Question 4.9: 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, and x_{2} = 1. Using the central difference method with a step size h = 0.5, we write Laplace’s equation as
\frac{d^{2} V}{dx^{2}} |_{x_{1}} = \frac{V_{2} – 2 V_{1} + V_{0}}{0.5^{2}} = 0The boundary conditions imply V_{0} = V(x_{0} = 0) = 0 and V_{2} = V(x_{2} = 1) = 3. Hence
V_{1} = 0.5 (V_{0} + V_{2}) = 1.5which demonstrates the principle of the average value for the middle point x_{1}. The second iteration with the smaller steps size h = 0.25 is applied to fi ve points in the same interval. The boundary conditions, which are now V_{0} = 0 and V_{4} = 3, lead to three simultaneous equations
V_{1} = 0.5 (V_{0} + V_{2}); V_{2} = 0.5 (V_{1} + V_{3}); V_{3} = 0.5 (V_{2} + V_{4});
In this case, the boundary conditions specify V_{0} and V_{4}, and the voltage V_{2} was calculated in the previous iteration. The solutions 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
A comparison of these computed values is in agreement with the analytical solution V(x) = 3x obtained in Example 4-6. The MATLAB calculation produces the following results:
| V= | 0 | NaN | 1.5000 | NaN | 3.0000 |
| V= | 0 | 0.7500 | 1.5000 | NaN | 3.0000 |
| V= | 0 | 0.7500 | 1.5000 | 2.2500 | 3.0000 |
The voltage distribution is shown below.
