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# 物理代写|电磁学代写Electromagnetism代考|Phys132 Point Charge on the Axis of a Dielectric Cylinder

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## 物理代写|电磁学代写Electromagnetism代考|Point Charge on the Axis of a Dielectric Cylinder

We want to find the potential caused by a point charge $Q$ at the axis of a dielectric, circular cylinder $\left(\varepsilon_1\right)$. For the remaining space, we assume a different dielectric $\left(\varepsilon_2\right)$ (see Fig. 3.17).
This is by no means a trivial problem, but it can be solved with the results obtained from the previous example. We need to solve Laplace’s equation in region 2, what is certainly possible with an approach similar to (3.185):
$$\varphi_2=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z) K_0(k r) g_2(k) d k .$$
The factor in front of the integral is just for convenience and could as well be regarded as belonging to $g_2(k)$. In region 1, we have to solve Poisson’s equation for a point charge. This may be done by superposition of the point charge potential according to (3.203) and the general solution of Laplace’s equation according to (3.185), i.e., by
$$\varphi_1=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z)\left[K_0(k r)+I_0(k r) g_1(k)\right] d k$$
Remember that the general solution of the inhomogeneous (Poisson) equation is obtained from the superposition of the specific solution of the inhomogeneous equation and the general solution of the homogeneous (Laplace) equation. Significant is now, to know the potential of the point charge in the specific form that fits this problem. Another way to view the trial functions of (3.205) and (3.206) is by superposition of the point charge potential and the potential of bound surface charges at the cylinder surface (wall). The following boundary conditions apply for $r=r_0$ :

\begin{aligned} & \left(\frac{\partial \varphi_1}{\partial z}\right){r=r_0}-\left(\frac{\partial \varphi_2}{\partial z}\right){r=r_0}=0 \ & \varepsilon_1\left(\frac{\partial \varphi_1}{\partial r}\right){r=r_0}-\varepsilon_2\left(\frac{\partial \varphi_2}{\partial r}\right){r=r_0}=0 . \end{aligned}
This gives
$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \sin (k z)\left[K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)\right] d k=0$$
and
$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \cos (k z)\left[-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k)\right] d k=0$$
or
\begin{aligned} K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k) & =0 \ -\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k) & =0 \end{aligned}

## 物理代写|电磁学代写Electromagnetism代考|Dirichlet’s Boundary Value Problem and the Fourier-Bessel Series

Consider the cylinder shown in Fig. $3.20$ with radius $r_0$ and height $h$. We want to find the potential inside the charge-free cylinder with the following boundary conditions.
$$\begin{array}{ll} \varphi=\varphi_h(r) & \text { for } z=h \ \varphi=0 & \text { on the remaining surface. } \end{array}$$
This is the cylindrical analogue to the example in Section 3.5.2.1. Again, $m=0$ because of the rotational symmetry. For the radial part $R$, we may choose either $J_0$ or $I_0$, but not $N_0$ or $K_0$ since those let the potential at the axis diverge. The potential has to also vanish for $r=r_0 . J_0$ has zeros for real arguments, $I_0$ does not. The convenient choice is thus $J_0$, For the $z$-dependency, we may choose an approach, for example, based on (3.158) where only the $\sinh$ is an option because of the restriction $\varphi=0$ for $z=0$. We suggest that solving the problem is possible by superposing expressions of the form:
$$J_0(k r) \sinh k z \text {. }$$
Hereby, the condition
$$J_0\left(k r_0\right)=0$$

## 物理代写|电磁学代写Electromagnetism代考|Point Charge on the Axis of a Dielectric Cylinder

$$\varphi_2=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z) K_0(k r) g_2(k) d k .$$

$$\varphi_1=\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} \cos (k z)\left[K_0(k r)+I_0(k r) g_1(k)\right] d k$$

$$\left(\frac{\partial \varphi_1}{\partial z}\right) r=r_0-\left(\frac{\partial \varphi_2}{\partial z}\right) r=r_0=0 \quad \varepsilon_1\left(\frac{\partial \varphi_1}{\partial r}\right) r=r_0-\varepsilon_2\left(\frac{\partial \varphi_2}{\partial r}\right) r=r_0=0 .$$

$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \sin (k z)\left[K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)\right] d k=0$$

$$\frac{Q}{2 \pi^2 \varepsilon_1} \int_0^{\infty} k \cos (k z)\left[-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k)\right] d k=0$$

$$K_0\left(k r_0\right)+I_0\left(k r_0\right) g_1(k)-K_0\left(k r_0\right) g_2(k)=0-\varepsilon_1 K_1\left(k r_0\right)+\varepsilon_1 I_1\left(k r_0\right) g_1(k)+\varepsilon_2 K_1\left(k r_0\right) g_2(k) \quad=0$$

## 物理代写|电磁学代可Electromagnetism代荣|Dirichlet’s Boundary Value Problem and the Fourier-Bessel Series

$$\varphi=\varphi_h(r) \quad \text { for } z=h \varphi=0 \quad \text { on the remaining surface. }$$

$$J_0(k r) \sinh k z .$$

$$J_0\left(k r_0\right)=0$$

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