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Poisson's equation to find V

  1. Oct 13, 2008 #1

    E&M

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

    Two infinite parallel plates separated by a distance s are at potential 0 and V_0
    a) Use Poisson's eqn to find potential V in the region between the plates where the space charge density is rho = rho_0(x/s). The distance x is measured from the plate at 0 potential.
    b)What are the charge densities in the plate?

    2. Relevant equations
    [tex]\nabla[/tex][tex]^2{}[/tex] V = - rho/ epsilon_0


    3. The attempt at a solution
    For this problem, I started with number of ways but none of them seem to be working. What will be V in this case?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2008 #2

    gabbagabbahey

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    Why don't you show me what you've tried?
     
  4. Oct 13, 2008 #3

    E&M

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    V will be in only x direction and it won't be in y and z direction. double partial derivative of V along x will be equal to -rho/epsilon_0. I was thinking of using differential equation but I am not sure.
     
  5. Oct 13, 2008 #4

    gabbagabbahey

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    You will need to use a differential equation. Luckily if V is only a function of x, then

    [tex]\frac{\partial ^2V}{\partial x^2}=\frac{d^2V}{dx^2}[/tex]

    And so you will have an ordinary differential equation instead of a partial differential equation.

    Give integrating it a shot and show me what you get.
     
  6. Oct 13, 2008 #5

    E&M

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    i got V = -(rho_0 * x^3)/(s* epsilon_0*6) and for charge densities I got rho = 0 at the plate with V = 0 and rho = rho_0 for the plate with V = V_0. does this look right?
     
  7. Oct 13, 2008 #6

    gabbagabbahey

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    Your V looks close, but there is a small error. On your first integration, you must include an integration constant:

    [tex]\frac{d^2V(x)}{dx^2}=\frac{-\rho_0x}{s \epsilon _0} \Rightarrow \int_{x'=0}^{x'=x} \frac{d^2V(x')}{dx'^2} dx'= \int_0^x \frac{-\rho_0x'}{s \epsilon _0}dx'[/tex]

    [tex] \Rightarrow V'(x)-V'(0)=\frac{-\rho_0x^2}{2s \epsilon _0} \Rightarrow V'(x)=\frac{dV(x)}{dx}=\frac{-\rho_0x^2}{2s \epsilon _0}+C[/tex]

    where [itex]C=V'(0)[/itex] is the constant of integration

    What do you get for V(x) when you don't forget the constant?
     
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