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

## 物理代写|电磁学代写electromagnetism代考|Forces on a Dielectric

We wish to calculate the force acting on a piece of dielectric due to the presence of an electric field. The variation in the electrostatic energy when the unit volume of dielectric undergoes the displacement $\delta \mathrm{s}$ is given
$$\delta U=-\int d \tau \delta \mathbf{s} \cdot \mathbf{f}$$

where $\mathbf{f}=$ force per unit volume acting on the dielectric. We shall assume that the true charges present in the dielectric do not move, so that only the dielectric will possibly be displaced.
The electrostatic energy is given by
$$U=\frac{1}{8 \pi} \int d \tau \frac{D^2}{K}$$
We shall call
\begin{aligned} &\mathbf{D}^{\prime}(\mathbf{x})=\mathbf{D}(\mathbf{x})+\delta \mathbf{D}(\mathbf{x}) \ &K^{\prime}(\mathbf{x})=K(\mathbf{x})+\delta K(\mathbf{x}) \end{aligned}
and
$$U^{\prime}=\frac{1}{8 \pi} \int d \tau \frac{D^{\prime 2}}{K^{\prime}}$$
But
\begin{aligned} \frac{\left(\mathbf{D}^{\prime}\right)^2}{K^{\prime}} &=\frac{(\mathbf{D}+\delta \mathbf{D})^2}{K+\delta K} \ &=\frac{D^2}{K}-\frac{D^2}{K^2} \delta K+2 \frac{\mathbf{D}}{K} \cdot \delta \mathbf{D}+\text { quadratic terms } \end{aligned}
Then
$$U^{\prime}=\frac{1}{8 \pi} \int d \tau\left(\frac{D^2}{K}-\delta K \frac{D^2}{K^2}+\frac{2}{K} \delta \mathbf{D} \cdot \mathbf{D}\right)$$
and
$$\delta U=U^{\prime}-U=\frac{1}{8 \pi} \int d \tau\left(-\delta K \frac{D^2}{K^2}+\frac{2}{K} \delta \mathbf{D} \cdot \mathbf{D}\right)$$

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

A capacitor in its simplest form consists of two flat conducting plates of area, say, $A$, separated by a distance $d$ (see Fig. 2.23). If the potentials on the plates $a$ and $b$ are $V_a$ and $V_b$, respectively, with $V_a>V_b$, then the electric field is directed from plates $a$ to $b$, perpendicularly to the plates; it is constant inside the capacitor, if we neglect the edge effects, and its value is
$$E=\frac{V_a-V_b}{d}$$
The surface charge density on the plate $a$ is
$$\sigma=\frac{E}{4 \pi}=\frac{V_a-V_b}{4 \pi d}$$
and the total charge on the same plate is
$$Q=\sigma A=A \frac{V_a-V_b}{4 \pi d}$$
The total charge on the plate $b$ is $-Q$.

The ratio of the charge $Q$ to the potential difference is called the capacitance of the capacitor and is designated $C$.
$$C=\frac{Q}{V_a-V_b}=\frac{A}{4 \pi d}$$
In the ESU system of units, the capacitance is measured in centimeters. A capacitor with plate area $A=200 \mathrm{~cm}^2$ and separation $d=1 \mathrm{~cm}$ has a capacitance $C=15.9 \mathrm{~cm}$.

A capacitor may take other more complex geometrical forms, such as the one consisting of two conducting spherical shells of radii $a$ and $b$ (see) Fig. 2.24). If a charge $Q$ is distributed over the surface of the internal shell, the charge $-Q$ is distributed over the inside surface of the outer shell. The surface charge density on the internal shell is
$$\sigma=\frac{Q}{4 \pi a^2}$$
The electric field at the surface of the internal shell is
$$E=4 \pi \sigma=\frac{Q}{a^2}$$

# 电磁学代考

## 物理代写|电磁学代写electromagnetism代考|Forces on a Dielectric

$$\delta U=-\int d \tau \delta \mathbf{s} \cdot \mathbf{f}$$

$$U=\frac{1}{8 \pi} \int d \tau \frac{D^2}{K}$$

$$\mathbf{D}^{\prime}(\mathbf{x})=\mathbf{D}(\mathbf{x})+\delta \mathbf{D}(\mathbf{x}) \quad K^{\prime}(\mathbf{x})=K(\mathbf{x})+\delta K(\mathbf{x})$$

$$U^{\prime}=\frac{1}{8 \pi} \int d \tau \frac{D^{\prime 2}}{K^{\prime}}$$

$$\frac{\left(\mathbf{D}^{\prime}\right)^2}{K^{\prime}}=\frac{(\mathbf{D}+\delta \mathbf{D})^2}{K+\delta K} \quad=\frac{D^2}{K}-\frac{D^2}{K^2} \delta K+2 \frac{\mathbf{D}}{K} \cdot \delta \mathbf{D}+\text { quadratic terms }$$

$$U^{\prime}=\frac{1}{8 \pi} \int d \tau\left(\frac{D^2}{K}-\delta K \frac{D^2}{K^2}+\frac{2}{K} \delta \mathbf{D} \cdot \mathbf{D}\right)$$

$$\delta U=U^{\prime}-U=\frac{1}{8 \pi} \int d \tau\left(-\delta K \frac{D^2}{K^2}+\frac{2}{K} \delta \mathbf{D} \cdot \mathbf{D}\right)$$

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

$$E=\frac{V_a-V_b}{d}$$

$$\sigma=\frac{E}{4 \pi}=\frac{V_a-V_b}{4 \pi d}$$

$$Q=\sigma A=A \frac{V_a-V_b}{4 \pi d}$$

$$C=\frac{Q}{V_a-V_b}=\frac{A}{4 \pi d}$$

$$\sigma=\frac{Q}{4 \pi a^2}$$

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