Chapter 3 - Potentials

Reference "Introduction to Electrodynamics" by David Griffiths.

The goal of electrostatics is to find the electric field given some stationary, immobile charge distribution.

Coulomb's Law is the way we do this for simple charge configurations, but for more complex charge configurations it's often easier to work with potential $V$. For areas of nonzero charge density $\rho \ne 0$ such as point, surface or volume charges, we use Poisson's equation: $${\nabla }^{2}V=-\frac{1}{{ϵ}_0}\rho$$ Outside of these charge regons (such as in regular space), this reduces to Laplace's equation: $${\nabla }^{2}V=0$$ $$\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}+\frac{{\partial }^{2}V}{\partial z^2}=0$$ with solutions called harmonic functions.

Laplace's equation is fundamental to the study of electrostatics according to Griffiths.

Laplace's Equation

In three dimensions, visualization is challenging - but the same two properties apply, with the first this time being

1. The value of $V(x,y,z)$ is the average $V$ over a spherical surface of radius $R$ centered at that point, with $$V(x,y,z)=\frac{1}{4\pi R^2}{\oint }_{\text{sphere}}Vda$$
2. $V$ can have no local maxima or minima, with extreme values only permitted at the boundaries (i.e. surface of the sphere).

Point 2 is particularly relevant in each of these circumstances - the value of $V$ at some point is the average of the surrounding $V$ on some surrounding boundary.

Uniqueness Theorems

The proof that a proposed set of boundary conditions will suffice takes the form of a uniqueness theorem (alternatively, a criterion to determine whether a solution to the Laplace or Poisson equations is unique).

Uniqueness Theorem 1: the solution to Laplace's equation ${\nabla }^{2}V=0$ in a volume $\mathcal{V}$ is unique if $V$ is specified on a boundary surface $\mathcal{S}$ enclosing the volume $\mathcal{V}$.

The surface does not have to be Gaussian - it can look totally eldritch and crazy and the theorem would hold.

Uniqueness Theorem 2: given a volume $\mathcal{V}$ surrounded by conductors and containing some charge density $\rho$, the electric field is uniquely determined if the total charge on each conductor is given.

Method of Images

Griffiths 3.2.

Say we have some charge $q$ a distance $d$ above an infinite grounded plane; what is the potential $V$ in the region above the plane? Our boundary conditions state

$$\{\begin{array}{ll}V=0& z=0\\ V\to 0& \text{far from the charge}\end{array}$$ We could solve Poisson's equation for this region, but a much easier technique is to use the Method of Images. Wikipedia says on the subject:

The method of image charges is used in electrostatics to simply calculate or visualize the distribution of the electric field of a charge in the vicinity of a conducting surface.

It is based on the fact that the tangential component of the electrical field on the surface of a conductor is zero, and that an electric field E in some region is uniquely defined by its normal component over the surface that confines this region (the uniqueness theorem).

Start by removing the conductor, and placing an opposite charge $-q$ at $-d$:

Then, the potential is easy to calculate: $$V(x,y,z)=\frac{1}{4\pi {ϵ}_0}\left[\frac{q}{\sqrt{x^2+y^2+(z-d{)}^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d{)}^2}}\right]$$ which obeys both of the boundary conditions of the original problem:

1. $V=0$ when $z=0$
2. $V\to 0$ for $x^2+y^2+z^2\gg d$

Thus by uniqueness theorem 1, is the solution to our original problem.

Uniqueness theorem 1 means that if a solution satisfies Poisson's equation in the region of interest and assumes the correct value at the boundaries, it must be right.

The Method of Images can be used in any scenario where we have a stationary charge distribution near a grounded conducting plane.

Induced Surface Charges

The surface charge density induced on a conductor is $$\sigma =-{ϵ}_0\frac{\partial V}{\partial n}$$ where $\partial V/\partial n$ is the normal derivative of $V$ at the surface. In the above case, this is is in the $z$ direction - if we take this partial derivative of our above calculated voltage, then $$\sigma (x,y)=\frac{-qd}{2\pi (x^2+y^2+d^2{)}^{3/2}}$$

As expected from a positive charge, the induced surface charge is negative and greatest at $x=y=0$.

The total induced charge is $$\begin{array}{rl}Q=\int \sigma da& ={\int }_0^{2\pi }{\int }_0^{\infty }-\frac{qd}{2\pi (r^2+d^2{)}^{3/2}}rdrd\varphi \\ & =\frac{qd}{\sqrt{r^2+d^2}}{|}_0^{\infty }\\ & =-q\end{array}$$ Yes!

Force & Energy

Since the induced charge on the conductor is $-q$ and our charge is $+q$, it is attracted to the plane with a force given by Coulomb's Law: $$\vec{F}=-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{(2d{)}^2}\hat{z}$$ While force is the same in our mirror problem, energy is not. For two point charges, $W=\frac{1}{2}{\sum }_{i=1}^n{q}_{i}V(\overrightarrow{r_i})$, such that $$W=-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{2d}$$ But for a single charge and conducting plane (continuous charge distribution), energy is half of this. $$W=-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{4d}$$

The work to bring two point charges towards one another does work on both of them, while to bring a point charge toward a grounded conductor has us only doing work on one charge - only half the work is necessary.

Separation of Variables

Griffiths' 3.3

Separation of variables is a way to solve ODEs and PDEs by rewriting equations such that each of the two variables occur on different sides of the equation.

Separable equations must be able to be written in the form $$\frac{dx}{dy}=g(x)h(y)$$ We can rearrange the terms to get $$dy\frac{1}{h(y)}=g(x)dx$$ integrate, $$\int \frac{1}{h(y)}dy=\int g(x)dx$$ and add some constant term to one side to represent all our constants of integration. $$H(y)=G(x)+C$$

In the context of electrostatics, separation of variables is very useful when solving 2D Laplace equations, such as $$\frac{{\partial }^{2}V}{\partial x^2}+\frac{{\partial }^{2}V}{\partial y^2}=0$$ We need solutions in the form of $$V(x,y)=X(x)Y(y)$$ This can be accomplished through some mathematical trickery to find our separated variables ... $$\frac{1}{X}\frac{d^{2}X}{d{x}^2}+\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}=0$$ ... which is of the form $f(x)+g(y)=0$. Thus, both $f(x)$ and $g(y)$ must be constant (we can't hold one constant and change the other with this solution still holding).

So, $$\frac{1}{X}\frac{d^{2}X}{d{x}^2}=C_1\text{and}\frac{1}{Y}\frac{d^{2}Y}{d{y}^2}=C_2{C}_1+C_2=0$$ Converting each equation into an ODE, $$\frac{d^{2}X}{d{x}^2}=k^{2}X\frac{d^{2}Y}{d{y}^2}=-k^{2}Y$$ ... we converted a PDE into two ODEs, which are much easier to solve. Our solutions will be a constant coefficient set: $$X(x)=c_1{e}^{kx}+c_2{e}^{-kx},Y(y)=c_3\sin (ky)+c_4\cos (ky)$$ We can find our constants based on our boundary conditions now.

Multipole Expansion

Griffiths 3.4

Say you have some charge $Q=Q_{\text{net}}$ that you can see in space, far away. $Q$ is almost like an exoplanet in a sense: we know it exists, we know its total mass, and maybe we know stuff like how far away it is.

But, we don't know what the surface looks like; what it's composed of, be it rock or ice or water. Just its net mass.

Similarly, our charge distribution might be crazy complicated, and it's net charge $Q_{\text{net}}$ only describes a tiny part of the story, albeit an important one - the multipole expansion is used to describe this fuller story, and is defined in terms of voltage as

$$\begin{array}{rl}V(\vec{r})& =\frac{1}{4\pi {ϵ}_0}\left[\frac{1}{r}(\text{monopole moment})+\frac{1}{r^2}(\text{dipole moment})+\frac{1}{r^3}(\text{quadrupole moment})\dots \right]\end{array}$$

Mathematically, it is $$V(\vec{r})=\frac{1}{4\pi {ϵ}_0}\sum _{n=0}^{\infty }\frac{1}{r^{n+1}}\int (r^{\prime }{)}^n{P}_n(\cos \alpha )\rho ({\vec{r}}^{\prime })d{\tau }^{\prime }$$ where $\alpha$ is the angle between $\vec{r}$ and $\overrightarrow{r^{\prime }}$, $\vec{r}$ is the reference point (from origin) and ${\vec{r}}^{\prime }$ the charge (from origin).

Monopole and dipole terms

At large $\vec{r}$, a charge distribution just looks like a net charge (like an exoplanet). Thus, the monopole moment $Q$ is just the net charge: $$\begin{array}{r}Q\equiv \int \rho d\tau V_{\text{mon}}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\frac{Q}{r}\end{array}$$ The dipole moment $\vec{p}$ describes the individual distribution of charges: $$\vec{p}=\sum _{i=1}^n{q}_i{\overrightarrow{r_i}}^{\prime }\equiv \int {\vec{r}}^{\prime }\rho (\overrightarrow{r^{\prime }})d{\tau }^{\prime }{V}_{\text{dip}}(\vec{r})=\frac{1}{4\pi {ϵ}_0}\frac{\vec{p}}{r^2}$$

${\vec{r}}^{\prime }$ points from the origin to some charge distribution. $\vec{r}$ is from the origin to some reference point.

Origin of coordinates

Changing the origin will never change the monopole moment $Q$, but will change the dipole moment $\vec{p}$ as long as the total charge $Q\ne 0$.

For example: if our system has $+q$ and $-q$ as its point distribution, $Q=+q+(-q)=0$ and the dipole moment is origin-independent.

Electric field of moments: the electric field is defined as $E=-\nabla V$.

To find the electric field i.e. caused by the dipole moment, find the voltage term for that dipole moment, then take the negative gradient of it.