### 物理代写|电磁学代写electromagnetism代考|PHYC20014

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## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Point Charge

Consider an isolated positive point charge $q$, as indicated in Fig. 3.4. Note that such a charge produces an clcctric ficld that is dirccted radially outward from the charge, as discussed in Chap. 1:

$$\mathbf{E}=k_{e} \frac{q}{r^{2}} \hat{\mathbf{r}}$$
Consider an arbitrary path from some point $A$ to the point $B$, and a small displacement vector along that path $d \mathbf{s}$ at the position $\mathbf{r}$ relative to the charge $q$, as shown in Fig. 3.4. Using Eq. (3.7), the potential difference is
\begin{aligned} \phi_{B}-\phi_{A} &=-\int_{A}^{B} \mathbf{E} \cdot d \mathbf{s} \ &=-\int_{A}^{B} k_{e} \frac{q}{r^{2}} \hat{\mathbf{r}} \cdot d \mathbf{s} \ &=-\int_{A}^{B} k_{e} \frac{q}{r^{2}} \cos \theta d s \ &=-\int_{A}^{B} k_{e} \frac{q}{r^{2}} d r \ &=k_{e} q\left(\frac{1}{r_{B}}-\frac{1}{r_{A}}\right) \end{aligned}
where $r_{A}$ and $r_{B}$ are the distances of $A$ and $B$ relative to $q$, and $\hat{\mathbf{r}}$ is a unit vector along the radial direction. In Eq. (3.23), $\theta$ is angle between the small displacement vector $d \mathbf{s}$ along the path between $A$ and $B$ and the radial small displacement vector $d \mathbf{r}$, as in Fig. 3.4.

Equation (3.23) implies that the electric potential of any arbitrary charge $q$ at a distance $r$ from the charge is given as
$$\phi(r)=k_{e} \frac{q}{r}$$
which is a function of the distance $r$ from the charge $q$.

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a System of Point Charges

Let $q_{1}, q_{2}, \ldots, q_{N}$ be a set of $N$ static discrete charges, positive or negative, as shown in Fig. 3.5. Based on superposition principle, the electric potential resulting from those point charges at some point $P$, with position vector $r$ with respect to the origin of the reference frame, is
$$\phi(\mathbf{r})=k_{e} \sum_{i=1}^{N} \frac{q_{i}}{\left|\mathbf{r}-\mathbf{r}{i}\right|}$$ where $\mathbf{r}{i}$ is the position vector of the $i$ th charge with respect to the origin $O$, as indicated in Fig. 3.5.

Equation $(3.25)$ indicates that the total potential at any point $P$ of a set of $N$ point charges is the sum of the potentials due to the individual charges.

In particular, for a system of two charged particles, we denote by $\phi_{1}$ the electric potential created by the charge $q_{1}$ at a point $P$ at distance $r$ from the charge $q_{1}$, which is taken to be at the origin $O$ of a coordinate system:
$$\phi_{1}=k_{e} \frac{q_{1}}{r}$$
The work done by an external agent to move the second charge $q_{2}$ from infinity to $P$ without accelerating it (i.e., the kinetic energy remains constant) is

$$W=q_{2} \phi_{1}$$
Therefore, from Eq. (3.27), the work is equal to the interaction potential energy
The $U_{12}$ of the particles, when they are separated by a distance $r_{12}$ :
$$U_{12}=k_{e} \frac{q_{1} q_{2}}{r_{12}}$$
For two particles with the same sign of charges, $U_{12}$ is positive, which is in agreement with what we know from the previous discussion that positive work has to be done by an external agent on the system to move the two charges near one another. That is also in agreement with the view that charges with the same sign repel each other (see Chap. 1). On the other hand, if the charges have opposite sign, $U_{12}$ is negative; that is, the external agent does a negative work to move the charges near each other against the attractive force between charges (in this case, it is the field which does the positive work).

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Continuous Charge Distribution

We can partition the volume into macroscopically small charge elements $d q$, as shown in Fig. 3.6. Then, we consider the potential due to macroscopically small charge element $d q$ by treating this element as a point charge:
$$d \phi=k_{e} \frac{d q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}$$
To obtain the total potential at point $P$, we integrate to include contributions from all elements of the charge distribution:
$$\phi=k_{e} \int \frac{d q}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}$$
If we assume that $d q=\rho d V$, where $\rho$ is the charge density inside the macroscopically small volume $d V=d \mathbf{r}^{\prime}$, then
$$\phi(\mathbf{r})=k_{e} \int_{V} \rho\left(\mathbf{r}^{\prime}\right) \frac{d V}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}$$
which is a function of $\mathbf{r}$.
For a continuous charge distribution, the potential interaction energy can be written as
$$U=\frac{k_{e}}{2} \int_{V} \int_{V} \frac{\rho(\mathbf{r}) \rho\left(\mathbf{r}^{\prime}\right)}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|} d \mathbf{r} d \mathbf{r}^{\prime}$$
Using Eq. (3.39), we can write Eq. (3.40) in the following convenient form: $U=\frac{1}{2} \int_{V} \rho(\mathbf{r}) \phi(\mathbf{r}) d \mathbf{r}$

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Point Charge

φ乙−φ一个=−∫一个乙和⋅ds =−∫一个乙ķ和qr2r^⋅ds =−∫一个乙ķ和qr2因⁡θds =−∫一个乙ķ和qr2dr =ķ和q(1r乙−1r一个)

φ(r)=ķ和qr

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a System of Point Charges

$$\phi(\mathbf{r})=k_{e} \sum_{i=1}^{N} \frac{q_{i}}{\left| \mathbf{r}-\mathbf{r} {i}\right|}$$ 其中 $\mathbf{r} {i}一世s吨H和p○s一世吨一世○n在和C吨○r○F吨H和一世吨HCH一个rG和在一世吨Hr和sp和C吨吨○吨H和○r一世G一世nO$，如图 3.5 所示。

φ1=ķ和q1r

（3.27） ，功等于相互作用势能在12粒子，当它们分开一段距离时r12 :

## 物理代写|电磁学代写electromagnetism代考|Electric Potential of a Continuous Charge Distribution

dφ=ķ和dq|r−r′|

φ=ķ和∫dq|r−r′|

φ(r)=ķ和∫在ρ(r′)d在|r−r′|

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