges, you must compute $U = \frac{1}{2}\sum_i q_i V_i$ or equivalently $U = \frac{1}{2}\int \rho V\, d^3\mathbf{r}$. The factor of $\frac{1}{2}$ avoids double-counting.

Mistake 3: Assuming the electric field is continuous across a charged surface The normal component of $\mathbf{E}$ is discontinuous across a surface charge density $\sigma$: $E_{2n} - E_{1n} = \sigma/\varepsilon_0$. Only the tangential component $E_t$ is continuous. Students who assume full continuity obtain incorrect boundary conditions for dielectric interfaces.

Cross-References

• Maxwell’s Equations: Electrostatics provides the static electric field solutions that form two of Maxwell’s four equations.
• Potentials and Gauge Transformations: The electric potential introduced here is the scalar potential used in the general gauge theory framework.
• Electrodynamics: Electrostatics is the zero-velocity limit of the full electrodynamics theory developed in the next chapter.

References

1. Griffiths, D. J. (2017). Introduction to Electrodynamics (4th ed.). Cambridge University Press.
2. Purcell, E. M. & Morin, D. J. (2013). Electricity and Magnetism (3rd ed.). Cambridge University Press.
3. Jackson, J. D. (1998). Classical Electrodynamics (3rd ed.). John Wiley & Sons.
4. Zangwill, A. I. (2013). Modern Electrodynamics. Cambridge University Press.
5. Feynman, R. P., Leighton, R. B. & Sands, M. (2011). The Feynman Lectures on Physics, Vol. II: The New Millennium Edition — Mainly Electromagnetism and Matter. Basic Books.