# E&M: Using Laplace's Equation to solve for a conducting slit

• eep
In summary, the problem involves solving for the potential using Laplace's equation for a set up with three parallel conductors at different potentials and insulated from each other at their corners. The solution involves separating the variables and applying boundary conditions to find the general solution for X(x) and Y(y). However, the solution for X(x) may be doubted due to the behavior of X as x goes to infinity. Further clarification may be needed about the specific set up and boundaries of the conductors.
eep
E&M: Using Laplace's Equation to solve for a conducting "slit"

## Homework Statement

The set up is as follows: You have a conductor at potential 0 along the y-axis at x=0. You have another conductor at potential V=Vo running along the x-axis at y=0. You have a third conductor at potential V=Vo running along the x-axis at y=a. The "touching" corners of the conductors are all insulated from one another. Determine what the potential is by using Laplace's equation.

## Homework Equations

$$\nabla^2V = 0$$

$$V(x,y) = X(x)Y(y)$$

$$\frac{1}{X}\frac{{d^2}X}{d^2x} = k^2$$

$$\frac{1}{Y}\frac{{d^2}X}{d^2y} = -k^2$$

## The Attempt at a Solution

So we're basically just trying to solve this by separation of variables. I know Y(y) must oscillate as the potential is Vo at two points. After applying boundry conditions in the Y-direction, and taking the simplest form, I get

$$Y(y) = {V_o}cos(\frac{n{\pi}y}{a})$$

However in the X-direction I only having one boundry condition, giving me

$$X(x) = 2sinh(\frac{n{\pi}x}{a})$$

The general solution is a sum over N of X*Y, however as x goes to infinity, X goes to infinity which makes me doubt my solution for X. If the conductor at x=0 was at potential Vo, and the conductors at y=0,a were at 0, I could apply the boundry condition that V goes to 0 as x goes to infinity, but I don't think I can apply that in this case. Suggestions?

Are yo sure you are suppsoed to be finding the potentail outside the pipe??

I don't think you need to impose the condition that the V-> 0 for x-> infty in that case...

Wouldn't it be $V\rightarrow V_{0} \ as \ x\rightarrow \pm\infty$ and V=0 at x=0? So you'd need to find what different values of n give, ie n=0, so you get something sort of like: $X(x)=V_{0}(1-\sum_n \exp{(-n\pi x/a)})$

eep said:

## Homework Statement

The set up is as follows: You have a conductor at potential 0 along the y-axis at x=0. You have another conductor at potential V=Vo running along the x-axis at y=0. You have a third conductor at potential V=Vo running along the x-axis at y=a. The "touching" corners of the conductors are all insulated from one another. Determine what the potential is by using Laplace's equation.

## Homework Equations

$$\nabla^2V = 0$$

$$V(x,y) = X(x)Y(y)$$

$$\frac{1}{X}\frac{{d^2}X}{d^2x} = k^2$$

$$\frac{1}{Y}\frac{{d^2}X}{d^2y} = -k^2$$

## The Attempt at a Solution

So we're basically just trying to solve this by separation of variables. I know Y(y) must oscillate as the potential is Vo at two points. After applying boundry conditions in the Y-direction, and taking the simplest form, I get

$$Y(y) = {V_o}cos(\frac{n{\pi}y}{a})$$

However in the X-direction I only having one boundry condition, giving me

$$X(x) = 2sinh(\frac{n{\pi}x}{a})$$

The general solution is a sum over N of X*Y, however as x goes to infinity, X goes to infinity which makes me doubt my solution for X. If the conductor at x=0 was at potential Vo, and the conductors at y=0,a were at 0, I could apply the boundry condition that V goes to 0 as x goes to infinity, but I don't think I can apply that in this case. Suggestions?

I'm inclined to think that you want constant potential (zero electric field) anywhere you are infinitely far from any wire. As x→∞ in the vicinity of y=0, the potential should not be altered by the zero potential wire at x=0. So as x→∞ the potential should be that of two infinite parallel wires at potential V=Vo. That's probably a logarithmic behavior as y→∞ for very large x since the electric field out there should decay as 1/y.

Last edited:
Yeah, I was thinking that infinitely far away from x=0, the existence of the plate at x=0 shouldn't matter. I'll ask my professor about it on Monday, I'm just studying for finals and came up with this problem. Thanks.

eep said:

## Homework Statement

The set up is as follows: You have a conductor at potential 0 along the y-axis at x=0. You have another conductor at potential V=Vo running along the x-axis at y=0. You have a third conductor at potential V=Vo running along the x-axis at y=a. The "touching" corners of the conductors are all insulated from one another. Determine what the potential is by using Laplace's equation.

## Homework Equations

$$\nabla^2V = 0$$

$$V(x,y) = X(x)Y(y)$$

$$\frac{1}{X}\frac{{d^2}X}{d^2x} = k^2$$

$$\frac{1}{Y}\frac{{d^2}X}{d^2y} = -k^2$$

## The Attempt at a Solution

So we're basically just trying to solve this by separation of variables. I know Y(y) must oscillate as the potential is Vo at two points. After applying boundry conditions in the Y-direction, and taking the simplest form, I get

$$Y(y) = {V_o}cos(\frac{n{\pi}y}{a})$$

However in the X-direction I only having one boundry condition, giving me

$$X(x) = 2sinh(\frac{n{\pi}x}{a})$$

The general solution is a sum over N of X*Y, however as x goes to infinity, X goes to infinity which makes me doubt my solution for X. If the conductor at x=0 was at potential Vo, and the conductors at y=0,a were at 0, I could apply the boundry condition that V goes to 0 as x goes to infinity, but I don't think I can apply that in this case. Suggestions?
1. Let U=V-V_0. Then the left will be at -V_0, and the top and bottom will be at 0.
2. The solution for this is Y(y)=sin(n pi y/a) and X(x)=exp(-n pi x/a).

Meir Achuz said:
1. Let U=V-V_0. Then the left will be at -V_0, and the top and bottom will be at 0.
2. The solution for this is Y(y)=sin(n pi y/a) and X(x)=exp(-n pi x/a).

I guess I misinterpreted the problem. I should have paid more attention to the title. I was thinking in terms of wires, but I guess the conductors are parallel plates. I was also thinking the conductors extended infinitely in the x direction, since the problem says nothing about them being of some particular length. Did I miss something? What does this thing look like?

____________________________________________________________________
|
|
|
|
|
____________________________________________________________________

Last edited:
The lines should all be solid. The 3 planes are infinite in the z dilrection and the 2 horizontal planes are infinite in the + x direction.

Meir Achuz said:
The lines should all be solid. The 3 planes are infinite in the z dilrection and the 2 horizontal planes are infinite in the + x direction.

I would have thought the plane at x = 0 extended to infinity also, with the horizontal planes going from -∞ to +∞. Of course you would need some small holes drilled through the horizontal planes to connect the pieces of the vertical plane together, or some sort of mechanism to keep the whole thing at zero potential. But I'm content with your interpretation.

Is your XY = U, or XY = V? I assume your solution is for the region between the plates, but if I'm wrong about that I need to look at this problem more carefully to clear some cob webs out of the old brain.

OlderDan said:
I would have thought the plane at x = 0 extended to infinity also, with the horizontal planes going from -∞ to +∞. Of course you would need some small holes drilled through the horizontal planes to connect the pieces of the vertical plane together, or some sort of mechanism to keep the whole thing at zero potential. But I'm content with your interpretation.

Is your XY = U, or XY = V? I assume your solution is for the region between the plates, but if I'm wrong about that I need to look at this problem more carefully to clear some cob webs out of the old brain.
This is a standard EM problem. The solution is only for the region between the horizontal plates, to the right of the vertical plate.
If you use U=V-V_0, then the problem simplilfies to U=-V_0 on the left plate wlith the two horizontal plates grounded. Then U=XY that I gave.
After you find U, just go back to V=U+V_0 for your final answer.

Meir Achuz said:
This is a standard EM problem. The solution is only for the region between the horizontal plates, to the right of the vertical plate.
If you use U=V-V_0, then the problem simplilfies to U=-V_0 on the left plate wlith the two horizontal plates grounded. Then U=XY that I gave.
After you find U, just go back to V=U+V_0 for your final answer.

OK. Thanks.

Thanks Meir Achuz, it was bothering me!

## 1. What is Laplace's Equation and how is it used in E&M?

Laplace's Equation is a partial differential equation that describes the behavior of electric and magnetic fields in a given region. It is used in E&M to solve for the distributions of these fields in conductors and other materials.

## 2. How does Laplace's Equation apply to conducting slits?

In the case of conducting slits, Laplace's Equation can be used to solve for the electric field in the region near the slit. This is important for understanding the behavior of currents and charges in the presence of a conducting slit.

## 3. What is the significance of solving for the electric field in a conducting slit?

The electric field in a conducting slit can give insight into the behavior of charges and currents in the presence of the slit. It can also help in designing and analyzing devices that utilize conducting slits, such as antennas and waveguides.

## 4. What are some methods for solving Laplace's Equation in conducting slits?

There are several methods for solving Laplace's Equation in conducting slits, including separation of variables, the method of images, and the use of Green's functions. Each method has its own advantages and is chosen based on the specific problem at hand.

## 5. Are there any limitations to using Laplace's Equation in conducting slits?

One limitation is that Laplace's Equation does not take into account the effects of time-varying fields, which may be important in some cases. Additionally, the assumptions made in solving for the electric field may not always accurately represent the real-world behavior of conducting slits.

Replies
5
Views
330
Replies
10
Views
3K
Replies
1
Views
2K
Replies
3
Views
893
Replies
2
Views
1K
Replies
11
Views
2K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
1
Views
639
Replies
7
Views
2K