### 物理代写|电磁学代写electromagnetism代考|PHYC20011

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## 物理代写|电磁学代写electromagnetism代考|Electric Potential

For a finite displacement of charge from $A$ to some $B$, the change in potential energy of the system charge-field $\Delta U$ is
$$\Delta U=U_{B}-U_{A}=\int_{A}^{B} d U=-q_{0} \int_{A}^{B} \mathbf{E} \cdot d \mathbf{s}$$
The integral in Eq. (3.5) is called path integral or line integral, and it is performed along the path that $q_{0}$ follows as it moves from $A$ to $B$. Since electric force is conservative force, then the integral does not depend on the path taken from $A$ to $B$. The quantity $\frac{U}{q_{0}}$ is independent of $q_{0}$, but it depends only on $\mathbf{E}$.
By definition, the ratio $\frac{U}{q_{0}}$ is called electric potential $\phi$ :
$$\phi=\frac{U}{q_{0}}$$
Equation (3.6) implies that electric potential, $\phi$, is a scalar quantity.
Potential difference is the difference of electric potential between two points $A$ and $B$ :

\begin{aligned} \Delta \phi &=\phi_{B}-\phi_{A} \ &=\frac{\Delta U}{q_{0}} \ &=-\int_{A}^{B} \mathbf{E} \cdot d \mathbf{s} \end{aligned}
Note that potential difference, $\Delta \phi$, is different physical quantity than change in potential energy, $\Delta U$ :
$$\Delta U=q_{0} \Delta \phi$$

## 物理代写|电磁学代写electromagnetism代考|Potential Difference in a Uniform Electric Field

Consider the motion of a point charge $q$ in a uniform electric field $\mathbf{E}$, as shown in Fig. 3.1. As a result of interaction with the electric field, the charge moves from $A$ to $\bar{B}$, assuming that it is a positive test charge.
First, we express the electric potential difference between points $A$ and $B$ :
$$\Delta \phi=\phi_{B}-\phi_{A}=-\int_{A}^{B} \mathbf{E} \cdot d \mathbf{s}$$
We also define $\mathbf{E}=-E \mathbf{j}$, where $\mathbf{j}$ is a unit vector along $y$-axis. Moreover, $d \mathbf{s}=-d s \mathbf{j}$. Then, Eq. $(3.14)$ can also be written as
$$\Delta \phi=-\int_{A}^{B} E(\mathbf{j} \cdot \mathbf{j}) d s=-E \int_{A}^{B} d s=-E d$$
where ( $-)$ sign indicates that $\phi_{B}<\phi_{A}$. Therefore, the electric field lines always point in the direction of decreasing electric potential.

Since the electric potential $\phi$ is a characteristic of the electric field $\mathbf{E}$, for the same field $\mathbf{E}$ as in Fig. 3.1, and an arbitrary test charge $q_{0}$ (positive or negative) moving

from $A$ to $B$, the change in potential energy of the charge-field system is
$$\Delta U=q_{0} \Delta \phi=-q_{0} E d$$
If $q_{0}>0$, then $\Delta U<0$; that is, a positive charge loses its potential energy when it moves in the direction of an electric field, and the field does work on the charge. Therefore, when released from the rest in an external electric field, a positive charge particle gains acceleration in the direction of the external electric field.

On the other hand, if $q_{0}<0$, then $\Delta U>0$; that is, a negative charge increases its potential energy when it displaces along the electric field direction. Therefore, if a negative charge releases from rest in the field $\mathbf{E}$, it accelerates in the opposite direction to the external electric field.

In the following, we consider a more general case. Assume a charged particle moves freely between any two points in a uniform electric field directed along the $x$-axis, as indicated in Fig. 3.2. Let $s$ be the displacement vector between $A$ and $\bar{B}$, then
\begin{aligned} \Delta \phi &=\phi_{B}-\phi_{A} \ &=-\int_{A}^{B} \mathbf{E} \cdot \mathbf{s} \ &=-\mathbf{E} \cdot \int_{A}^{B} d \mathbf{s} \ &=-\mathbf{E} \cdot \mathbf{s} \end{aligned}
The change in the potential energy of a charge $q_{0}$ is
$$\Delta U=q_{0} \Delta \phi=-q_{0} \mathbf{E} \cdot \mathbf{s}$$

## 物理代写|电磁学代写electromagnetism代考|Equipotential Surface

Consider again the potential difference between the points $A$ and $B$, see Fig. $3.3$, as
$$\Delta \phi=\phi_{B}-\phi_{A}=-\mathbf{E} \cdot \mathbf{s}=-E s \cos \theta=-E d$$
On the other hand, the potential difference between the points $C$ and $A$ is
$$\Delta \phi=\phi_{C}-\phi_{A}=-E d$$
Combining Eqs. (3.18) and (3.19), we obtain the following potential difference relationship:
$$\phi_{B}-\phi_{A}=\phi_{C}-\phi_{A}$$
Thus, $\phi_{B}=\phi_{C}$.
By definition, the equipotential surface is called any surface consisting of a continuous distribution of points having the same electric potential.

Using Eq. (3.8), since $\phi_{B}=\phi_{C}$ for any two points $B$ and $C$ in an equipotential surface
$$\Delta U=q_{0}\left(\phi_{C}-\phi_{B}\right)=0$$
Hence, no work is done in moving a test charge between any two points on an equipotential surface.

The equipotential surfaces of a uniform electric field consist of a family of planes that are all perpendicular to the field.

## 物理代写|电磁学代写electromagnetism代考|Electric Potential

Δ在=在乙−在一个=∫一个乙d在=−q0∫一个乙和⋅ds

φ=在q0

Δφ=φ乙−φ一个 =Δ在q0 =−∫一个乙和⋅ds

Δ在=q0Δφ

## 物理代写|电磁学代写electromagnetism代考|Potential Difference in a Uniform Electric Field

Δφ=φ乙−φ一个=−∫一个乙和⋅ds

Δφ=−∫一个乙和(j⋅j)ds=−和∫一个乙ds=−和d

Δ在=q0Δφ=−q0和d

Δφ=φ乙−φ一个 =−∫一个乙和⋅s =−和⋅∫一个乙ds =−和⋅s

Δ在=q0Δφ=−q0和⋅s

## 物理代写|电磁学代写electromagnetism代考|Equipotential Surface

Δφ=φ乙−φ一个=−和⋅s=−和s因⁡θ=−和d

Δφ=φC−φ一个=−和d

φ乙−φ一个=φC−φ一个

Δ在=q0(φC−φ乙)=0

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