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Chapter 3

Electric Potential

Scalar potential V, Poisson's and Laplace's equations, energy, and multipole expansion.

3.1 The Scalar Potential

Since $\nabla \times \mathbf{E} = 0$ in electrostatics, the field is conservative and we can write $\mathbf{E} = -\nabla V$ for a scalar function $V(\mathbf{r})$called the electric potential (or voltage):

$$\boxed{\mathbf{E} = -\nabla V}, \qquad V(\mathbf{r}) = -\int_{\mathcal{O}}^{\mathbf{r}} \mathbf{E} \cdot d\boldsymbol{\ell}$$

The reference point $\mathcal{O}$ is arbitrary (usually taken at infinity). For a point charge $q$ at the origin:

$$V(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}$$

For a collection of charges, superposition applies directly to the scalar potential (much simpler than adding vectors!):

$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\sum_i \frac{q_i}{|\mathbf{r} - \mathbf{r}_i|} = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\mathbf{r}')}{|\mathbf{r} - \mathbf{r}'|}\,d\tau'$$

3.2 Poisson's & Laplace's Equations

Substituting $\mathbf{E} = -\nabla V$ into Gauss's law $\nabla \cdot \mathbf{E} = \rho/\epsilon_0$:

$$\boxed{\nabla^2 V = -\frac{\rho}{\epsilon_0}} \qquad \text{(Poisson's equation)}$$

In source-free regions ($\rho = 0$): $\nabla^2 V = 0$ (Laplace's equation)

3.2.1 Uniqueness Theorems

  • First uniqueness theorem: The solution to Laplace's equation in a volume is uniquely determined by the boundary conditions (values of $V$ on the boundary).
  • Second uniqueness theorem: The electric field in a volume is uniquely determined if the charge distribution inside and the potential on the boundary are specified.

3.2.2 Separation of Variables (Spherical Coordinates)

For problems with azimuthal symmetry, Laplace's equation in spherical coordinates admits solutions of the form $V(r, \theta) = R(r)\Theta(\theta)$. The solutions are:

$$V(r, \theta) = \sum_{\ell=0}^{\infty} \left(A_\ell r^\ell + \frac{B_\ell}{r^{\ell+1}}\right) P_\ell(\cos\theta)$$

where $P_\ell$ are the Legendre polynomials: $P_0 = 1$, $P_1 = \cos\theta$,$P_2 = \frac{1}{2}(3\cos^2\theta - 1)$, etc. Coefficients are determined by boundary conditions.

3.3 Energy in the Electric Field

The work needed to assemble a continuous charge distribution against its own Coulomb repulsion is stored as energy in the electric field:

$$W = \frac{\epsilon_0}{2}\int_{\rm all\,space} E^2\,d\tau = \frac{1}{2}\int \rho V\,d\tau$$

This defines the energy density of the electric field:

$$u_E = \frac{\epsilon_0}{2} E^2 = \frac{\epsilon_0}{2}|\nabla V|^2 \qquad [\text{J/m}^3]$$

3.3.1 Multipole Expansion

For a localized charge distribution, the potential at large distances can be expanded in powers of $1/r$:

$$V(\mathbf{r}) = \frac{1}{4\pi\epsilon_0}\left[\frac{Q}{r} + \frac{\mathbf{p}\cdot\hat{r}}{r^2} + \frac{1}{2}\sum_{i,j} Q_{ij}\frac{\hat{r}_i\hat{r}_j}{r^3} + \cdots\right]$$

Monopole

$$Q = \int \rho\,d\tau$$

Falls off as $1/r$

Dipole

$$\mathbf{p} = \int \mathbf{r}'\rho\,d\tau'$$

Falls off as $1/r^2$

Quadrupole

$$Q_{ij} = \int (3r'_i r'_j - r'^2 \delta_{ij})\rho\,d\tau'$$

Falls off as $1/r^3$

Simulation: Quadrupole Potential

Computes and visualizes the electric potential and field of a quadrupole charge distribution, and verifies $\mathbf{E} = -\nabla V$ numerically.

Electric Potential & Quadrupole

Visualizes V and E for a quadrupole, and verifies E = -∇V along the x-axis.

Click Run to execute the Python code

First run will download Python environment (~15MB)

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