Question 3.6: Find the potential distribution between two surfaces if V(x...
Find the potential distribution between two surfaces if V(x = 0) = 0 and V(x = 1) = 3. There is no charge distribution in the space 0 ≤ x ≤ 1.
Let us use only three points for the first iteration: x_{0}=0, x_{1}=0.5 , x _{2}=1 . Using the Central difference method with a step size h=1 / 2 , we write Laplace’s equation as
\left.\frac{d^{2} V}{d x^{2}}\right|_{x 1}=\frac{V_{2}-2 V_{1}+V_{0}}{0.5^{2}}=0
The boundary conditions imply V _{0}= V \left( x _{0}=0\right)=0 and V _{2}= V \left( x _{2}=1\right)=3 . Hence
V_{1}=0.5\left(V_{0}+V_{2}\right)=1.5
which demonstrates the principle of the average value for the middle point x1 .
The second iteration with the smaller steps size h=1 / 4 is applied to five points in the same interval. The same boundary conditions are now V0 = 0 and V4 = 3 leads to three simultaneous equations
V _{1}=0.5\left( V _{0}+ V _{2}\right) ; V _{2}=0.5\left( V _{1}+ V _{3}\right) ; V _{3}=0.5\left( V _{2}+ V _{4}\right)
In this case, the boundary conditions specify V0 and V4 and the voltage V2 was calculated in the previous iteration.
The solution for the three intermediate points and the two end points are
V _{0}=0 ; V _{1}=0.75 ; V _{2}=1.50 ; V _{3}=2.25 ; V _{4}=3 \text {. }
A comparison of these computed values is in agreement with the analytical solution obtained in Example 3-2. The MATLAB calculation produces the following results.
\begin{array}{|l|l|l|l|l|l|}\hline V = & 0 & NaN & 1.5000 & NaN & 3.0000 \\\hline V = & 0 & 0.7500 & 1.5000 & NaN & 3.0000 \\\hline V = & 0 & 0.7500 & 1.5000 & 2.2500 & 3.0000 \\\hline\end{array}
The voltage distribution is shown below.
