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Using Laplace Equation in one dimension to solve for a charge-free slab

  1. Oct 25, 2012 #1
    1. The problem statement, all variables and given/known data
    An empty (charge-free) slab shaped region with walls parallel to the yz-plane extends from x=a to x=b; the (constant) potential on the two walls is given as Va and Vb , respectively. Starting with LaPlace's equation in one dimension, derive a formula for the potential at any point in the region


    2. Relevant equations


    d2V/dx2=0

    3. The attempt at a solution

    I get the whole concept how they get the average potential since if we were to integrate twice laplace equation we would get two integration constant

    d2V/dx2=0

    dV/dx= a
    V = ax + b

    and if we were to average from V(x+a) and V(x-a) we would get

    V (x) =1/2[V (x + a) + V (x-a)] and you can use for any point within

    how did the answer to this problem came out to be

    V(x)=[Vb-Va)x+Vab-Vba]/b-a

    ??
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 25, 2012 #2
    You are using the symbols a and b to represent two different things. If you represent the potential by

    V = Cx + D

    and use the two boundary conditions to solve for C and D, you get

    V = Va + (Vb - Va) (x -a)/(b -a)

    which is the same as the answer you were trying to match.
     
  4. Oct 26, 2012 #3
    So if at we have Q+ at a and Q- at b, from a to b Vb=0 and at a the potential will be Va or we have a distance of d from a to b, Va=Qd/Aε0.

    So V(x)=C(x)+D
    at V(0)=0+D
    D=Vb

    but you got D as Va?

    How did start solving for C? What boundary conditions?
     
  5. Oct 26, 2012 #4
    Sorry, I totally forgot that is a charge free slab. Therefore I'm still stuck on the boundary conditions

    How come your results is equal to the answer? Va is not part of it
     
  6. Oct 26, 2012 #5
    Va is the voltage at coordinate x=a.

    V = Cx + D
    Va = Ca + D
     
  7. Oct 26, 2012 #6
    This is no longer a physics problem. It has been boiled down to strictly a math problem. The question is, are you capable of solving the following second order ordinary differential equation:

    d2V/ dx2 = 0

    subject to the following split boundary conditions:

    V = Va at x = a

    V = Vb at x = b

    Well,....can you?
     
  8. Oct 30, 2012 #7
    Yes I can

    Since we have the boundary conditions

    V = Va at x = a

    V = Vb at x = b

    Va=Ca+D

    Vb=Cb+D

    Set D=Va-Ca

    Solve for C on Vb

    Then solve for D. Then just plug in Vx

    Thanks a lot!!
     
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