1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Laplace equation on two semi-infinite conductor planes

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data

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

    Whats the potencial close to the origin?

    2. Relevant equations

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

    3. The attempt at a solution

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



    Need some help, please!
     
    Last edited: Sep 18, 2012
  2. jcsd
  3. Sep 18, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Re: Laplace equation on tow semi-infinite conductor planes

    Show us what you tried.
     
  4. Sep 18, 2012 #3

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    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].
     
  5. Sep 18, 2012 #4
    Re: Laplace equation on tow semi-infinite conductor planes

    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...
     
    Last edited: Sep 18, 2012
  6. Sep 18, 2012 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    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].
     
  7. Sep 18, 2012 #6
    Re: Laplace equation on tow semi-infinite conductor planes


    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!
     
  8. Sep 18, 2012 #7
    Re: Laplace equation on tow semi-infinite conductor planes

    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?
     
  9. Sep 18, 2012 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    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.
     
  10. Sep 18, 2012 #9
    Re: Laplace equation on tow semi-infinite conductor planes

    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!
     
  11. Sep 18, 2012 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    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.
     
  12. Sep 18, 2012 #11
    Re: Laplace equation on tow semi-infinite conductor planes

    You're right. I'll will impose that lim[itex]_{r→∞}[/itex] V(r,[itex]\theta[/itex]) = 0
     
  13. Sep 18, 2012 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    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.
     
    Last edited: Sep 18, 2012
  14. Sep 18, 2012 #13
    Re: Laplace equation on tow semi-infinite conductor planes

    Or maybe,

    lim [itex]_{r→ -∞ }[/itex]V(r,θ) = 0
     
  15. Sep 18, 2012 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Re: Laplace equation on tow semi-infinite conductor planes

    No, r is non-negative by definition.
     
  16. Sep 19, 2012 #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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook