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I Poissons equation

  1. Sep 11, 2016 #1
    Poissons equation states that:

    2φ = -ρ/ε

    Now suppose that the charge density is actually only a function of one coordinate ρ = ρ(x) but constant in y and z. Is the problem then equivalent to solving:

    d2φ/dx^2 = -ρ(x)/ε

    or what will the effect of the partial derivatives of y and z be in this case?
  2. jcsd
  3. Sep 11, 2016 #2
    No that's not generally true. Consider for example ##\rho(x) = x## then ##\varphi(\vec r) = -\frac{y^2}{2\epsilon}x## would be a solution.

    To solve a problem like this can be quite complicated and depends on the boundary and initial conditions. You can learn how to solve boundary problems like this in a book/course about Fourier analysis.
  4. Sep 11, 2016 #3
    Any charge, whatever its dependence on the coordinates may be, affects the potential at any point in space. The Poisson equation that you wrote is not a vector equation. So you cannot just take the x dependence on either side and say that they are equal. Each partial derivative on the left is related to the entire right hand side, and so is their sum.
  5. Sep 11, 2016 #4
    Okay but I have a numerical problem where I am given ρ(x,y,z) = ρ(x). To solve for the electrostatic potential I then discretize ρ(x) on a grid of n points and approximate the second order derivative D = d2/dx2 as a matrix in the standard way using finite differences. I then calculate the electrostatic potential as:

    φ = [d2/dx2]-1(-ρ/ε), where [d2/dx2]-1 is the inverted matrix of D written above.

    Will this in general not give the correct solution? I guess not since, I am assuming that Poissons equation can be written as:

    d2φ/dx2 = -ρ/ε
  6. Sep 11, 2016 #5
    Is this an exercise? Maybe it should be in the exercise section then with all information provided. I don't know anything about numerical methods but if you assume the equation of that form doesn't mean you necessarily get a wrong solution. You simply only get a subset of all possible solutions. But imo the question doesn't make much sense without having boundary conditions.
  7. Sep 11, 2016 #6
    The boundary condition is that the potential should vanish at infinity. It is not an exercise. Rather part of a script I am writing for solving Poissons equation.
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