电磁学代写|Electromagnetism（Electricity and Magnetism）代考|高质保分代写电磁学作业

Riemann surface
matlab

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Solution:
(a) The charge density is given by Maxwell’s equation
$$
\rho=\nabla \cdot \mathbf{D}=\varepsilon_0 \nabla \cdot \mathbf{E} .
$$
As $\nabla \cdot u \mathbf{v}=\nabla u \cdot \mathbf{v}+u \nabla \cdot \mathbf{v}$,
$$
\nabla \cdot \mathbf{E}=A\left[\nabla\left(e^{-b r}\right) \cdot \frac{\mathbf{e}r}{r^2}+e^{-b r} \nabla \cdot\left(\frac{\mathbf{e}_r}{r^2}\right)\right] . $$ Making use of Dirac’s delta function $\delta(\mathbf{r})$ with properties $$ \begin{aligned} \delta(\mathbf{r}) & =0 \quad \text { for } \quad \mathbf{r} \neq 0, \ & =\infty \text { for } \quad \mathbf{=}=0, \ \int_V \delta(\mathbf{r}) d V= & \text { if } V \text { encloses } \quad r=0, \ & =0 \text { if otherwise, } \end{aligned} $$ we have $$ \nabla^2\left(\frac{1}{r}\right)=\nabla \cdot \nabla\left(\frac{1}{r}\right)=\nabla \cdot\left(-\frac{e_r}{r^2}\right)=-4 \pi \delta(r) $$ Thus $$ \begin{aligned} \rho & =\varepsilon_0 A\left[-\frac{v\epsilon^{-b r}}{r^2} \mathbf{e}_r \cdot e_r+4 \pi e^{-b r} \delta(\mathbf{r})\right] \
& =-\frac{\varepsilon_0 A b}{r^2} e^{-b r}+4 \pi \varepsilon_0 A \delta(\mathbf{r}) .
\end{aligned}
$$

Hence the charge distribution consists of a positive charge $4 \pi \varepsilon_0 A$ at the origin and a spherically symmetric negative charge distribution in the surrounding space, as shown in Fig. 1.1.

Fig. 1.1
(b) The total charge is
$$
\begin{aligned}
Q & =\int_{\text {all space }} \rho d V \
& =-\int_0^{\infty} \frac{\varepsilon_0 A b e^{-b r}}{r^2} \cdot 4 \pi r^2 d r+\int_{\text {all space }} 4 \pi \varepsilon_0 A \hat{i}(\mathbf{r}) d V^{\prime} \
& =4 \pi \varepsilon_0 A\left[e^{-b r}\right]_0^{\infty}+4 \pi \varepsilon_0 A \
& =-4 \pi \varepsilon_0 A+4 \pi \varepsilon_0 A=0 .
\end{aligned}
$$
It can also be obtained from Gauss’ flux theorem:

$$
\begin{aligned}
Q & =\lim {r \rightarrow \infty} \oint_S \varepsilon_0 \mathbf{E} \cdot d \mathbf{S} \ & =\lim {r \rightarrow \infty} \frac{\varepsilon_0 A e^{-b r}}{r^2} \cdot 4 \pi r^2 \
& =\lim _{r \rightarrow \infty} 4 \pi \varepsilon_0 A e^{-b r}=0,
\end{aligned}
$$
in agreement with the above.

(a) The electric field surrounding the point charge $q$ is
$$
\mathrm{E}(r)=\frac{q}{4 \pi \varepsilon_0} \frac{1}{r^2}(1-\sqrt{\alpha r}) \mathbf{e}r, $$ where $r$ is the distance between a space point and the point charge $q$, and $e_r$ is a unit vector directed from $q$ to the space point.

Fig. 1.2 (b) As in Fig. 1.2, for the closed path $L$ we find $$ d \mathbf{l} \cdot \mathbf{e}{\boldsymbol{r}}=d l \cos \theta=d r
$$
Problems $\&$ Solutions on Electromagnetism
and
$$
\begin{aligned}
\oint_L \mathbf{L} \cdot d l & =\oint \frac{q}{4 \pi \varepsilon_0} \frac{1}{r^2}(1-\sqrt{\alpha r}) d r \
& =\frac{q}{4 \pi \varepsilon_0}\left[-\oint_L d\left(\frac{1}{r}\right)+2 \sqrt{\alpha} \oint_L d\left(\frac{1}{\sqrt{r}}\right)\right]=0
\end{aligned}
$$
From Coulomb’s law $\mathbf{F}{12}=\frac{q 12}{4 \pi \varepsilon_0 r{12}^2} \mathbf{e}{\mathbf{r}{12}}$, we can obtain the electric field of the point charge
$$
\mathbf{L}(r)=\frac{q}{4 \pi \varepsilon_0 r^2} \mathbf{e}_r
$$
Clearly, one has
$$
\oint_L \mathrm{~L} \cdot d 1=0
$$
So the Coulomb result is the same as that of this problem.

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