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Homework Help: Laplacian for hyperbolic plates

  1. Dec 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Find a complete set of conditions on the constants a, b, c, n such that, for Cartesian coordinates (x, y, z), V = axn + byn + czn is a solution of Laplace’s equation ##∇^2V = 0##. A mass filter for charged particles consists of 4 electrodes extended along the z direction, whose surfaces describe hyperbolae in the xy plane: ##x^2 − y^2 = d^2## for one pair, and ##x^2 − y^2 = −d^2## for the other pair, where d is the distance from the axis to the nearest point on an electrode surface. [Each hyperbola has two branches and thus describes two (diagonally opposite) electrodes.] A positive voltage ##V_0## is applied to the pair at ##x^2 − y^2 = d^2##, and ##−V_0## is applied to the other pair. Taking the electrode length along the z axis to be effectively infinite, find the electric potential in the region between the electrodes.


    2. Relevant equations


    3. The attempt at a solution
    So the boundary conditions i get are
    ##V(x,\sqrt{x^2-d^2},z)=V_0##
    ##V(x,\sqrt{x^2+d^2},z)=-V_0##
    ##\nabla^2V=0##

    for solutions of the form ##V=\summation a_nx^n+b_ny^n+c_nz^n ## ##c_n=0##

    When I use all of these boundary contains i find that we must only have odd powers of x and that

    ##\summation a_n x^n +b_n(x^2-d^2)^{\frac{n}{2}}=V_0##
    ##\summation a_n x^n +b_n(x^2+d^2)^{\frac{n}{2}}=-V_0##

    but I can't see how to formulate a solution given these as my coefficients will depend on x .

    Many thanks
     
  2. jcsd
  3. Dec 23, 2017 #2

    TSny

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    Begin with the first task:
     
  4. Dec 23, 2017 #3
    ##ax^{n-2}+by^{n-2} +cz^{n-2}=0 ## ?
    But this would imply
    ##ax^{n-2}=k_x##
    ##by^{n-2}=k_y##
    ##cz^{n-2}=k_z##
    ?
     
  5. Dec 23, 2017 #4

    TSny

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    OK
    Is ##k_x## some constant? If so, how can you make ##ax^{n-2}## a constant?
     
  6. Dec 23, 2017 #5
    Only if we have n=0?
     
  7. Dec 23, 2017 #6

    TSny

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    If n = 0 then ##ax^{n-2} = ax^{-2} = a/x^2##.
     
  8. Dec 23, 2017 #7
    sorry n=2?
     
  9. Dec 23, 2017 #8

    TSny

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    Right.

    There is no need to introduce the notation ##k_x##, ##k_y##, ##k_z## .

    You found that you need the relation ##ax^{n-2}+by^{n-2} +cz^{n-2}=0 ## to be satisfied. And this must be satisfied for all points (x, y, z) in the region where Laplace's equation holds. Thus, the left hand side of this relation must be independent of the choice of (x, y, z). This can only happen for one value of n.
     
  10. Dec 23, 2017 #9
    Right Then for the next part i have ##V_0=ax^2+b(x^2-d^2)## and ##-V_0=ax^2+b(x^2+d^2)##

    ##2V_0=-2bd^2##
    ##b=-\frac{V_0}{d^2}##

    and considering the point (d,0)
    ##a=\frac{V_0}{d^2}## ?
     
    Last edited: Dec 23, 2017
  11. Dec 23, 2017 #10

    TSny

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    There's more required for the first part. You've found that n must equal 2. Are there any restrictions on a, b, and c?
     
  12. Dec 23, 2017 #11
    a+b+c=0 ?
     
  13. Dec 23, 2017 #12

    TSny

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    Yes. So, at this point, how would you write the function V(x, y, z) in the region where Laplace's equation holds?
     
  14. Dec 23, 2017 #13
    V=ax^2+by^2-(a+b)z^2 ?
     
  15. Dec 23, 2017 #14

    TSny

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    OK.

    Now, moving on to the particular problem with the hyperbolic electrodes, can you say anything about how V(x, y, z) should depend on z?
     
  16. Dec 23, 2017 #15
    No dependence on z
     
  17. Dec 23, 2017 #16

    TSny

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    Good.
     
  18. Dec 23, 2017 #17
    ##V=ax^2-ay^2##

    ##2V_0=-2bd^2##
    ##b=-\frac{V_0}{d^2}##

    and considering the point (d,0)
    ##a=\frac{V_0}{d^2}## ?
     
  19. Dec 23, 2017 #18

    TSny

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    That's a bit of a jump for me. Does the independence of V on z imply any restriction on the constants ##a## and ##b## ?

    At this point, what's the simplest way to write V(x, y, z) in the region where Laplace's equation holds. By simplest, I mean write V with the least number of constants.
     
  20. Dec 23, 2017 #19
    a=-b
    ##V=a(x^2-y^2)##
     
  21. Dec 23, 2017 #20

    TSny

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    Yes. I now see that you had written ##V=ax^2-ay^2## in post #18, which I overlooked. Good. And you also correctly found the value of ##a##.

    So, what's your final answer for V(x, y, z)?
     
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