Potential boundary conditions, but no charge distribution

We know how to find the field...

  • From some known charge distribution: $\myv E = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\myv r')}{\rr^2}\uv \rr\,d\tau'$,
  • Or from a potential specified everywhere...$\myv E = -\myv \grad V$.

What if we don't have either one of these??

It turns out we can solve a differential equation to find $V$...

In the (hokey) charge on conductors picture, we know the potentials and shapes of the conductors. What we don't know:

  • How is the charge distributed on the conducting surfaces?
  • What's the electric potential $V(\myv r)$ between the conductors?

Mathematically, this involves solving...

Laplace's equation

If the region between conductors is empty of charge, then that means at every point in the region... $$\myv{\grad}\cdot \myv E = \rho/\epsilon_0 = 0.$$

But our boundary conditions are in terms of the potentials, not of the fields.

Write this in terms of potentials instead: Substituting $\myv E = -\myv \grad V$ into the equation above... $$\myv{\grad}\cdot\left( -\myv \grad V)\right) = -\grad^2 V = 0.$$ Multiplying both sides by an innoccuous -1, this is Laplace's Equation:

$$\grad^2 V = 0 = \frac{\del^2 V}{\del x^2} + \frac{\del^2 V}{\del y^2} +\frac{\del^2V}{\del z^2}.$$

This is the equation that we need to solve, subject to certain boundary conditions.

The solutions to Laplace's equations are called harmonic functions.

For a function of one variable, the second derivative is a measure of "curvature". To generalize to higher dimensions, $\grad^2 V$ is a measure of the curvature averaged over all directions.

Solutions in one dimension

[The field between two, practically infinite, flat conductors, with no charge between them, with surface normals in the $\uv x$- direction.]

Translational symmetry means that $\myv E = E(x)\uv x$. Show that this leads to... Laplace's equation in one dimension: $$\Rightarrow\frac{d^2V}{dx^2} = 0.$$

The first derivative of a 1-d function is its slope, and the second derivative is the local "curvature" at that point. This equation says that $V$ has zero curvature, that is, solutions are straight lines: $$V=mx + b$$.

A solution that matches the boundary conditions (represented by two blue dots) is shown at right:

Some trivial-sounding characteristics of this solution that are also shared with higher-dimensional solutions:

  • a minimum or maximum can only occur at the boundary--never in between.
  • V(x) is everywhere the average value of the function evaluated at points equidistant in $x$. That is: $V(x)=\frac{1}{2}[V(x+R) + V(x-R)]$.

This last statement is the 1-d version of Gauss's harmonic function theorem, which states that $V(x)$ is the average of the values of $V$ on a "sphere" of radius $R$ centered at $x$.

Two dimensions

Now Laplace's equation is $$\frac{\del^2 V}{\del x^2}+\frac{\del^2 V}{\del y^2} = 0$$.

We can think of "V" as representing the height of a landscape as a function of position $x$ (East/West position) and $y$ (North/South position). The "curvature" point of view means that at every position, the landscape is either flat, or the curvature in the $x$- direction is equal and opposite of the $y$-curvature:

This defines what is known as a minimal surface.

A soap film suspended from a wire is another example of a physical surface that is a minimal surface.

Again no local minima or maxima are allowed, except at the boundary. For, at a minimum (or a maximum) the curvature in both the $x$- and $y$-directions would have to be the same.

A ping-pong ball placed on a such a surface would have to roll off: it could find no resting place.

The 2-d version of Gauss's harmonic function theorem involves an integral over a circle centered at $(x,y)$: $$V(x,y) = \frac{1}{2\pi R} \oint_{\text{circle}} V dl$$ That is, $V$ is the average value of the $V$ at all the points on a 2-d sphere (a circle) centered at $(x,y)$.

Computer solutions

This is the basis for the method of relaxation: On a computer you can make random guesses at a mesh of points at positions in space. You can find a solution to Laplace's equation by replacing each point by the average of all equidistant points. After enough iterations, the values converge to a solution.

1-d solution in a spreadsheet

Let's actually do this in a spreadsheet...

In 1-d the Laplace equation is $$\frac{d^2V}{dx^2}=0$$ Solve this for boundary conditions $V(0)=$20 V and $V(10)=$0 V:

  • Set up a row of column position labels: 11 column labels representing x=0...10
  • in the next row, set the first column (x=0) to 20 and the last column (x=10) to 0.
  • Put random numbers in the columns in-between (let's say, between -50 and +100). The numbers in this row represent some arbitrary 'guess' at the potential, which we will now refine, one iteration (row) at a time.
  • In the next row down, copy the values for V(x=0) and V(x=10) from the row above. In betweeen, column $i$ from (1...9) will be calculated by averaging the values for $V_{i-1} and V_{i+1} from the row above. Write (click) a spreadsheet formula to do that.
  • Now, copy/paste this row about 20 times. How close are you to a nice monotonic decrease in potential after 20 iterations? (Voltage should drop down evenly by 2 V in each column after a while.)

Three dimensions

It's harder to draw a picture of a field satisfying Laplace's equation in 3 dimensions.

Again, no minima or maxima are allowed except at the boundaries. An interesting consequence of this is that no electrostatic field can hold a charged particle still. (This is too bad if you'd like to confine a plasma, for example..).