Electric Potentials and Fields

未分类
Author:GUO Xiao

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Abstract

This article will solve the electric potential and electric field of figure 5.7 and compare 3 different methods to solve this problem.

Background

In regions of space that don't contain any electric charges,the electric potential obeys Laplace's equation

$$\Delta V=0$$ or $$\frac{\partial^2V}{\partial x^2} +\frac{\partial^2V}{\partial y^2}+\frac{\partial^2V}{\partial z^2}=0$$
This is partial differential equation (PDE).To solve this equation, we will use relaxation method.

Problem

Exercise 5.7
When boundary condition is given as shown in following figure:

We can solve the distribution of electric potential by numerical method.

Numerical Calculation

$$\frac{\partial^2V}{\partial x^2}\approx\frac{V(i+1,j,k)+V(i-1,j,k)-2V(i,j,k)}{(\Delta x)^2}$$
For two dimensional problem, Laplace's equation can be written as
$$V(i,j)=\frac{1}{4}(V(i+1,j)+V(i-1,j)+V(i,j+1)+V(i,j-1))$$

Solution

program code
potential.py This program used Jacobi method.

Different methods

Jacobi method

$$V_{new}(i,j)=\frac{1}{4}(V_{old}(i+1,j)+V_{old}(i-1,j)+V_{old}(i,j+1)+V_{old}(i,j-1))$$

Gauss-Seidel method

$$V_{new}(i,j)=\frac{1}{4}(V_{old}(i+1,j)+V_{new}(i-1,j)+V_{old}(i,j+1)+V_{new}(i,j-1))$$

Simultaneous over-relaxation method (SOR method)

$$V^*(i,j)=\frac{1}{4}(V_{old}(i+1,j)+V_{new}(i-1,j)+V_{old}(i,j+1)+V_{new}(i,j-1))$$
$$\Delta V(i,j)=V^*(i,j)-V_{old}(i,j)$$
$$V_{new}(i,j)=\alpha\Delta V(i,j)+V_{old}(i,j)$$
The best choice for $\alpha$ is $$\alpha\approx\frac{2}{1+\frac{\pi}{L}}$$
SOR.py
This is a program to solve this problem, which uses SOR method and prints almost same results with Jacobi method.

Comparison

Next, I use this porgram problem5.7.py to obtain the data which describes the relationship between numbers of iterations and L.

The comparison of 3 different methods at the same accuracy:

We can see that under the same L, the convergent speeds of SOR method is faster than SOR method, which is faster than that of Jacobi method. (The smaller number of iterations is, the faster convergent speed is.)
Concretely,

Jacobi method

Fit equation:$N=a*L^2$
$a=0.2849$
$R^2=0.9834$
The number of iterations N is about proportional to $L^2$.

SOR method

Fit equation:$N=a*L$
$a=1.531$
$R^2=0.9949$
The number of iterations N is approximately proportional to $L$.

Calculation Error

Because most elements of the array of V are very small, the difference $\Delta V$ between $V_{old}$ and $V_{new}$ will approach 0 very much.
This a part of array $V_{new}-V_{old}$.

Due to error of calculation, all elements of the array $\Delta V$ are easily regarded as 0 by computer. This may lead to some bugs.
This problem can be solved by beginning calculation from a random array, not all-zero array.

Conclusions

• I solved the electric potential and field of exercise 5.7.
• Comparing 3 different methods, the convergent speeds:SOR method > GS method > Jacobi method.
• At the same accuracy, using Jacobi method, the number of iterations N is about proportional to $L^2$;while using SOR method, the number of iterations N is about proportional to $L$.

Acknowledgement

There is a little bug shortly before.Thanks to Chen Yangyao classmate, who helped me to modify my faults.