这是一份bath巴斯大学PH40073作业代写的成功案

Sometimes the determination of the gradient is straightforward. Given
$$
f(x, y, z)=\frac{x y}{z},
$$
we easily find
$$
\frac{\partial f}{\partial x}=\frac{y}{z}, \quad \frac{\partial f}{\partial y}=\frac{x}{z}, \quad \frac{\partial f}{\partial z}=-\frac{x y}{z^{2}}
$$
leading to $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$

PH40073 COURSE NOTES ：

Dropping the factor $q / 4 \pi \varepsilon_{0}$, and taking
$$
E_{x}=\frac{x}{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}},
$$
we get
$$
\frac{\partial E_{x}}{\partial x}=\frac{1}{r^{3}}-\frac{3}{2} \frac{2 x^{2}}{r^{5}}=\frac{r^{2}-3 x^{2}}{r^{5}}
$$
Adding to this the analogous expressions for $\partial E_{y} / \partial y$ and $\partial E_{z} / \partial z$, we get
$$
\nabla \cdot \mathbf{E}=\frac{r^{2}-3 x^{2}}{r^{5}}+\frac{r^{2}-3 y^{2}}{r^{5}}+\frac{r^{2}-3 z^{2}}{r^{5}}=0 .
$$