Q7.3-7P - Introduction to Electrodynamics

Question

Suppose 

 $\mathbf{E}(\mathbf{r}\mathbf{,}\mathbf{t})=\frac{1}{\mathbf{4}\pi {\mathbf{\epsilon }}_0}\frac{q}{{\mathbf{r}}^2}\theta (\mathbf{r}-\upsilon \mathbf{t})\widehat{\mathbf{r}}$;  $\mathbf{B}(\mathbf{r}\mathbf{,}\mathbf{t})=\mathbf{0}$

(The theta function is defined in Prob. 1.46b). Show that these fields satisfy all of Maxwell's equations, and determine  $\rho$ and $\mathbf{J}$. Describe the physical situation that gives rise to these fields.

Step-by-Step Solution

Answer

The value offirst Maxwell’s equation for the given functions of  $B$ and $E$ are .$\rho =-q{\delta }^3\left(r\right)\theta \left(t\right)+\frac{q}{4\pi r^2}\delta (\upsilon t-r)$

The value of second Maxwell’s equation for the given functions of  $B$ and $E$  are $\nabla \cdot \overrightarrow{B}=0$.

The value of Third Maxwell’s equation for the given functions of  $B$ and  $E$ are $\nabla \times \overrightarrow{E}=0$ .

The value of fourth Maxwell’s equation for the given functions of $B$  and $E$ are  $\overrightarrow{J}=\left(\frac{q}{4\pi r^2}\right)\left(\delta (\upsilon t-r)\right)\widehat{r}$.

Step 1: Write the given data from the question.

Consider the given electrical field is $\text{E}\left(\text{r,t}\right)=-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\theta (\upsilon t-r)\widehat{r}$.

Consider the given magnetic field is $\text{B}\left(\text{r,t}\right)=0$.

Consider the Maxwell’s equations are given by

 $\begin{array}{l}\nabla \cdot \overrightarrow{E}=\frac{\rho }{{\epsilon }_0}\\ \nabla \cdot \overrightarrow{B}=0\\ \nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t}\\ \nabla \times \overrightarrow{B}={\mu }_0\overrightarrow{J}+{\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t}\end{array}$

Step 2: Determine the formulaof Maxwell’s equation for the given functions of and

Write the formula of first Maxwell’s equation for the given functions of $B$ and $E$.

$\nabla \cdot \overrightarrow{\mathbf{E}}=\frac{\rho }{{\mathbf{\epsilon }}_0}$ …… (1)

Here, $\rho$ is a charge and ${\epsilon }_0$  is absolute permittivity.

Write the formula of second Maxwell’s equation for the given functions of $B$ and $E$.

 $\nabla \cdot \overrightarrow{\mathbf{B}}$

Here,  $\nabla$ is derivative and  $\overrightarrow{\mathbf{B}}$ is magnetic field.

Write the formula of third Maxwell’s equation for the given functions of $B$ and $E$.

$\nabla \times \overrightarrow{\mathbf{E}}$ …… (2)

Here, $\nabla$  is derivative,  $\overrightarrow{\mathbf{B}}$ is electrical field.

Write the formula of fourth Maxwell’s equation for the given functions of  and $E$.

$\nabla \times \overrightarrow{\mathbf{B}}={\mu }_0\overrightarrow{\mathbf{J}}+{\mu }_0{\mathbf{\epsilon }}_0\frac{\partial \mathbf{{\rm E}}}{\partial \mathbf{t}}$ …… (3)

Here,${\mu }_0$  is permeability and $\overrightarrow{\mathbf{J}}$  is current density, $\mathbf{E}$  is permittivity and   is electric field.

Step 3:Determine thevalue Maxwell’s equation for the given functions of B and E

Determine the value of first Maxwell’s equation for the given functions of $B$ and .$E$

Substitute $\theta (\upsilon t-r)\nabla$  for   and $(-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r})-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r}\cdot \nabla \left|\theta (\upsilon t-r)\right|$  into equation (1).

 $\begin{array}{c}\nabla \cdot E=\theta (\upsilon t-r)\nabla \cdot (-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r})-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\widehat{r}\cdot \nabla \left|\theta (\upsilon t-r)\right|\\ =-\frac{q}{{\epsilon }_0}{\delta }^3\left(r\right)\theta (\upsilon t-r)-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}(\widehat{r}\cdot \widehat{r})\frac{\partial }{\partial r}\theta (\upsilon t-r)\end{array}$

From problem 1.45, we are given the next ${\delta }^3\left(r\right)\theta (\upsilon t-r)-{\delta }^3\left(r\right)\theta \left(t\right)$ and $\frac{\partial }{\partial r}\theta (\upsilon t-r)--\delta (\upsilon t-r)$. So, plug these expressions into equation (1) to get.

$\begin{array}{c}\nabla \cdot \text{E}=-\frac{q}{{\epsilon }_0}{\delta }^3\left(r\right)\theta (\upsilon t-r)-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}(\widehat{r}\cdot \widehat{r})\frac{\partial }{\partial r}\theta (\upsilon t-r)\\ \rho ={\epsilon }_0\nabla \cdot \text{E}--q{\delta }^3\left(r\right)\theta \left(t\right)+\frac{q}{4\pi r^2}\delta (\upsilon t-r)\end{array}$

 

Determine the value of second Maxwell’s equation for the given functions of $B$ and .$E$

Plug the expression of $B$, so we get it by 

Substitute  $\nabla \cdot \left(0\right)$  for  $\nabla$ and  $\nabla \cdot \left(0\right)=0$

Determine the value of third Maxwell’s equation for the given functions  of  $B$ and $E$.

Given that the electric field is independent of $\theta$  and  $\varphi$ the vector product will be zero.

Substitute $\varphi$  for $\frac{\partial \overrightarrow{B}}{\partial t}$ into equation (2).

 $\nabla \times \overrightarrow{E}=0$

Determine the value of fourth Maxwell’s equation for the given functions of $B$ and $E$.

To obtain the current density, plug in the formula for $B$  and  $E$ as follows:

Substitute $(-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\theta (\upsilon t-r))\widehat{r}$  for  $\overrightarrow{E}$ into equation (3).

 $\begin{array}{l}\nabla \times 0={\mu }_0\overrightarrow{J}+{\mu }_0{\epsilon }_0\frac{\partial }{\partial t}(-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\theta (\upsilon t-r))\widehat{r}\\ 0=\overrightarrow{J}+{\epsilon }_0\frac{\partial }{\partial t}(-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\theta (\upsilon t-r))\widehat{r}\\ \overrightarrow{J}=-{\epsilon }_0\frac{\partial }{\partial t}(-\frac{1}{4\pi {\epsilon }_0}\frac{q}{r^2}\theta (\upsilon t-r))\widehat{r}\\ =-{\epsilon }_0(-\frac{1}{4\pi {\epsilon }_0})\frac{q}{r^2}\frac{\partial }{\partial t}\left(\theta (\upsilon t-r)\right)\widehat{r}\end{array}$

Solve further as

 $\overrightarrow{J}=\left(\frac{q}{4\pi r^2}\right)\left(\delta (\upsilon t-r)\right)\widehat{r}$.