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# 物理代写|电磁学代写Electromagnetism代考|PHYS102 Resistance and Ohm’s Law

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## 物理代写|电磁学代写Electromagnetism代考|Resistance and Ohm’s Law

Note that the electric field inside a conductor is zero, which is valid only if the conductor is in static equilibrium. However, when charges move in a conductor, they produce a current as a result of an electric field, which is maintained by connecting the conductor to a battery. The charges move because of the electric field, and hence a non-electrostatic situation exists in the conductor.

Let $I$ be the current flowing in a conductor of cross-sectional area $A$. The ratio of the current $I$ with cross-sectional surface area $A$ defines the current density $J$ in the conductor or the current per unit area:
$$J=\frac{I}{A}$$
Since the current $I=n q v_d A$, the current density is
$$J=n q v_d$$
Note that Eq. (5.8) implies that $J$ has SI units of $\mathrm{A} / \mathrm{m}^2$. Furthermore, Eq. (5.9) is valid only if the current density is uniform, and only if the surface of the cross-sectional surface is perpendicular to the direction of the current flow.
The current density vector is defined as
$$\mathbf{J}=n q \mathbf{v}_d$$
Equation (5.10) indicates that current density is in the direction of charge motion for positive charge carriers $(q>0)$ and opposite the direction of motion for negative charge carriers $(q<0)$. Therefore, it is similar to the current $I$; however, current $I$ is not a vector quantity but the current density $\mathbf{J}$ is a vector.

## 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).

## 物理代写|电磁学代写Electromagnetism代考|Resistance and Ohm’s Law

$$J=\frac{I}{A}$$

$$J=n q v_d$$

$$\mathbf{J}=n q \mathbf{v}_d$$

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

$$E=\frac{\Delta V}{d}$$

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