这是一份UCL伦敦大学学院PHAS00010作业代写的成功案
In vector notation, the gradient of $V$ is written as either grad $V$ or $\vec{\nabla} V$. Then
$$
\vec{E}=-\vec{\nabla} V
$$
In general, the potential function can depend on $x, y$, and $z$. The Cartesian components of the electric field are related to the partial derivatives of the potential with respect to $x, y$, or $z$. For example, the $x$ component of the electric field is given by
$$
E_{x}=-\frac{\partial V}{\partial x}
$$
partial V}{\partial x}
$$
Similarly, the $y$ and $z$ components of the electric field are related to the potential by
$$
E_{y}=-\frac{\partial V}{\partial y}
$$
and
$$
E_{z}=-\frac{\partial V}{\partial z}
$$
Thus, Equation 23-15 in Cartesian coordinates is written
$$
\vec{E}=-\vec{\nabla} V=-\left(\frac{\partial V}{\partial x} \hat{i}+\frac{\partial V}{\partial y} \hat{j}+\frac{\partial V}{\partial z} \hat{k}\right) \quad 23-17
$$
PHAS00010 COURSE NOTES ：
Outside the spherical shell, the electric field is radial and is the same as if all the charge $Q$ were a point charge at the origin:
$$
\overrightarrow{\boldsymbol{E}}=\frac{k Q}{r^{2}} \hat{\boldsymbol{r}}
$$
where $\hat{\boldsymbol{r}}$ is a unit vector directed away from the center of the sphere. The change in the potential for some displacement $d \vec{\ell}$ outside the shell is then
$$
d V=-\overrightarrow{\boldsymbol{E}} \cdot d \vec{\ell}=-\frac{k Q}{r^{2}} \hat{\boldsymbol{r}} \cdot d \vec{\ell}=-\frac{k Q}{r^{2}} d r
$$
where the product $\hat{\boldsymbol{r}} \cdot d \overrightarrow{\boldsymbol{\ell}}$ is equal to $d r$ (the component of $d \vec{\ell}$ in the direction of $\hat{\boldsymbol{r}}$ ). Integrating along a path from the reference point at infinity, we obtain
$$
V_{p}=-\int_{\infty}^{\vec{r}{P}} \overrightarrow{\boldsymbol{E}} \cdot d \vec{\ell}=-\int{\infty}^{r_{p}} \frac{k Q}{r^{2}} d r=-k Q \int_{\infty}^{r_{p}} r^{-2} d r=\frac{k Q}{r_{p}}
$$