Question 4.10: Repeat Example 4-9 with a uniform charge distribution ρν = -...
Repeat Example 4-9 with a uniform charge distribution ρ_{ν} = – 4 ε_{0} in the space 0 ≤ x ≤ 1. Find the potential distribution between two surfaces if V(x = 0) = 0 and V(x = 1) = 3.
Using the central difference method, we write Poisson ‘s equation as
\frac{d^{2} V}{dx^{2}} |_{x_{1}} = \frac{V_{2} – 2 V_{1} + V_{0}}{0.5^{2}} = 4for the first iteration. The boundary conditions imply V_{0} = V(x = 0) = 0 and V_{2} = V(x = 1) = 3. Hence
\frac{V_{2} – 2 V_{1} + V_{0}}{0.5^{2}} = 4 (3 – 2 V_{1} + 0) = 4 \Rightarrow V_{1}= 1The second iteration with the boundary conditions V_{0} = V(x = 0) = 0 and V_{4} = V(x = 1) = 3 leads to
4(V_{2} – 2V_{1} +V_{0}) = 4; 4(V_{3} – 2V_{2} +V_{1}) = 4; 4(V_{4} – 2V_{3} +V_{2}) = 4;
The solutions for the three intermediate points and the two end points are
V_{0} = 0; V_{1} = 0.375 ; V_{2} = 1 ; V_{3} = 1.875 ; V_{4} = 3
The term V_{2} = 1 from the previous iteration.
Applying the same analytical technique as the one we used in Example 4-6, we find for the exact solution the following explicit expression: V(x) = 2x^{2} + x. Once again, the approximate solution found numerically agrees with the analytical solution. The output of the MATLAB program is
| V= | 0 | NaN | 1 | NaN | 3 |
| V= | 0 | 0.3750 | 1.0000 | NaN | 3.0000 |
| V= | 0 | 0.3750 | 1.0000 | 1.8750 | 3.0000 |
The plot of the results is shown below.
