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Griffiths Problem 3.15

  1. Mar 10, 2008 #1
    1. The problem statement, all variables and given/known data
    This question refers to Griffiths E and M book.

    2. Relevant equations

    3. The attempt at a solution
    Obviously this is a 3D problem and we should use Example 3.5 as a model. However, since the region of interest is bounded in all 3 directions doesn't that mean that all of C_1, C_2, and C_3 need to be negative? But then C_1+C_2+C_3=0 is impossible! The constants C_i that I am referring to are on page 135.
  2. jcsd
  3. Mar 11, 2008 #2

    [tex]\nabla^2 V = 0[/tex]

    After separation of variables:

    [tex]\frac{1}{F}\frac{d^2 F}{dx^2} + \frac{1}{G}\frac{d^2 G}{dy^2} + \frac{1}{H}\frac{d^2 H}{dz^2} = 0[/tex]

    Argument: How can a function of x and y and z added together be zero always?
  4. Mar 11, 2008 #3
    That is my point! If
    [tex]\frac{1}{F}\frac{d^2 F}{dx^2}=C_1[/tex]
    [tex]\frac{1}{G}\frac{d^2 G}{dy^2}=C_2[/tex]
    [tex] \frac{1}{H}\frac{d^2 H}{dz^2} =C_3[/tex]

    Then C_1 +C_2+C_3 must be 0 which means that not all the C_i can be negative, which is absurd because we need trig functions not exponentials since the region is bounded in all three directions!?!?
  5. Mar 11, 2008 #4
    Absolutely, positive constants will give the trivial solution.

    Why not say -C1-C2-C3=0? Does it really matter whether the constants we make are negative or positive, aren't they allowing the functions to *sum* to zero in either case?
  6. Mar 11, 2008 #5


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    Why not exponentials? It seems like the potential will start at Vo at z = a, then decay (exponentially) down the z axis.
  7. Mar 11, 2008 #6
    Try to apply the boundary conditions to exponentials. It may be easier if you write them in terms of sinh and cosh in this case.
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