物理代写|电磁学代写electromagnetism代考|PHY53040

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物理代写|电磁学代写electromagnetism代考|Parallel-Plate Capacitors

Now, let us consider a capacitor composed of two parallel conductor plates of equal area $A$, which are at a distance $d$, see also Fig.4.3. One of the plates carries a charge $+Q$, and the other $-Q$. Note that charges of like sign repel one another and that charges of opposite signs attract one another (see also Chap. 1). As a battery is charging a capacitor, electrons flow into the negative plate and out of the positive plate (see Fig. 4.2).

Note that the electric field between the plates of a parallel-plate capacitor is uniform near the center but nonuniform near the edges. When the capacitor plates are large, the accumulated charges can distribute themselves over a substantial area, and hence the amount of charge stored on each plate $Q$, for a given potential difference $\Delta V$, increases as the plate area increases to ensure a constant surface charge density $\sigma$. A simple argument can be used for that: because the electric field just outside the conductor is perpendicular to the surface of the conductor and with magnitude $E=\sigma / \epsilon_0$, where $E$ is proportional to constant $\Delta V$, then $\sigma$ is constant. Thus, we expect the capacitance $C$ to be proportional to the plate area $A$.

Above we derived a relationship between the electric field between the plates and magnitude of potential difference, given as
$$
E=\frac{\Delta V}{d}
$$
From Eq. (4.9), we see that when $d$ decreases, $E$ increases, for fixed $\Delta V$. If we move the plates closer together (that is, $d$ decreases), We also consider the situation before the charges have moved in response to that change, such that no charges have moved. Hence, the electric field between the plates is the same but extends over a shorter distance between plates. That situation corresponds to a new capacitor with a potential difference between the plates that is different from the terminal voltage of the battery. Now, across the wires connecting the battery to the capacitor exists a potential difference (see also Fig. $4.2$ for an illustration).

Based on the arguments that we discussed for a situation in Fig. 4.2, that potential difference creates an electric field in the wires that drives more charges onto the plates, which in turn increases the potential difference between the plates of the capacitor. When it becomes equal to the potential difference between the terminals of the battery.

物理代写|电磁学代写electromagnetism代考|Parallel Combination

Figure $4.5$ presents a combination of two capacitors connected in parallel. Also, we show a circuit diagram for this combination of capacitors, as often seen in an electric circuit. Note from Fig. $4.5$ that the left plates of the capacitors connect to the positive terminal of the battery using conducting wires; therefore, those plates, after equilibrium of the electric potential establishes, are at the same electric potential as the positive terminal of the battery. For the same reason, the right plate connecting to the negative terminal of the battery has equal electric potential with the negative terminal after the equilibrium of the electric potential establishes. As a result, the potential differences across each capacitor connected in parallel are the same and equal to the voltage applied to the battery; that is, $\Delta V_1=\Delta V_2=\Delta V$.

Applying the model described above in Fig. 4.2, when two capacitors are initially connected in a circuit, as shown in Fig. 4.5, electrons migrate between the wires and the plates. As a result, the left plates charge positively, and the right plates

negatively. In other words, the internal chemical energy stored in the battery is the source of that migration; that is, the internal chemical energy of the battery converts into electric potential energy associated with the surface charges in the plates of the capacitors at a separation $d$. During the process of the electrons migration, the voltage across the capacitors becomes equal to that across the battery terminals and then charge transfer stops. When that establishes in the circuit, the capacitors load to their maximum charge capacity.

In the following, we show a few steps to calculate the equivalent capacitance, $C_{e q}$, of the combinations of $C_1$ and $C_2$. For that, we denote by $Q_1$ and $Q_2$ the maximum charges on each capacitor, respectively, and by $Q$ the total charge stored by the two capacitors:
$$
Q=Q_1+Q_2
$$
$Q$ is also the charge stored in the capacitor $C_{e q}$. The voltages applied across each capacitor are the same, see also Fig. 4.5, and hence the charges in each capacitor are
$$
\begin{aligned}
&Q_1=C_1 \Delta V \
&Q_2=C_2 \Delta V
\end{aligned}
$$

物理代写|电磁学代写electromagnetism代考|PHY53040

电磁学代考

物理代写|电磁学代写electromagnetism代考|Parallel-Plate Capacitors

现在,让我们考虑一个由两块面积相等的平行导体板组成的电容器,它们之间的距离为$d$,也见图4.3。其中一块板带有$+Q$的电荷,另一块板带有$Q$的电荷。请注意,相同符号的电荷会相互排斥,相反符号的电荷会相互吸引(也见第1章)。当电池对电容器充电时,电子流入负极板,并从正极板流出(见图4.2)。

请注意,平行板电容器的板间电场在中心附近是均匀的,但在边缘附近是不均匀的。当电容器板块较大时,累积的电荷可以分布在相当大的面积上,因此在给定的电位差$Delta V$下,每个板块上储存的电荷量$Q$会随着板块面积的增加而增加,以确保表面电荷密度$sigma$不变。对此可以用一个简单的论证:因为导体外面的电场垂直于导体的表面,其大小为$E=/sigma/ epsilon_0$,其中$E$与常数$Delta V$成正比,那么$sigma$就是常数。因此,我们期望电容$C$与板面积$A$成正比。

以上我们得出了板块之间的电场和电位差大小之间的关系,给定为
$$
E=frac{Delta V}{d}。
$$
从公式(4.9)中,我们看到,当$d$减少时,$E$增加,对于固定的$Delta V$。如果我们将板块移近(即$d$减小),我们也考虑到电荷在响应该变化而移动之前的情况,如没有电荷移动。因此,板块之间的电场是相同的,但在板块之间延伸的距离更短。这种情况相当于一个新的电容器,板间的电位差与电池的端电压不同。现在,在连接电池和电容器的导线上存在着一个电位差(参见图4.2$的说明)。

根据我们在图4.2中讨论的论点,该电位差在导线中产生一个电场,促使更多的电荷进入极板,这反过来又增加了电容器极板间的电位差。当它变得与电池两端的电位差相等时。

物理代写|电磁学代写electromagnetism代考|Parallel Combination

图4.5$展示了两个电容器并联的组合。同时,我们还展示了这种电容器组合的电路图,这在电路中经常见到。从图4.5中注意到,电容器的左侧板用导电线连接到电池的正极;因此,这些板在电动势平衡建立后,与电池的正极处于相同的电动势。出于同样的原因,连接到电池负极的右板在电势平衡建立后与负极的电势相等。因此,并联的每个电容器上的电位差是相同的,并且等于施加在电池上的电压;也就是说,$/Delta V_1=/Delta V_2=/Delta V$。

应用上述图4.2的模型,当两个电容器最初连接在一个电路中时,如图4.5所示,电子在导线和板之间迁移。结果是,左边的板子会充以正电,而右边的板子会充以负电。换句话说,储存在电池中的内部化学能是这种迁移的来源;也就是说,电池的内部化学能转化为与电容器板中的表面电荷有关的电势能,其间隔为$d$。在电子迁移的过程中,电容器两端的电压变得与电池两端的电压相等,然后电荷转移停止。当电路中出现这种情况时,电容器就会加载到其最大充电能力。

在下文中,我们将展示几个步骤来计算$C_{e q}$的等效电容,$C_1$和$C_2$的组合。为此,我们用$Q_1$和$Q_2$分别表示每个电容器上的最大电荷,而用$Q$表示两个电容器储存的总电荷。
$$
Q=Q_1+Q_2
$$
$Q$也是存储在电容器$C_{e q}$中的电荷。施加在每个电容器上的电压是相同的,也见图4.5,因此每个电容器中的电荷是
$$
\begin{aligned}
&Q_1=C_1 δ V\\
&Q_2=C_2 δV
\end{aligned}
$$

物理代写|电磁学代写electromagnetism代考

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