Questions regarding something in Feynman's lectures.

1. Jun 21, 2010

SrEstroncio

This isn't quite homework help, although it might seem a little like it.

I was reading my copies of the Feynman's lectures the other day and in volume 1, chapter 2, section 2, "Physics before 1920" he mentioned something which confused me a little: when explaining electric charges and introducing the electric field he sets up a mind experiment in which you shake a charged comb and a little piece of paper, some distance away, moves accordingly, but with a certain delay; then he explains that we then notice something even more interesting, that as we shake the comb we see that the effect of it's charge on the piece of paper falls off more slowly than the inverse of the square of the distance, that is, that vibrating electric charges produce a field which is more far-reaching than the force associated with simple electrical attraction or repulsion.
He follows this up with a metaphor in which he compares the two charges with pieces of cork on a pool (the water being the electric field): we can move one piece of cork by means of moving the other one towards it, thus causing a perturbation in the water which will move the second piece of cork away, even though the effect will be smaller the more distance between the pieces of cork there is; he then adds that by jiggling one piece of cork we can obtain waves whose influence extends very much farther out than the effect of a single movement.

Is he trying to say that only vibrating electric charges produce an electric field? or that the electric field's effect falls of more slowly than the inverse of the square of the distance? I really don't know a whole lot about EM but from what I recall from Resnick's "Physics", charged particles emit a field whether or not they're vibrating, and it's effect also falls off inversely as the square of the distance. I got a little confused and was wondering if someone could clarify what the lectures meant or something.

Sorry for my poor composition skills, by the way.

2. Jun 22, 2010

lilphil1989

He's saying that the radiation field generated by an oscillating charge drops off less quickly than the inverse square.

A stationary charge generates the regular r-2 coulomb field.

3. Jun 22, 2010

pgardn

Charges that are moving also produce a B (magnetic) field. Maybe this helps with EMagnetic wave part?

4. Jun 22, 2010

SrEstroncio

I believe he's only concerned with the electrical aspects of said oscilating charge, as he briefly explains and dismisses the magnetic part of if as the cause of said effect.

Could you elaborate on this?

5. Jun 23, 2010

lilphil1989

I could try!

If you're reading Feynmann's lectures, I would imagine that you're acquainted with Coulomb's law - a stationary charge generates a static electric field proportional to the magnitude of the charge which drops off with the square of the distance:

$$E \propto$$ $$\frac{q}{r^2}$$

I think the easiest way to see the generation of a radiation field is to look up the calaculation of the electric field generated by a Hertzian dipole. I can recommend Griffith's "Introduction to Electrodynamics" for this, although I would imagine that Feynmann covered it too.

6. Jun 23, 2010

cesiumfrog

Ah, now it makes sense. From a stationary charge, the force (the electric field) on a test charge falls off proportional to r^2, as dictated by the geometry of the field lines.

From a vibrating charge, the power incident on a test charge (the intensity of the EM wave) also falls off proportional to r^2, as dictated by the geometry of the energy radiation. But this is comparing peaches to avocados, we should return to discussing force: the amplitude of the wave (the electric field or the force the wave produces on a stationary test charge) only falls off proportional to r.

7. Jun 25, 2010

denni89627

Interestingly this is why we abandoned the original Edison DC grid for the more efficient and easier to propagate Tesla AC system. To achieve similar results a DC system would need gigantic conductors and copper is expensive.

Tesla = Genius

8. Jun 26, 2010

cesiumfrog

Uh, I don't think that's right. DC is actually more efficient to propagate. Because it doesn't radiate energy away to space, and because it stays high voltage all the time (whereas AC is low voltage for part of its cycle, increasing the resistive losses). The reason we use AC is because it was easy to transform AC voltages (between transmission lines and household devices) using primitive technology, whereas the technology for efficient DC power voltage conversion is more recent.