Electromagnetic theory: potential problem

[meanyack]

Homework Statement

To see the problem, please click the link below: potential_problem.jpg

Homework Equations

\nabla^{}2V=0

Seperation of variables: V(x,y)=X(x)Y(y)

The Attempt at a Solution

For simplicity I divide problem into two parts:

1st part- Click the link below: potential_problem_part1.jpg

I tried to use separation of variables, yet, didn't decide which one would be sin(kx),cos(kx) and exp(\pmkx).
Here is another problem : There is no way to determine "k",and thus use sum of "fourier sine series".

2nd part- Click the link below: potential_problem_part2.jpg
The potential at x=0 is a step function i.e

V(x=0,y) =\left\{\begin{array}{cc}0,&\mbox{ if }x<0\\V_{0} , & \mbox{ if } x>0\end{array}\right
so, how to find the functions for this boundary?

 

[jdwood983]

Why would you separate the problem as you did? Just use Laplace's equation & separation of variables for the problem as it is given.

 

[diazona]

Why would you separate the problem as you did? Just use Laplace's equation & separation of variables for the problem as it is given.

Agreed... in fact I just did so to check and it seems reasonably straightforward (except that you may have to look up a Fourier transform).

 

[meanyack]

Why would you separate the problem as you did? Just use Laplace's equation & separation of variables for the problem as it is given.

If you don't seperate, you can't find the constant k=n*pi/(2*a) because, in original shape boundaries are different so you can't choose a cyclic function ie f(x)=f(x+T).

Actually, it's not an easy question, because you will be stuck when you look try to solve first part. To understand, give a try

 

[meanyack]

Agreed... in fact I just did so to check and it seems reasonably straightforward (except that you may have to look up a Fourier transform).

Did you mean Fourier transform or Fourier series expansion? You know, they are different

 

[jdwood983]

If you don't seperate, you can't find the constant k=n*pi/(2*a) because, in original shape boundaries are different so you can't choose a cyclic function ie f(x)=f(x+T).

Actually, it's not an easy question, because you will be stuck when you look try to solve first part. To understand, give a try

I'm not saying it's an easy problem, it is a difficult one. But what you have done, by separating the problem, is create two different problems (really 3 problems, but let's not get too technical here). You cannot combine the two solutions and expect the same result as if you had done it properly the first time.

It will be a Fourier series integral you will need to have in your solution.

In the original problem, what are your boundary conditions?

 

[meanyack]

The boundary conditions are on the picture
Let me write them again
v(x=0,y)=0
v(x,y=0)=0
v(x,y=a)=0
while x goes to inf, V(x,y)=Y(y)

Edit: You said " You cannot combine the two solutions and expect the same result as if you had done it properly the first time." but, yes we can. Actually, we have done it in class and superpositioning two different problems which gives the first problem is relatively easy. If you want I can post it

 

[jdwood983]

The boundary conditions are on the picture
Let me write them again
v(x=0,y)=0
v(x,y=0)=0
v(x,y=a)=0
while x goes to inf, V(x,y)=Y(y)

Ummm, no...look at the picture and try again.

Edit: You said " You cannot combine the two solutions and expect the same result as if you had done it properly the first time." but, yes we can. Actually, we have done it in class and superpositioning two different problems which gives the first problem is relatively easy. If you want I can post it

While this is true and possible, it is not always possible. In most cases, you cannot (and should not) assume that two totally different problems can be superpositioned to solve a third problem. I have done several problems myself that allow you to use superposition as well as several problems that don't allow you to use superposition. In general, you should not immediately assume you can use superposition, unless someone (the author, the professor) tells you that you can use it.

 

[gabbagabbahey]

While this is true and possible, it is not always possible. In most cases, you cannot (and should not) assume that two totally different problems can be superpositioned to solve a third problem. I have done several problems myself that allow you to use superposition as well as several problems that don't allow you to use superposition. In general, you should not immediately assume you can use superposition, unless someone (the author, the professor) tells you that you can use it.

Could you provide an example where the electrostatic potential does not satisfy the superposition principle?

 

[gabbagabbahey]

The boundary conditions are on the picture
Let me write them again
v(x=0,y)=0
v(x,y=0)=0
v(x,y=a)=0
while x goes to inf, V(x,y)=Y(y)

I'd probably solve the problem with V(L,y)=Y(y) first, and then take the limit as L\to\infty.

 

[meanyack]

I'd probably solve the problem with V(L,y)=Y(y) first, and then take the limit as L\to\infty.

Actually, I was wondering that. This can help me... thanks

 

[jdwood983]

Could you provide an example where the electrostatic potential does not satisfy the superposition principle?

I don't know any electrostatic potential problems nor did I make the claim that there are electric potential problems that don't follow the superposition principle; I simply stated that in general, you should not assume the superposition principle holds because I've taken enough non-linear dynamics courses to know that you shouldn't.

 

[meanyack]

Hey guys, our instructor posted the answer to the question. I never thought spliting the problem like in answer... The method is quite impressive. If you wonder, here is the link
http://www.fen.bilkent.edu.tr/%7Ebulutay/315/hw3_s1.gif
http://www.fen.bilkent.edu.tr/%7Ebulutay/315/hw3_s2.gif