# 物理代写|电磁学代写electromagnetism代考|PHY53040

## 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}

# 电磁学代考

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

$$E=frac{Delta V}{d}。$$

$$Q=Q_1+Q_2$$
$Q$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}

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