You can calculate the electric potential over every point in a defined space by solving Laplace's equation. To do this in a computer program you set up an 2-d array/ matrix and loop the internal points applying the following formula:

$$\phi(x,y)= \frac{\phi(x_{i-1},y_j)+\phi(x_{i+1},y_j)+\phi(x_i,y_{j-1})+\phi(x_i,_{j+1})}{4}$$

Meaning that the electric potential at $\phi(x,y)$ is just the average of the 4 neighbouring points. Included in the array are areas of constant potential i.e. the algorithm does not alter them. These are the boundary conditions (the outer points of the grid/ matrix, set to 0) and the two plates (at a user defined location, set to -1 and 1). Now with the grid set up as it is currently: evenly spaced in both x and y direction, only a parallel plate capacitor can be modelled. If i try to put the plate at an incline it will just look like a stair case.

My question is, how can i set this up in terms of a computer program in order to calculate the electric potential over all points with one plate rotated at an arbitrary angle $\theta$?



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What you're describing is a very specific way to solve the Laplace equation: using the Jacobi iteration to solve the five-point stencil when using finite differences. But there are much more general methods that would be difficult to derive from your starting point. A more promising approach is to start with the Poisson equation and ask how it can be approximated with finitely many points. One way would be to apply the finite element method using an unstructured grid that allows you not only to consider inclined plates but, in fact, any kind of geometry. This then gives you a finite dimensional linear system that you can, if you want, invert using the Jacobi iteration, but again, if you start with the linear system you will realize that there are many other methods that allow you to solve it: Direct solvers such as Gaussian elimination or LU decomposition, or if you want to exploit the fact that the system is sparse, solvers like Conjugate Gradients or multigrid.


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