Laplace's Equation and Boundary Condition Problem

[HansBu]

Homework Statement

Determine the potential and electric field at any point due to an infinite conducting plane lying in x-z plane. Half of the plane (x > 0) is maintained at +Vo while the other half (x < 0) is maintained at -Vo.

Relevant Equations

A relevant equation would include Laplace's Equation for Electrostatic Problems.

I really have no idea as to how to attack the problem in the first place. I am here to ask for some generous help on how to start. The figure is shown below for reference.

 

[HansBu]

What I am having trouble here is looking for the set of boundary conditions.

 

[hutchphd]

I will say a few "magic" words as hints:
The boundary at y=0 divides the space into y>0 and y<0. You will need a solution for each space that goes to zero for y large and matches the "step" in x at y=0
Clearly you can ignore z because of translational symmetry.
Separation of variables in the cartesian coordinates x,y.

 

[PhDeezNutz]

Are you familiar with Green's functions and integrating them to find potentials?

$\Phi = \frac{1}{4 \pi \epsilon_0} \int \rho \left(\vec{r}'\right) G\left(\vec{r}, \vec{r'} \right) \, dV' + \frac{1}{4 \pi} \int \Phi_s \frac{\partial G}{\partial n'} \, da'$

I would think the charge density would be zero on the surface because the potential has no derivatives (on the surface) so you only have to worry about the second integral. (Maybe someone can clarify)

Look up the normal derivative of the Green's function for a plane and break the second integral into two parts (one for each half of the plane).

After you set it up look at some integral tables.

 

[hutchphd]

After you set it up look at some integral tables.

Because you do not know the charge distribution this would seem ill-advised to me.

 

[PhDeezNutz]

Because you do not know the charge distribution this would seem ill-advised to me.

I assume you are referring to the first integral.

Doesn't the charge density in the first integral refer to the charge density in the volume of interest (i.e. above the plane)? Wouldn't that be zero unless otherwise stated?

This is my new reasoning for why the first integral vanishes.

 

[hutchphd]

Yes I think that would in fact be correct. I think separation of variables is a more direct route for the OP. But essentially the same surface integral will show up in that solution I believe. Let's let @HansBu work through it and we can show the equivalence afterwards .(or we can get confused!)

 

[PhDeezNutz]

Yes I think that would in fact be correct. I think separation of variables is a more direct route for the OP. But essentially the same surface integral will show up in that solution I believe. Let's let @HansBu work through it and we can show the equivalence afterwards .(or we can get confused!)

I’ve never solved it using separation of variables. Would be interesting to see. I might try it later.

 

[hutchphd]

I think it is not too difficult. As I recall Griffiths does some similar problems. But I haven't worked it through.

 

[vela]

What I am having trouble here is looking for the set of boundary conditions.

You need to mathematically express the potential where you know its value. Where is it equal to $V_0$? Where is it equal to $-V_0$? The fact that the planes are conducting may yield another condition on the derivative of $\phi$.

 

[HansBu]

From the problem, I can deduce the following boundary conditions:

1. As x approaches infinity, V = Vo
2. As x approaches negative infinity, V = -Vo

 

[hutchphd]

The picture is much more specific than that. What is V at (x,y)=(1,0) for instance?
What course is this for?

 

[vela]

From the problem, I can deduce the following boundary conditions:

1. As x approaches infinity, V = Vo
2. As x approaches negative infinity, V = -Vo

In addition to what @hutchphd said, is this for all values of $y$ and $z$? Remember this is a problem in three dimensions, so you should also specify the other two coordinates in defining a boundary.

 

[HansBu]

The picture is much more specific than that. What is V at (x,y)=(1,0) for instance?
What course is this for?

Clearly, the potential has no dependence on y. Hence, I am to arrive at a potential function in term of x and z only. Am I right?

 

[hutchphd]

No. I believe the problem is unchanged by translation along z. So variation only in x and y. It would help me to know your educational level and the course for which this is assigned. There are various ways to solve it.

 

[HansBu]

No. I believe the problem is unchanged by translation along z. So variation only in x and y. It would help me to know your educational level and the course for which this is assigned. There are various ways to solve it.

I am a bs physics 2 student and only knowledgeable on Laplace's Equation on Cartesian Coordinate System for electrostatic problems. Usually, the method is restricted only on variable separation and my only main problem here is the boundary condition. I am sorry for the trouble.

 

[hutchphd]

This is not trouble. Please believe me.
The problem is best solved in cartesian coordinates. So do you see that the z axis is not important (it is not in the boundary values)?. That makes this a 2D problem in (x,y).
Lets look only at y>=0 for now . What are the known boundary conditions?

NB I may disappear it is past 11:00 pm here...

 

[HansBu]

Have you seen the figure? The plane is parallel to the xz-plane not on the xy-plane. I think you missed it and I cannot see the logic as to why we should consider x and y only instead of x and z

 

[PhDeezNutz]

Alright let's do this. I'm going to teach you something about Green's Functions (which to me is the only way to solve this because the final answer I get doesn't seem to be separable, in fact I think it's not right to ask a question like this in an undergrad course).

Hopefully MathJax doesn't screw up

A Green's Function helps you turn your boundary conditions into integrals in order to solve the potential.

Without going through derivation of the theory the final result is

$\Phi \left(\vec{r} \right) = \frac{1}{4 \pi \epsilon_0} \int \rho \left( \vec{r}' \right) G \left( \vec{r}, \vec{r}' \right) \,d^3 r' - \int \Phi_{surface} \frac{1}{4 \pi} \frac{\partial G\left( \vec{r}, \vec{r}'\right)}{\partial n'} \,d^2 r'$

Where $r$ and $\vec{r}$ refer to position coordinates and $r'$ and $\vec{r}'$ refer to "source coordinates" (i.e. the plane). $\frac{\partial G\left( \vec{r}, \vec{r}'\right)}{\partial n'}$ is the normal derivative to the plane.

Both the Green's Function and the Green's function normal derivative are evaluated at the surface in the above integral.

The Green's function for a plane is

$G \left( \vec{r}, \vec{r'} \right) = \frac{1}{\left[\left( x - x'\right)^2 + \left( y - y'\right)^2 + \left(z - z' \right)^2\right]^{\frac{1}{2}}} - \frac{1}{\left[\left( x - x'\right)^2 + \left( y + y'\right)^2 + \left(z - z' \right)^2\right]^{\frac{1}{2}}}$

Evaluate this Green's function at the plane. What is it equal to? What does that mean for the first integral above?

What is the normal derivative (in terms of source coordinates) (compute it) of $G$? Evaluate this at the plane.

Break the second integral into two parts for the two different regions (both $+V_0$ and $-V_0$) and integrate over the two planes. Add the results together and then add this result to the first integral (with $G$).Integral tables are your friend here. You may initially think that doing these integrals over semi-infinite planes would either diverge or cancel out but I assure you they do not.