这是一份southampton南安普敦大MATH1055W1-01作业代写的成功案例
If a square plane plate is considered as an example, we should remember that the potential in a finite and very thin plane plate can be evaluated by (Balanis, 1990):
$$
V(x, y, z=0)=\frac{1}{4 \pi \varepsilon} \int_{-a}^{a} \mathrm{~d} x^{\prime} \int_{-b}^{b} \mathrm{~d} y^{\prime} \frac{\rho\left(x^{\prime}, y^{\prime}\right)}{\left[\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}\right]^{1 / 2}}
$$
Thus, after applying the method of the moments, knowing the function of the approximated solution $f(x, y)$, the expansion function $g(x, y)$ and the weighed function $W(x, y)$, the potential in a square plane plate will be estimated by the inner product of these functions:
$$
V(x, y)=\langle g, W, f\rangle \frac{1}{R}=\int_{-a}^{a} \frac{g(x, y) W(x, y) f(x, y)}{R(x, y)} \mathrm{d} x
$$
where
$$
R(x, y)=\sqrt{\left(x-x^{\prime}\right)^{2}+\left(y-y^{\prime}\right)^{2}}
$$
Dividing the plate in equal segments and applying the weighed function as being the Dirac delta function, we had that $W_{m}=\delta\left(x-x_{m}\right) \delta\left(y-y_{m}\right)$, being the inner product in the point given by:
$$
V(x, y, z=0)=\left\langle W_{m}, f, L g\right\rangle
$$
MATH1055W1-01 COURSE NOTES ：
$$
H_{z} \sim 1+k_{0,1} W_{\mathrm{TE}}
$$
where
$$
k_{0, \perp} \sim \omega\left[\varepsilon_{0} \varepsilon_{\mathrm{r}} \mu_{0} \mu_{\mathrm{r}}-\frac{1}{2}\left(\frac{\gamma}{\omega}\right)^{2}\right]
$$
is the quasi-static wavenumber.
On the other hand, in the T.M. case it is shown that
$$
E_{2} \sim 1+k_{0, \perp} W_{\mathrm{TM}}
$$
where
$$
\Delta W_{\mathrm{TM}}=0, \quad \text { in } Y \backslash C
$$