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Laplace equation on two semi-infinite conductor planes

  • #1

Homework Statement



Two semi-infinite conductor planes (like this ∠ ) have an angle β at a constant potential.

Whats the potencial close to the origin?

Homework Equations



∇²V = (1/r)(∂/∂r){ r ( ∂V/∂r ) } + (1/r²)(∂²V/∂[itex]\theta[/itex]²)

The Attempt at a Solution



Trying for laplacian equation on polar coordinates. But I cant solve-it?



Need some help, please!
 
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Answers and Replies

  • #2
vela
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Show us what you tried.
 
  • #3
TSny
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You might try a separation-of-variables type solution: [itex]V = R(r)\Phi(\alpha)[/itex] and determine separate differential equations for [itex]R(r)[/itex] and [itex]\Phi(\alpha)[/itex].
 
  • #4


After some brain spanking, got to this:

Starting with u(r, θ) = ϕ(r) ψ(θ) and putting it in the Laplace Equation(polar)

ϕ''(r) ψ(θ) + (1/r) ϕ'(r) ψ(θ) + (1/r²)ϕ(r) ψ''(θ) = 0

Dividing by (1/r²) ϕ(r)ψ(θ) , I have:

r² ϕ''(r) ϕ(r) + rϕ'(r)ϕ(r) = ψ''(θ)ψ(θ) = λ

Then:

* r² ϕ''(r) + rϕ'(r) − λϕ(r) = 0 (*)
* ψ''(θ) + λψ(θ) = 0 (**)

I think the (**) comes to ψ(θ) = A cos µθ + B sin µθ.

Is it correct until here?

The (*) equation, I failed to solve yet...
 
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  • #5
TSny
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* r² ϕ''(r) + rϕ'(r) − λϕ(r) = 0 (*)
* ψ''(θ) + λψ(θ) = 0 (**)

I think the (**) comes to ψ(θ) = A cos µθ + B sin µθ.

Is it correct until here?
Look's good. How is [itex]\mu[/itex] related to λ?
The (*) equation, I failed to solve yet...
Try a solution of the form [itex]\phi(r) = r^{\alpha}[/itex] and see if it will solve the equation for appropriate values of [itex]\alpha[/itex].
 
  • #6


Look's good. How is [itex]\mu[/itex] related to λ?


Try a solution of the form [itex]\phi(r) = r^{\alpha}[/itex] and see if it will solve the equation for appropriate values of [itex]\alpha[/itex].

I've forgot to write this: λ = µ²

I''ll try to solve using your suggestion. I'll post it here as soon as I can...


Thanks!
 
  • #7


Lol.. it was very easy to solve the ODE... I think I can re write the last equations to:


* r² ϕ''(r) + rϕ'(r) − λ²ϕ(r) = 0 (*)
* ψ''(θ) + λ²ψ(θ) = 0 (**)

So, I'll have

ψ(θ) = A cos λθ + B sin λθ
ϕ(r) = r[itex]^{λ}[/itex]


My solution, so, would be V(r,θ) = r[itex]^{λ}[/itex] ( A cos λθ + B sin λθ )

The boundaries are

V(r, 0) = V(r,[itex]\alpha[/itex]) = V[itex]_{0}[/itex]

But I think I need one more... Can u tell me wich is?
 
  • #8
TSny
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ψ(θ) = A cos λθ + B sin λθ
ϕ(r) = r[itex]^{λ}[/itex]


My solution, so, would be V(r,θ) = r[itex]^{λ}[/itex] ( A cos λθ + B sin λθ )

The boundaries are

V(r, 0) = V(r,[itex]\alpha[/itex]) = V[itex]_{0}[/itex]

But I think I need one more... Can u tell me wich is?
You should have found that ϕ(r) = r[itex]^{-λ}[/itex] would also be a solution. You should be able to argue that this solution has to be thrown out. (Maybe you've already done that.)

For there to be a unique solution for V, I think there would have to be an additional boundary condition that specifies the behavior of V for large r. With the two boundary conditions that you have stated, you should nevertheless be able to find an approximate expression for V near the origin. The expression will have an undetermined constant, but it will at least show you the behavior of V near the origin. The undetermined constant is due to not knowing the boundary condition for V for large r.
 
  • #9


You should have found that ϕ(r) = r[itex]^{-λ}[/itex] would also be a solution. You should be able to argue that this solution has to be thrown out. (Maybe you've already done that.)

For there to be a unique solution for V, I think there would have to be an additional boundary condition that specifies the behavior of V for large r. With the two boundary conditions that you have stated, you should nevertheless be able to find an approximate expression for V near the origin. The expression will have an undetermined constant, but it will at least show you the behavior of V near the origin. The undetermined constant is due to not knowing the boundary condition for V for large r.
Yes. I have just find out that I should have ϕ(r) = r[itex]^{−λ}[/itex].

Thanks again!


PS: I need to finish this and other questions in 1 hour at least. Back to work!
 
  • #10
TSny
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Well, I overlooked another possible solution to the differential equation for ϕ(r) when λ happens to be zero. This solution is not of the form [itex]r^{\alpha}[/itex], but when you find it you will be able to give a reason for discounting this solution.
 
  • #11


You're right. I'll will impose that lim[itex]_{r→∞}[/itex] V(r,[itex]\theta[/itex]) = 0
 
  • #12
TSny
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You're right. I'll will impose that lim[itex]_{r→∞}[/itex] V(r,[itex]\theta[/itex]) = 0
You would have to set the limiting value at infinity be the value of V for the conducting planes if the planes are not at zero potential. Now, imposing this boundary condition will be severe. Your solution for V will then be pretty trivial.

Nevertheless, if the conducting planes are truly infinite in extent, then I don't see any other reasonable boundary condition! I believe the intent of the question was probably to determine the behavior of the potential near the origin for "large" plates that don't necessarily extend to infinity and for which no boundary conditions are specified for large r.
 
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  • #13


You would have to set the limiting value at infinity be the value of V for the conducting planes if the planes are not a zero potential. Now, imposing this boundary condition will be severe. Your solution for V will then be pretty trivial.

Nevertheless, if the conducting planes are truly infinite in extent, then I don't see any other reasonable boundary condition! I believe the intent of the question was probably to determine the behavior of the potential near the origin for "large" plates that don't necessarily extend to infinity and for which no boundary conditions are specified for large r.
Or maybe,

lim [itex]_{r→ -∞ }[/itex]V(r,θ) = 0
 
  • #14
TSny
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Or maybe,

lim [itex]_{r→ -∞ }[/itex]V(r,θ) = 0
No, r is non-negative by definition.
 
  • #15
Yes. You are right.


Guess that what I first meant to say is that

lim r→−∞ V(r, α/2 ± ε) = 0 , ε ≥ 0

In some area around the two planes where r→∞, V→0;


The other option would be V→∞ when r→∞. I dont think so, but I cant strongly argue the reason.
 

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