## 物理代写|电磁学代写electromagnetism代考|Field Energy in a Dielectric with Constant K

Assume that we have a dielectric liquid in which charges can move. If a true charge $Q$ is present in the liquid at $\mathbf{x}Q$, $$\nabla \cdot \mathbf{D}(\mathbf{x})=4 \pi \rho{\text {true }}(\mathbf{x})=4 \pi Q \delta\left(\mathbf{x}-\mathbf{x}Q\right)$$ If the dielectric is homogeneous, that is, if the dielectric constant is not a function of the position, $$\mathbf{D}(\mathbf{x})=K \mathbf{E}(\mathbf{x})$$ and $$\mathbf{E}(\mathbf{x})=-\nabla_x \frac{Q}{K\left|\mathbf{x}-\mathbf{x}{Q \mid}\right|}$$

The field $\mathbf{E}$ is due to all charges, including true charges and polarization charges.

If we consider a test charge $q$ at $\mathbf{x}$, the field $\mathbf{E}(\mathbf{x})$ is equal to the force acting on the test charge, divided by the value of the charge. The force has the magnitude
$$|F(\mathbf{x})|=q|\mathbf{E}(\mathbf{x})|=\frac{q Q}{K\left|\mathbf{x}-\mathbf{x}Q\right|^2}$$ The dielectric has the effect of reducing the force by $K$. Therefore, for a number of charges, the electrostatic energy is given by $$U=\frac{1}{2} \sum{\substack{s, t \ s \neq t}} \frac{q_s q_t}{K\left|\mathbf{x}_s-\mathbf{x}_t\right|}$$
Consider now a density of true charges $\rho(\mathbf{x})$ embedded in a medium of dielectric constant $K$. We have in this case the following relations:
\begin{aligned} \nabla \cdot \mathbf{D}(\mathbf{x}) &=4 \pi \rho(\mathbf{x}) \ \rho(\mathbf{x}) &=\frac{1}{4 \pi} \nabla \cdot \mathbf{D}(\mathbf{x}) \ \phi(\mathbf{x}) &=\int \frac{\rho\left(\mathbf{x}^{\prime}\right)}{K\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d \tau^{\prime} \end{aligned}
and
\begin{aligned} U &=\frac{1}{2} \iint d \tau d \tau^{\prime} \frac{\rho(\mathbf{x}) \rho\left(\mathbf{x}^{\prime}\right)}{K\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} \ &=\frac{1}{2} \int d \tau\left(\frac{1}{4 \pi} \boldsymbol{\nabla} \cdot \mathbf{D}\right) \phi(\mathbf{x})=\frac{1}{8 \pi} \int d \tau(\nabla \cdot \mathbf{D}) \phi(\mathbf{x}) \ &=-\frac{1}{8 \pi} \int d \tau \mathbf{D} \cdot \nabla \phi=\frac{1}{8 \pi} \int d \tau \mathbf{E} \cdot \mathbf{D} \end{aligned}

## 物理代写|电磁学代写electromagnetism代考|Field Energy in a Dielectric for Which

We shall assume the presence of a continuous charge distribution $\rho(\mathbf{x})$. We shall set
$\delta \mathbf{s}(\mathbf{x})=$ displacement of the charge element located at $\mathbf{x} \quad(2.16 .1)$
The force acting on the true charge in $d \tau$ is given by
$$\rho(\mathbf{x}) d \tau \mathbf{E}(\mathbf{x})$$
The amount of work done by the forces on the charges due to the displacements $\delta s$ is given by
$$\delta W=\int d \tau \rho(\mathbf{x})[\mathbf{E}(\mathbf{x}) \cdot \delta \mathbf{s}(\mathbf{x})]$$
The change in field energy is given by
$$\delta U=-\int d \tau \rho(\mathbf{x})[\mathbf{E}(\mathbf{x}) \cdot \delta \mathbf{s}(\mathbf{x})]$$
We do not change the charges, we just displace them; we have then to take into account the continuity equation
$$\frac{\partial \rho(\mathbf{x}, t)}{\partial t}+\nabla \cdot[\rho(\mathbf{x}, t) \mathbf{v}(\mathbf{x})]=0$$
where we have explicitly shown the dependence of $\rho$ on $t$, and $\mathbf{v}(\mathbf{x})$ is the velocity of the charge element located at $\mathrm{x}$ :
$$\mathbf{v}(\mathbf{x}) \delta t=\delta \mathbf{s}(\mathbf{x})$$
We can then write
$$\delta \rho=\delta t \frac{\partial \rho}{\partial t}=-\nabla \cdot(\rho \mathbf{v} \delta t)=-\nabla \cdot \rho \delta \mathbf{s}$$

# 电磁学代考

## 物理代写|电磁学代写electromagnetism代考|Field Energy in a Dielectric with Constant K

$$\nabla \cdot \mathbf{D}(\mathbf{x})=4 \pi \rho \operatorname{true}(\mathbf{x})=4 \pi Q \delta(\mathbf{x}-\mathbf{x} Q)$$

$$\mathbf{D}(\mathbf{x})=K \mathbf{E}(\mathbf{x})$$

$$\mathbf{E}(\mathbf{x})=-\nabla_x \frac{Q}{K|\mathbf{x}-\mathbf{x} Q| \mid}$$

$$|F(\mathbf{x})|=q|\mathbf{E}(\mathbf{x})|=\frac{q Q}{K|\mathbf{x}-\mathbf{x} Q|^2}$$

$$U=\frac{1}{2} \sum s, t s \neq t \frac{q_s q_t}{K\left|\mathbf{x}_s-\mathbf{x}_t\right|}$$

$$\nabla \cdot \mathbf{D}(\mathbf{x})=4 \pi \rho(\mathbf{x}) \rho(\mathbf{x}) \quad=\frac{1}{4 \pi} \nabla \cdot \mathbf{D}(\mathbf{x}) \phi(\mathbf{x})=\int \frac{\rho\left(\mathbf{x}^{\prime}\right)}{K\left|\mathbf{x}-\mathbf{x}^{\prime}\right|} d \tau^{\prime}$$

$$U=\frac{1}{2} \iint d \tau d \tau^{\prime} \frac{\rho(\mathbf{x}) \rho\left(\mathbf{x}^{\prime}\right)}{K\left|\mathbf{x}-\mathbf{x}^{\prime}\right|}=\frac{1}{2} \int d \tau\left(\frac{1}{4 \pi} \boldsymbol{\nabla} \cdot \mathbf{D}\right) \phi(\mathbf{x})=\frac{1}{8 \pi} \int d \tau(\nabla \cdot \mathbf{D}) \phi(\mathbf{x})=-\frac{1}{8 \pi} \int d \tau \mathbf{D}$$

## 物理代写|电磁学代写electromagnetism代考|Field Energy in a Dielectric for Which

$$\rho(\mathbf{x}) d \tau \mathbf{E}(\mathbf{x})$$

$$\delta W=\int d \tau \rho(\mathbf{x})[\mathbf{E}(\mathbf{x}) \cdot \delta \mathbf{s}(\mathbf{x})]$$

$$\delta U=-\int d \tau \rho(\mathbf{x})[\mathbf{E}(\mathbf{x}) \cdot \delta \mathbf{s}(\mathbf{x})]$$

$$\frac{\partial \rho(\mathbf{x}, t)}{\partial t}+\nabla \cdot[\rho(\mathbf{x}, t) \mathbf{v}(\mathbf{x})]=0$$

$$\mathbf{v}(\mathbf{x}) \delta t=\delta \mathbf{s}(\mathbf{x})$$

$$\delta \rho=\delta t \frac{\partial \rho}{\partial t}=-\nabla \cdot(\rho \mathbf{v} \delta t)=-\nabla \cdot \rho \delta \mathbf{s}$$

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