Find potential within a pipe using Laplace's equation

In summary: Still not seeing how you get that. Please show all the steps.Starting from -Csin(ak)=Csin(ak), I can divide both sides by sin(ak) to get -C=C. Then I can divide both sides by -C to get 1 = -1. This is a contradiction, so C must be equal to 0.Starting from -Csin(ak)=Csin(ak), I can divide both sides by sin(ak) to get -C=C. Then I can divide both sides by -C to get 1 = -1. This is a contradiction, so C must be equal to 0.In summary, there is a contradiction in the equation -Csin(ak
  • #1
Bestphysics112
24
2

Homework Statement


Hello,

I'm trying to solve laplaces equation to find a solution for the potential in a pipe with the given boundary conditions:

at x=b, V=V_0

at x= -b, V = -V_0

at y=a, V=0

at y=-a, V=0

(Assume this configuration is centered on the origin, pipe as dimensions -b<=x<=b -a<=x<=a, infinite in z direction)

I have general form of solution (with exponential and trig combinations, I found this in griffith's section on separation of variables). I'm stumped on how to obtain more information from the boundary conditions. Here is what I'm referring to

https://imgur.com/a/wpKAY

For example, If the potential is 0 at y=a (which we are given), what exactly does this tell us for the coefficients? In the text, the example deals with instances in which one of the boundaries is the origin (x and y at 0), so I can eliminate a coefficient. But with the boundary at a non zero x and y coordinate, what should I do? Please don't give the full solution, but please give me a hint of sorts. Sorry if my english is poor, If my question doesn't make sense please say so and I will try to fix

Homework Equations


https://imgur.com/a/wpKAY (Potential expression obtained for cartesian coordinates.)

The Attempt at a Solution


Ok, here is my attempt at the solution. For the last two boundary conditions we are given, we can find that coefficient D=-D or C=-C. I'm not sure how to proceed. I've done similar problems where at least two of the boundary conditions lie on the origin (i.e. at x = 0 or y = 0). If it was like this, it is easy to see that a certain coefficient will be 0 and so forth. If, after plugging in y = +- a, I get the mentioned expressions, how can I determine what value C and D are? I asked someone and they said that C (or D) is zero but how can I determine which is 0? Because surely they both cannot be 0. A hint would be greatly appreciated.
 
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  • #2
Bestphysics112 said:
For the last two boundary conditions we are given, we can find that coefficient D=-D or C=-C.
How do you deduce that? What trig function would be zero at -a and +a?
 
  • #3
haruspex said:
How do you deduce that? What trig function would be zero at -a and +a?

My reason is as follows - If I plug y = a into (Csin(ky)+Dcos(ky)), I can solve for C. Then if I plug in y = -a and my value for C, I can get -Dcos(ak) = Dcos(ak) (because sin is odd function and cos is even function) I'm not sure if my reasoning is correct so feedback is appreciated. I'm not sure which trig function would be zero at -a and +a, can I get a hint? Is it cos because cos(0) = 1 so cos at the boundaries would be 0?
 
  • #4
Bestphysics112 said:
get -Dcos(ak) = Dcos(ak)
And from that you deduce D=-D? What is another possibility?
 
  • #5
haruspex said:
And from that you deduce D=-D? What is another possibility?

If I do the same but instead solve for C, I can find that C= -C.
 
  • #6
Bestphysics112 said:
If I do the same but instead solve for C, I can find that C= -C.
That does not answer my question. Read it again.
 
  • #7
haruspex said:
That does not answer my question. Read it again.

Another possibility is D=0? I'm unsure of what else it could be
 
  • #8
Bestphysics112 said:
Another possibility is D=0? I'm unsure of what else it could be
If xy=0, what are the possibilities?
 
  • #9
haruspex said:
If xy=0, what are the possibilities?

Either x = 0 or y = 0

Edit. I think I see what your point is. The other possibility would be that cos(ak) = 0, meaning k = n*pi/(2a)
 
  • #10
Bestphysics112 said:
Either x = 0 or y = 0
Right. Apply that principle to the equation I quoted in post #4.
 
  • #11
haruspex said:
If xy=0, what are the possibilities?

haruspex said:
Right. Apply that principle to the equation I quoted in post #4.
I think I see what your point is. The other possibility would be that cos(ak) = 0, meaning k = n*pi/(2a)
 
  • #12
Bestphysics112 said:
I think I see what your point is. The other possibility would be that cos(ak) = 0, meaning k = n*pi/(2a)
Right.
 
  • #13
haruspex said:
Right.
How does this fact help me in determining which (C or Do) is 0? Sorry for all the questions!
 
  • #14
Bestphysics112 said:
How does this fact help me in determining which (C or Do) is 0? Sorry for all the questions!
Why do you keep writing "C or D is zero"? For each you had an equation which (you thought) implied it was zero, so surely you thought you had shown they were both zero.
Now you have seen that D need not be zero. What about C?
 
  • #15
haruspex said:
Why do you keep writing "C or D is zero"? For each you had an equation which (you thought) implied it was zero, so surely you thought you had shown they were both zero.
Now you have seen that D need not be zero. What about C?

C need not be zero as well, so there must be another way to determine coefficients
 
  • #16
Bestphysics112 said:
C need not be zero as well,
Please post you complete reasoning for that.
 
  • #17
haruspex said:
Please post you complete reasoning for that.
If I follow the steps illustrated in post 3 but instead solve for d, I obtain -Csin(ak)=Csin(ak), so two possibility is C=0 or sin(ak)=0, so k=n*pi/a for this instance
 
  • #18
Can you sketch what you think the solution is going to look like in terms of the lines of constant V?
 
  • #19
Bestphysics112 said:
If I follow the steps illustrated in post 3 but instead solve for d, I obtain -Csin(ak)=Csin(ak), so two possibility is C=0 or sin(ak)=0, so k=n*pi/a for this instance
Still not seeing how you get that. Pleaee show all the steps.
 

1. What is Laplace's equation?

Laplace's equation is a mathematical equation that describes the potential within a region of space. It is commonly used in physics and engineering to solve problems involving electric fields, heat transfer, and fluid flow.

2. How is Laplace's equation used to find potential within a pipe?

Laplace's equation can be applied to find the potential within a pipe by considering the boundary conditions at the walls of the pipe. By solving the equation for the given boundary conditions, the potential can be determined at any point within the pipe.

3. What are the assumptions made when using Laplace's equation to find potential within a pipe?

The assumptions made when using Laplace's equation include the pipe being a perfect conductor, the potential being constant along the walls of the pipe, and the absence of any sources or sinks within the pipe.

4. Can Laplace's equation be used to find potential within a non-cylindrical pipe?

Yes, Laplace's equation can be applied to find potential within a non-cylindrical pipe as long as the boundary conditions and assumptions remain valid. However, the mathematical solution may be more complex and require numerical methods.

5. Are there any limitations to using Laplace's equation to find potential within a pipe?

Yes, there are some limitations to using Laplace's equation. It assumes a steady state condition and does not account for time-dependent effects. Additionally, it may not be applicable to highly irregular or non-symmetric pipes.

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