Understanding Maxwell's Equations

[robert25pl]

I want to make sure that I understand this good.
Given E and B are possible in a region of free space (J=0) only if [tex] \triangledown \times E=0 [/tex] and [tex] \triangledown \cdot B = 0 [/tex]

 

[dextercioby]

That [itex] \vec{B} [/itex] needs to be stationary (time independent)...Else [itex] \vec{E} [/itex] would not be a purely potential-derived field.

Daniel.

 

[robert25pl]

I have this two equations:

[tex]E=3\sin(3z-6t) \vec{k} [/tex]
[tex]B=- \frac{1}{15} \sin(3z-6t) \vec{j}[/tex]

So what should I do first?

 

[dextercioby]

Verify whether such a field configuration satisfies the eqn-s

[tex] \nabla\cdot\vec{B}=0 [/tex]

[tex] \nabla\times\vec{E}=-\frac{\partial \vec{B}}{\partial t} [/tex]

Isn't this what u're supposed to do?

Daniel.

 

[jtbell]

So what should I do first?

That depends on the question you're supposed to be answering.

 

[robphy]

"Free space" means [tex]\rho={\color{red}0}[/tex] and [tex]\vec\jmath={\color{red}\vec 0}[/tex].
So, in two of Maxwell's Equations, this means that
[tex] \nabla \cdot \vec E=\rho ={\color{red}\ 0} [/tex] and
[tex] \nabla \times \vec B=\vec \jmath + \frac{\partial \vec E}{\partial t}={\color{red}\ \vec 0} + \frac{\partial \vec E}{\partial t} [/tex].
Of course, [from the other two equations] we must always have
[tex] \nabla \cdot \vec B= 0 [/tex] and
[tex] \nabla \times \vec E= -\frac{\partial \vec B}{\partial t} [/tex]

[Insert your own conventional constants]

 

[Nylex]

What?! When we were taught Maxwell's equations in free space, we were told that: [tex]\nabla \cdot \vec E=\frac{\rho}{\epsilon_{0}}[/tex], as free space meant in air not in any sort of medium.

 

[robphy]

What?! When we were taught Maxwell's equations in free space, we were told that: [tex]\nabla \cdot \vec E=\frac{\rho}{\epsilon_{0}}[/tex], as free space meant in air not in any sort of medium.

So, maybe term I should have used is "source-free".

 

[robert25pl]

[tex] \nabla\cdot\vec{B}=0 [/tex] I verified that

[tex]\nabla\times\vec{E}=\left|\begin{array}{ccc}\vec{i }& \vec{j} &\vec{k}\\\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\ 0 & 0 & 3\sin (3z-6t) \end{array} \right| [/tex]


[tex] \triangledown \times E [/tex] gave me 0 and [tex] -\frac{\partial \vec{B}}{\partial t} =- \frac{2}{5} \cos(3z-6t) \vec{j}
[/tex]

I think this wrong?

 

[jdavel]

robert,

You say:

divB = 0, I agree

curlE = 0, I agree

-dB/dt = (-2/5)sin(3z - 6t)j, I think you've got a mistake here

 

[dextercioby]

Should be a "cos" there.

Daniel.

 

[robert25pl]

[tex] \frac{\partial \vec{B}}{\partial t} =-\frac{2}{5} \cos(3z-6t) \vec{j}[/tex]
So [tex] \nabla \times \vec E= -\frac{\partial \vec B}{\partial t} [/tex]
are nor equal and they are not possible in region of space?

 

[dextercioby]

That's that.It is not possible.They should identically solve every equation from Maxwell's system...

Daniel.