Poisson's Equation: Solving for φ with ρ(x)

[aaaa202]

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?

 

[Incand]

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.

 

[Chandra Prayaga]

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.

 

[aaaa202]

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

 

[Incand]

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.

 

[aaaa202]

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.