# Electric field

1. May 22, 2005

### robert25pl

Given E in a region of space ( $$\epsilon_{o} , \mu_{o}$$)
I should think of it as a free space or vacuum?

Last edited: May 22, 2005
2. May 22, 2005

### Dr.Brain

This is vacuum.

3. May 22, 2005

### robert25pl

I have to find $$B$$, $$\rho$$ and $$J$$

$$\nabla \times \vec E= -\frac{\partial \vec B}{\partial t}$$

$$\nabla \cdot D=\rho$$

In free space J = 0 so for J in vacuum I should use

$$\nabla \times H =J+ \frac{\partial D}{\partial t}$$

or something else?

4. May 23, 2005

### OlderDan

You probably want to use

$$D = \epsilon_{o} E$$

$$B = \mu_{o} H,$$

to eliminate D and H. Don't know if you will need it, but you can round out your set of equations with

$$\nabla \cdot B=0$$

5. May 23, 2005

### robert25pl

I think I got B, D, H and $$\rho$$ and using

$$\nabla \times H =J+ \frac{\partial D}{\partial t}$$

I get J

What is the difference between free space (J = 0) and vacuum?

6. May 23, 2005

### dextercioby

Vacuum means $\mu_{0},\epsilon_{0}$.If you're speaking about "free space",then u should assume no charge density $\rho=0$ and no charge transport $\vec{J}=0$.

So we can have sources in vacuum.But not in free space.

Free space is typically a vacuum in which electromagetic waves (radiation far away from the sources) propagate.

Daniel.

7. May 23, 2005

### robert25pl

Thanks, So in my problem I will get J not equal 0

8. May 23, 2005

### dextercioby

Daniel.

9. May 23, 2005

### robert25pl

In the region of space $$\epsilon_{o} , \mu_{o}$$
$$E = (6z\vec{i}+10y\vec{j})cos500t \vec{j}$$

Find $$B, \rho, J$$

10. May 23, 2005

### dextercioby

$$\nabla\cdot\vec{E}=\frac{\rho}{\epsilon_{0}}$$ (1)

$$\nabla\times\vec{E}=-\frac{\partial\vec{B}}{\partial t}$$ (2)

$$\nabla\times\vec{B}=\mu_{0}\vec{J}+\mu_{0}\epsilon_{0}\frac{\partial\vec{E}}{\partial t}$$ (3)

Daniel.