6.1 Scalar potential

Scalar potential describes the potential energy per unit charge at each point in an electric field. The electric field itself can be recovered from the scalar potential through differentiation, making $\phi$ one of the most powerful tools for solving electrostatic (and, later, electrodynamic) problems. This topic ties together Gauss's law, Laplace's and Poisson's equations, boundary conditions, and multipole expansions.

Definition of scalar potential

The scalar potential $\phi$ is a scalar field assigning a single number to every point in space. Physically, it equals the work per unit charge needed to bring a positive test charge from infinity (where $\phi = 0$ by convention) to that point against the electric field. Its SI unit is the volt (V), with 1 V = 1 J/C.

Because $\phi$ is defined through a path integral of a conservative field, only potential differences carry direct physical meaning. The absolute value of $\phi$ depends on the chosen reference point.

Relationship between electric field and scalar potential

The electric field is the negative gradient of the scalar potential:

$\vec{E} = -\nabla \phi$

• The negative sign means $\vec{E}$ points from high potential toward low potential.
• Where $\phi$ changes rapidly over a short distance, $|\vec{E}|$ is large; where $\phi$ is nearly flat, the field is weak.
• This relationship holds exactly in electrostatics. In time-varying situations an additional term involving the vector potential appears (covered below).

Calculation of scalar potential from electric field

Line integral method

The potential difference between points $a$ and $b$ is:

$\phi(b) - \phi(a) = -\int_a^b \vec{E} \cdot d\vec{l}$

Because the electrostatic field is conservative ($\nabla \times \vec{E} = 0$), this integral is path-independent: any path from $a$ to $b$ gives the same result. That path-independence is what makes $\phi$ a well-defined function in the first place.

Gradient operator

When $\phi$ is known analytically, you recover the field by computing the gradient. In Cartesian coordinates:

$\nabla \phi = \hat{x}\frac{\partial \phi}{\partial x} + \hat{y}\frac{\partial \phi}{\partial y} + \hat{z}\frac{\partial \phi}{\partial z}$

The gradient vector points in the direction of the steepest increase of $\phi$, so $\vec{E} = -\nabla\phi$ points in the direction of steepest decrease.

Electric potential energy

Definition and formula

Electric potential energy $U$ is the energy stored in a configuration of charges due to their mutual interactions. For a single charge $q$ sitting at a location where the potential is $\phi$:

$U = q\phi$

Relationship to work and scalar potential

When a charge $q$ moves from a point at potential $\phi_1$ to a point at potential $\phi_2$, the work done by the electric field is:

$W = -\Delta U = -q(\phi_2 - \phi_1) = q(\phi_1 - \phi_2)$

This depends only on the endpoints, not the path. The scalar potential is therefore the potential energy per unit charge: $\phi = U/q$.

Equipotential surfaces

Definition and properties

An equipotential surface is a locus of points sharing the same value of $\phi$. Key properties:

• $\vec{E}$ is everywhere perpendicular to the equipotential surface. If it weren't, there would be a component of $\vec{E}$ along the surface, implying a potential difference between neighboring points on it.
• No work is done when a charge moves along an equipotential surface, since $\Delta\phi = 0$.

Relationship to electric field lines

Electric field lines cross equipotential surfaces at right angles. Where equipotential surfaces are closely spaced, $\phi$ is changing rapidly and $|\vec{E}|$ is large. Where they are widely spaced, the field is weak. This gives you a quick visual way to read field strength from a potential map.

Boundary conditions for scalar potential

Conductor surfaces

At electrostatic equilibrium:

• The electric field inside a conductor is zero.
• The potential is constant throughout the conductor's volume and on its surface (the entire conductor is one equipotential).
• Just outside the surface, $\vec{E}$ is perpendicular to the surface with magnitude $\sigma/\epsilon_0$, where $\sigma$ is the local surface charge density.

Dielectric interfaces

At the boundary between two linear dielectrics with permittivities $\epsilon_1$ and $\epsilon_2$:

• Normal component: $\epsilon_1 E_{1n} = \epsilon_2 E_{2n}$ (assuming no free surface charge).
• Tangential component: $E_{1t} = E_{2t}$ (continuous across the interface).
• Potential: $\phi$ itself is continuous across the interface. A discontinuity in $\phi$ would imply an infinite electric field at the boundary, which is unphysical.

Laplace's equation for scalar potential

Derivation in free space

Start with Gauss's law in a charge-free region: $\nabla \cdot \vec{E} = 0$. Substitute $\vec{E} = -\nabla\phi$:

$\nabla \cdot (-\nabla\phi) = 0 \quad\Longrightarrow\quad \nabla^2\phi = 0$

This is Laplace's equation. Any solution to it is called a harmonic function. A key property: $\phi$ at any point equals the average of $\phi$ over any sphere centered on that point (the mean-value theorem), which means $\phi$ can have no local maxima or minima in a charge-free region.

Solutions in different coordinate systems

The explicit form of $\nabla^2\phi = 0$ depends on the coordinate system you choose. Pick the one that matches the symmetry of your problem.

• Cartesian:

$\frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2} = 0$

• Cylindrical $(s, \varphi, z)$:

$\frac{1}{s}\frac{\partial}{\partial s}\!\left(s\frac{\partial \phi}{\partial s}\right) + \frac{1}{s^2}\frac{\partial^2 \phi}{\partial \varphi^2} + \frac{\partial^2 \phi}{\partial z^2} = 0$

• Spherical $(r, \theta, \varphi)$:

$\frac{1}{r^2}\frac{\partial}{\partial r}\!\left(r^2\frac{\partial \phi}{\partial r}\right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial \theta}\!\left(\sin\theta\frac{\partial \phi}{\partial \theta}\right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 \phi}{\partial \varphi^2} = 0$

Separation of variables in each system yields the standard families of solutions (sines/cosines/exponentials in Cartesian, Bessel functions in cylindrical, Legendre polynomials and spherical harmonics in spherical).

Poisson's equation for scalar potential

Derivation with source charges

When charge density $\rho$ is present, Gauss's law reads $\nabla \cdot \vec{E} = \rho/\epsilon_0$. Substituting $\vec{E} = -\nabla\phi$ gives Poisson's equation:

$\nabla^2\phi = -\frac{\rho}{\epsilon_0}$

Laplace's equation is the special case $\rho = 0$. Poisson's equation is the bridge between a known charge distribution and the resulting potential.

Green's function method for solutions

The formal solution to Poisson's equation uses a Green's function $G(\vec{r},\vec{r}')$, defined as the potential produced by a unit point source at $\vec{r}'$:

$\nabla^2 G(\vec{r},\vec{r}') = -\delta^3(\vec{r}-\vec{r}')$

The potential for an arbitrary charge distribution is then built by superposition:

$\phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^3r'$

Here the free-space Green's function $G(\vec{r},\vec{r}') = \frac{1}{|\vec{r}-\vec{r}'|}$ has been used. When boundaries are present, the Green's function must be modified to satisfy the appropriate boundary conditions, which is what makes the method both powerful and nontrivial.

Multipole expansion of scalar potential

Monopole, dipole, and quadrupole terms

For a localized charge distribution observed at a field point far from the source, you can expand $\phi$ in powers of $1/r$:

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

• Monopole ($\sim 1/r$): depends on the total charge $Q$.
• Dipole ($\sim 1/r^2$): depends on the dipole moment $\vec{p} = \int \vec{r}'\,\rho(\vec{r}')\,d^3r'$.
• Quadrupole ($\sim 1/r^3$): depends on the traceless quadrupole moment tensor $Q_{ij}$. It does not require literally four charges in a square; any distribution with a nonzero quadrupole moment contributes.

Each successive term falls off faster with distance, so at large $r$ only the lowest nonvanishing term matters.

Far-field approximations

• If the distribution has a net charge, the monopole term dominates at large $r$ and the distribution looks like a point charge.
• For a neutral distribution ($Q = 0$), the dipole term is the leading contribution.
• If the dipole moment also vanishes, the quadrupole term takes over.

These approximations are extremely useful: they let you replace a complicated charge distribution with a few numbers ($Q$, $\vec{p}$, $Q_{ij}$) when you only care about the field far away.

Scalar potential in electrostatic systems

Capacitors and capacitance

A capacitor stores energy in the electric field between two conductors held at different potentials. Capacitance is defined as:

$C = \frac{Q}{\Delta V}$

where $Q$ is the magnitude of the charge on either plate and $\Delta V$ is the potential difference. $C$ depends on geometry and the dielectric filling (e.g., $C = \epsilon_0 A/d$ for an ideal parallel-plate capacitor with plate area $A$ and separation $d$).

Charge distributions and Coulomb's law

For a discrete set of point charges:

$\phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\sum_i \frac{q_i}{|\vec{r}-\vec{r}_i|}$

For a continuous distribution with volume charge density $\rho(\vec{r}')$:

$\phi(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec{r}')}{|\vec{r}-\vec{r}'|}\,d^3r'$

Both expressions follow from the superposition principle: the potential due to many charges is the algebraic sum of the potentials due to each charge individually.

Scalar potential in time-varying fields

Electrodynamic potentials

Once fields vary in time, a scalar potential alone is no longer sufficient. You need both $\phi$ and the vector potential $\vec{A}$. The fields are then:

$\vec{E} = -\nabla\phi - \frac{\partial \vec{A}}{\partial t}$

$\vec{B} = \nabla \times \vec{A}$

The extra $-\partial\vec{A}/\partial t$ term accounts for the electric field produced by a changing magnetic field (Faraday's law). Note that $\phi$ and $\vec{A}$ are not unique; they can be changed by a gauge transformation without affecting the physical fields $\vec{E}$ and $\vec{B}$.

Retarded potentials and Liénard-Wiechert potentials

Because electromagnetic disturbances travel at the speed of light, the potential at $(\vec{r}, t)$ depends on what the sources were doing at the retarded time:

$t_r = t - \frac{|\vec{r}-\vec{r}'|}{c}$

The retarded scalar potential is:

$\phi(\vec{r}, t) = \frac{1}{4\pi\epsilon_0}\int \frac{\rho(\vec{r}', t_r)}{|\vec{r}-\vec{r}'|}\,d^3r'$

For a single point charge $q$ moving along a trajectory $\vec{w}(t)$, this integral reduces to the Liénard-Wiechert potential:

$\phi(\vec{r}, t) = \frac{q}{4\pi\epsilon_0}\frac{1}{|\vec{r}-\vec{w}| - \frac{\vec{v}\cdot(\vec{r}-\vec{w})}{c}}\bigg|_{t_r}$

where $\vec{v}$ is the charge's velocity, and everything on the right is evaluated at the retarded time. The denominator reflects the relativistic compression or stretching of the "information sphere" due to the charge's motion.