# I Maxwell eqs. - integral formulation in dynamics

1. Apr 17, 2017

### lightarrow

The integral formulation of Maxwell equations, for example the one called "Gauss' theorem", see the first equation here:
https://en.wikipedia.org/wiki/Maxwell's_equations
is still valid in dynamics, with rapidly varying sources? We know that a rapidly varying charge here gives a flux of an electric field there not at the same time but with some time delay.

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lightarrow

2. Apr 17, 2017

### Staff: Mentor

Yes, it is still valid.

Could you describe your "rapidly varying sources" idea in a little more detail? We can discuss how the integral law applies.

3. Apr 17, 2017

### lightarrow

If the wavelenght of the signal generated by the sources is much lower than the distance from the sources and the points in which the fields are computed.
Let's say the total charge in the origin of coordinates vary at 300GHz and we want to compute the flux of E through a surface 10 cm apart.
Thanks.

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lightarrow

4. Apr 17, 2017

### Staff: Mentor

Hmm, don't forget the continuity equation. Charge is conserved, so you cannot just have charge at the origin vary without having current flowing in to the origin.

Can you revise your scenario with that in mind? Where does this charge at the origin come from?

Last edited: Apr 17, 2017
5. Apr 17, 2017

### lightarrow

Right, thank you. The scenario can be a variable current which enters, from one point, into the region V considered (where we compute the total charge Q) inside the surface S, and exits from another point.
Do I write Gauss' Law in the same manner as the static case, that is: flux(E)_S = Q/eps_0, where E and Q are computed at the same instant of time?

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lightarrow

6. Apr 17, 2017

### Staff: Mentor

@Dale spotted the difficult part of your scenario right away. It sounds like you still don't see it.

With currents in and out as you describe, how will you determine Q?

7. Apr 17, 2017

### Staff: Mentor

Yes, exactly the same manner.

Since the current crosses the surface any change in the flux can happen due to fields localized at the entry point.

8. Apr 17, 2017

### lightarrow

It's possible.
Integrating current in the time from t1 to t. After a tiny interval of time dt, so at time t+dt, the total charge be Q+dQ. But the flux of E far from there has not changed yet.

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lightarrow

9. Apr 17, 2017

### Staff: Mentor

It doesn't have to change far from there, it just has to change somewhere on the surface

10. Apr 17, 2017

### lightarrow

Sorry but I can't understand why the change of flux(E) should be due to that only: consider, e.g., a closed surface with one of a condenser's plates inside and the cable connecting that plate entering the surface.

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lightarrow

11. Apr 17, 2017

### lightarrow

Yes, that's true, however it's not easy to me understand what happens to the flux while the charge enters and goes more inside, far from the surface.
Thanks however.

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lightarrow

12. Apr 17, 2017

### Staff: Mentor

The contents of the surface doesn't matter in the least. All that matters is the charge inside. To simplify matters, consider a spherical surface as a point charge enters.

As the charge is far from the surface the field lines are nearly parallel and uniform, every line that enters clearly leaves. As the charge gets closer the field lines become more concentrated near the charge and diverge, but still every line that enters also exits. Immediately after the charge crosses the boundary the situation changes so that now all the field lines are exiting the surface, the lines are very concentrated near the charge and have rapidly shifted from inward to outward while the lines on the opposite side are nearly unchanged. As the charge moves towards the center the field lines spread out until they become uniform all over, but the net flux does not change as the same number of field lines continue to exit the surface.

13. Apr 18, 2017

### lightarrow

Very informative, thank you.

Can I finally ask about the 4° equation of Maxwell? In absence of any time variations of E, how I write the line integral of B(t) along a circumference (in the void) around an axial cable (with negligible resistance so E = 0 inside of it) in which flows a current i(t) very rapidly changing? It's still equal to mu_0 i(t), that is at the same t when B is computed?
Thanks again.

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lightarrow

14. Apr 18, 2017

### Staff: Mentor

This is not possible. "In the absence of any time variations of E" is incompatible with the rest of the scenario. See Jefimenko's equations. The rapidly changing current necessarily produces a varying E.

15. Apr 18, 2017

### lightarrow

Right: a variable current obviously generates around a variable B field which generate a variable E field and so on, Didn't think to it

Do you remember how are the E and B fields generated by an infinite long and straight antenna with a variable current? Is B still coaxial and E radial? This approximation of infinite long cable can be applied here? If it is and E is however radial, its time variation shouldn't contribute to the line integral of B around that circumference.

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lightarrow

16. Apr 18, 2017

### Staff: Mentor

Yes. If you take cylindrical coordinates with the wire along z then B points in the $\hat{\theta}$ direction and E points in the $\hat{r}$ direction.

You do have to be careful with infinite wires. They violate the continuity equation as usually expressed. So it is better to consider the midpoint of a coaxial cable close to the inner conductor.

That said, I don't have a handy explanation for the 4th equation, like I did for the first. I am sure there is one, but I don't know it.

17. Apr 18, 2017

### lightarrow

So it was not a trivial question, after all, I feel a little bitt less silly :-)
Thanks Dale.