Question for the beginning of class:

We hold a charge \(q\) a distance \(d\) above an infinite grounded conducting plane. What’s the potential in the region above the plane?

What you should expect from the environment in physics.

  • mutual respect
  • welcoming spaces
  • honoring difference…all the differences you can think of
  • integrity

We’ve got Coulomb’s law

\[\vec{E}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int\frac{\hat{R}}{R^2}\rho(\vec(r^\prime) d \tau^\prime\]

where \(R = \vec{r} - \vec{r}^\prime\)

and the integral for the potential

\[\vec{V}(\vec{r}) = \frac{1}{4\pi\epsilon_0} \int\frac{1}{R}\rho(\vec(r^\prime) d \tau^\prime\]

but sometimes neither of those is possible and you need to “recast the problem in differential form” (as Griffiths says) using Poisson’s equation

\[\nabla^2 V = - \frac{1}{\epsilon_0}\rho\]

If you’re using Poisson’s equation in a region where there’s no charge then Poisson’s equation becomes Laplace’s equation.

\[\nabla^2 V = 0\]

By the way, there can be lots of interesting field in this region, just not charge.

The solutions to Laplace’s equation are called harmonic functions. That’s where we’re going in this class…

In one dimension:

\[\frac{d^2}{dx^2} = 0\] \[V(x) = mx + b\]

And then if you know that \(V(0)=5\) and \(V(1)=4\) then you can figure out what \(m\) and \(b\) are.

So that’s completely specified.

So that’s really what we mean by solving Laplace’s equation with boundary conditions. I don’t know - I think it sounds really scarey until you point out it’s the thing you did in 7th grade where you had to figure out what the equation of the line was.

Some interesting things to notice:

  • The potential in the middle is the average of the potentials on either side.
  • There can be no minimum or maximum in this region, except at the end points.

Three dimensions

In three dimensions, what does it mean to specify the potential on the boundaries? It means to specify it on the surface.

First uniqueness theorem:

The solution to Laplace’s equation in some volume \(\mathscr{V}\) is uniquely determined if the \(V\) (the potential) is specified on the boundary surface \(S\).

You can get it by analogy to the one dimensional case, right?

The things we noticed before are still true:

The potential at the center of a sphere is the average of the potential around a sphere that surrounds it.

\[V(\vec{r}) = \frac{1}{4\pi R^2} \oint_{\rm sphere} V da\]

For this reason the potential is an “averaging” function.

There’s an interesting consequence of the first uniqueness theorem. If we can find a situation that produces the same potential on the boundary in question, then we can find the potential every where in the volume.

Classic Image Problem Your homework has two image problems on it, by the way.

Brings us to the question I asked at the beginning of class:

We hold a charge \(q\) a distance \(d\) above an infinite grounded conducting plane. What’s the potential in the region above the plane?

This uses the second uniqueness theorem (but similar to the first):

Second uniqueness theorem: In a volume \(\mathscr{V}\) surrounded by conductors and containing a specified charge density \(\rho\), the electric field is uniquely determined if the total charge on each conductor is given.

Griffiths picks the definition apart, and even proves it to you. Let’s just see how to use it. The basic idea is the same as we’ve been talking about. We’re only interested in the field above the plane. We know the charge, and we know the potential on the boundary (the plane) because it’s a conductor and therefore an equipotential.

So instead of doing this problem we can do a different problem that has the same charge as in the area we’re interested in, and the same potential on the boundary.

What’s the different problem?

Well, can you think of a relatively simple arrangement of charges that would have only this positive charge on this side of the plane, and \(V=0\) at the boundary. In other words, you can put whatever charges you want over here, as long as there’s an equipotential.

(We had a super cool conversation about this in class. We quickly got that the image charge should be opposite the real charge, but there was a big debate about sign it should have. It was awesome. The thing that caught people was that zero potential at the boundary didn’t mean that there was zero force/field out of the plane. The potential being zero in the place just means that there’s no force on charges IN the plane, but there can be a component of the field perpendicular to the plane. And in fact, in this case there has to be non-zero perpendicular component if you want the potential to be zero on the plane. Another way of saying this same thing is that the real charge induces a charge in the conducting plane. And the charge in the conducting plane produces a field.)

Notice that this is made possible (brought to you by) Laplace’s/Poisson’s equation. As long as you know what \(\nabla^2\) of the potential is, and you know it on the boundary, you can know it everywhere, and it’s uniquely determined by those criteria.

We need the other boundary too, but that’s just the usual V goes to 0 at infinity.

We’ll put the charges at \(z = +/- d\) so the potential will look like this:

\[V(x,y,z) = \frac{1}{4\pi\epsilon_0} \left[\frac{q}{\sqrt(x^2+y^2 + (z-d)^2)} - \frac{q}{\sqrt(x^2+y^2 + (z+d)^2)} \right]\]

Remember thankfully in the potential you don’t need vectors, and that thing in the denominator is just the difference between wherever you are and the point charge.

Does it satisfy our criteria?

  • Goes to zero at infinity
  • Is zero on the plan

So this is the potential everywhere that’s bounded by our boundaries.

Morale: If it satisfies Poisson’s equation in the region of interest, and assumes the correct value at the boundaries, then it must be right.

What’s the surface charge on the conductor?

We did that in class a couple days ago. It’s the derivative of the potential at the boundary.

\[\sigma = \epsilon_0 \frac{\partial V}{\partial n}\]

You can do that derivative (Griffiths shows it) and then you can integrate the surface charge over the whole plane to get the total charge. What do you think the total charge is?

(Answer: \(-q\))

Your homework problems #5 and #6.

#5 is two parallel conducting plates with a charge in between them. Talk with your neighbor and think about what set of image charges you’d need to produce two V=0 planes.

(Answer: It happens to be an infinite set of charges. Each one you have to add to produce zero potential on one of the plates. The one you just added forces you to add another one to make the potential zero on the other place. This goes on for infinity.)

#6 is a charge a distance \(a\) away from the center of a neutral (not-grounded) conducting sphere of radius \(R\).

The grounded sphere case

Griffiths does the case where the sphere is grounded (Example 3.2 page 128). It’s pretty interesting. It turns out that if you put an image charge a distance \(R^2/a\) away from the center of the sphere with value \(-\frac{R}{a}q\) then you get potential of zero on the sphere.

What’s the total charge on the sphere? (talk amongst yourselves)
You can integrate Gauss’ Law in the image situation and know that that \(\oint \vec{E}\cdot da = \frac{q^\prime}{\epsilon_0}\). That has to be true in the real situation just outside the boundary (because the field is the same in both the real situation and the analog/image situation). So then there must be a charge \(q^\prime\) on the sphere.

The ungrounded sphere case (the case you’re doing for homework)

So what do you do if the sphere isn’t grounded, but rather neutral and isolated? Talk amongst yourselves.

A Gaussian surface just outside the sphere must now integrate to zero, so the total charge on/in the sphere has to be zero by Gauss’s law. The conducting sphere has to be an equipotential by the rules we laid out in Boundary Conditions and Conductors. So now, to get from the grounded situation (Griffiths example) to the ungrounded situation (your homework #6) you need to do two things:

  • Add a charge that brings the total charge on/in the sphere to zero.
  • Place that charge such that the potential of the sphere goes from zero to some other constant.

What will you do?

Comment on Force and Energy.

Griffiths has a nice section on this (3.2.3 pg 126). The basic deal is that the method of images only gives you the potential in the region in question, so you have to be careful about how far you extend the analogy or method.

The image problem works to calculate the force

The negative gradient of the potential is the electric field, and the electric field multiplied by the point charge in that region is the force on that charge. All those things are the same in the analog problem as in the real problem, so you can calculate the force on a particle in the region in question by calculating the force on it due to the all the image charges.

The image problem doesn’t work to calculate the work

However, in order to calculate the work done to assemble the configuration of charges (or the energy of the configuration) you can’t use the image charges. You can find the electric field by taking the negative gradient of the potential. And then the integral of the square of the electric field is the work:

\[W = \frac{\epsilon_0}{2}\int E^2d\tau\]

There’s that factor of 2 we found in Potential, Work, Energy.

Another way to find the work is to integrate the force:

\[W = \int_\infty^\vec{r} \vec{F}(\vec{r}^\prime)\cdot d\vec{l}\]

where \(\vec{r}\) is the location you bring the charge to and \(\vec{F}(\vec{r})\) is

\[\vec{F}(\vec{r}) = -q\nabla V(\vec{r}\prime)|_{\vec{r}^\prime = \vec{r}}\]

and \(V(\vec{r})\) is the potential at the point \(\vec{r}\) where the real charge was when you calculated the potential using the method of images.

Note: I don’t really like the way I said that above. The point is that to calculate the work done you have to slowly bring the charge in from infinity, calculating the potential at its location along the way. In other words, the static potential of the final configuration is not the potential you use in the equation above. Rather you have to calculate the potential at the location of the charge as you bring it in. Frankly, squaring the electric field and integrating it seems a lot more straightforward.