Changes in electric field lines as a result of an oscillating charge

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  • #1
Costweist
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The last couple of days I’ve been troubled with a specific part of electromagnetism. How will electric field lines be affected by an oscillating charge? More specific, what will happen with the “amplitude” of a wave in an electrical field line as the wave propagate away from the charge?

1. Will the amplitude be unchanged as the distance increase?
2. Will the amplitude increase as the distance increase?

When such wave propagation is described in figures / illustrations I actually find both cases (hence the confusion). Below you will find a short video that simulates case no.1 (no increase in amplitude with distance) using excel. I suspect, however that this is an incorrect description..

 
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  • #2
PeroK
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What do you mean by amplitude here? You have a varying electric field strength at each point.
 
  • #3
Costweist
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What do you mean by amplitude here? You have a varying electric field strength at each point.

The term "amplitude" may be a misleading. I've tried to illustrate the two cases in the figure below:

Capture.JPG
 
  • #4
PeroK
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##A_1## and ##A_2## are distances between points in space, aren't they? Or, am I misled?
 
  • #5
Costweist
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##A_1## and ##A_2## are distances between points in space, aren't they? Or, am I misled?

Yes - the figure has two spatial dimensions
 
  • #6
PeroK
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Yes - the figure has two spatial dimensions
Those aren't amplitudes. The electric field is everywhere in space. The thing that varies is the field strength at each point. That oscillates in strength and direction as the charge oscillates.
 
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  • #7
Costweist
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Thank you for your response. I agree.

I was able to find a video of what I believe is case no 2 (where the field lines disturbances incrase in magnitude as disturbances move away from the charge).



In the video in the original post the magnitude of the field lines disturbances do not incrase as disturbances move away from the charge.
 
  • #8
PeroK
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I'm not sure I understand exactly what your question is. Whatever happens, the strength of the electric field reduces with distance from the point charge. The electric field lines represent the direction of the electric field at each point - and, you can indicate strength by using the thickness of the line.

If you are modelling radiation, then that is something different.
 
  • #9
Costweist
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I'm not sure I understand exactly what your question is. Whatever happens, the strength of the electric field reduces with distance from the point charge. The electric field lines represent the direction of the electric field at each point - and, you can indicate strength by using the thickness of the line.

If you are modelling radiation, then that is something different.

I trully apologize for being unclear. Below I’ve made a new attempt to illustrate my question (I'm only concerned with the electrical field lines).

Let’s say we have a charge that oscillates in vertical direction. The oscillation disturbs the electrical field creating a wave pattern. How will this pattern propagate in space? For simplicity I've only included the electrical field lines in the horizontal directions (although they off course are everywhere).

Capture3.JPG
 
  • #10
PeroK
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You may be missing the point that electric fields lines are only illustrative. There are no discrete lines - the field is continuous across space. Whatever lines you choose to draw are arbitrary. In particular, those apparent amplitudes are not anything physical. An electric field line is not a physical thing. It's just a diagrammatic tool to help you visualise the field direction at each point.

An alternative approach to viewing an electric field would be to draw a fixed grid of equally spaced points and draw a vector at each point to represent the field (strength and direction). Each of these vectors would change over time and that would be a description of the changing electric field.
 
  • #11
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If you oscillate a charge back and forth along a line it will produce an EM wave. The type of EM wave is called dipole radiation. The amplitude intensity of dipole radiation falls off as ##1/r^2##. So if it is ##1\text{ W/m}^2## at ##10\text{ m}## distance then at ##20\text{ m}## distance it will be ##\frac{1}{4}\text{ W/m}^2##
 
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  • #12
hutchphd
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If you oscillate a charge back and forth along a line it will produce an EM wave. The type of EM wave is called dipole radiation. The amplitude of dipole radiation falls off as ##1/r^2##. So if it is ##1\text{ W/m}^2## at ##10\text{ m}## distance then at ##20\text{ m}## distance it will be ##\frac{1}{4}\text{ W/m}^2##

Let's be careful here. The amplitude of the fields falls off as ##1/r##. The energy is then quadratic in the fields.

.
 
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  • #13
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Let's be careful here. The amplitude of the fields falls off as ##1/r##. The energy is then quadratic in the fields.
Absolutely! I was being sloppy, thanks for the clarification.
 
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  • #14
A.T.
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The term "amplitude" may be a misleading. I've tried to illustrate the two cases in the figure below:

View attachment 278173

The single field line that you measure there just represents how the direction of the E-field vector changes. But for the amplitude of the EM-wave you also have to consider the magnitude of the E-field vector, which is represented by the closeness between neighboring field lines. You cannot just look at a single field line for this.
 
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  • #15
Costweist
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Thank you for all the response and I really apologize if my question appeared unprecise. It is likely due to both poor English and inadequate understanding / theoretical background.

Nevertheless, from what I understand the E-field lines may be understood as curves drawn through space so that its tangent at any point is in the direction of the E-field vector at that point.

So if we are to consider these E-field lines separately (i.e. not radiation, EM-waves etc.) - how will any patterns in these E-field lines develop as they propagate from an accelerated point charge? For instance, in the case we have an oscillating point charge, the E-field lines will form a wave pattern. How will characteristics of these wave patterns (e.g., amplitude) develop with distance from the point charge? As far as I understand there are three possibilities:

1. The wave pattern continues unchanged with distance (case 1)
2. The wave amplitude increases with distance (case 2)
3. Something else happens :nb)

Capture4.JPG
 
  • #16
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How will characteristics of these wave patterns (e.g., amplitude) develop with distance from the point charge? As far as I understand there are three possibilities:
You cannot see the amplitude in field line diagrams. All they show is the direction of the field at each point along the field line (and if you are interested in a point off the field line you just draw another field line through that point).
 
  • #17
A.T.
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1. The wave pattern continues unchanged with distance (case 1)
2. The wave amplitude increases with distance (case 2)
3. Something else happens :nb)

View attachment 278224

Case 2 is more correct. But the yellow field line should come out from the side like in your first picture:

capture3-jpg.jpg


This video is OK I think:




Here is a similar one:




The later was made with this applet (you need a browser that still supports Flash):
https://phet.colorado.edu/sims/radiating-charge/radiating-charge_en.html
 
  • #18
Costweist
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Thank you - this was helpfull 👍
 
  • #19
hutchphd
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which is represented by the closeness between neighboring field lines.
Absolutely. Which is why this is not much used to represent radiation fields. The electric field lines in the radiation zone will be closed loops, not zaggy lines. Here is one example:
 
  • #20
A.T.
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The electric field lines in the radiation zone will be closed loops, not zaggy lines. Here is one example:


Yes, for a dipole source. But a single oscillating charge can also produce EM waves. See the video at 2:30.
 
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  • #21
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How will characteristics of these wave patterns (e.g., amplitude) develop with distance from the point charge? As far as I understand there are three possibilities:

1. The wave pattern continues unchanged with distance (case 1)
2. The wave amplitude increases with distance (case 2)
3. Something else happens
The amplitude decreases as I said above. The amplitude decreases as 1/r and the intensity decreases as 1/r^2.
 
  • #22
sophiecentaur
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You cannot see the amplitude in field line diagrams.
The separation of the field lines from place to place tells you the relative field strength, no?

It seems to me that the very first diagram is along the right lines (sorry - unintended pun there). If the charge at the centre moves then there will be a change in their direction. Move it fast enough (say up down) then the density of the near vertical lines will vary a bit and the direction of the near horizontal lines will change up down. The propagation delay will mean the lines are 'wavy'. So the modulation will be a combination of amplitude and direction. But the waviness is not a plane wave.
 
  • #23
hutchphd
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Yes, for a dipole source. But a single oscillating charge can also produce EM waves. See the video at 2:30.
The dipole has no monopole part and it is therefor easier to analyze the radiative (far field) piece. All classical electromagnetic waves in free space are strictly transverse. All the stuff not perpendicular to the radial vector is not radiation and falls off far field. (not EM waves in my vernacular)
 
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  • #24
A.T.
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The dipole has no monopole part and it is therefor easier to analyze the radiative (far field) piece.
Yes. The dipole case makes it easier to see the difference between the field line pattern and the wave pattern. The video you posted shows this nicely at 2:50.

In the monopole case the patterns are similar (due to the radial E component) which can lead to confusing them.
 
  • #25
sophiecentaur
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Absolutely. Which is why this is not much used to represent radiation fields. The electric field lines in the radiation zone will be closed loops, not zaggy lines. Here is one example:

The current flowing along a wire antenna really is the equivalent of a charge oscillating between the ends. Think of a wire with a capacitance at each end. This is what the OP is talking about.

That video is well presented but there a few creaky bits when he tries to introduce practical application of his correct basic message. He says that only a λ/2 dipole will work 'properly'. You can match any transmitter to any piece of wire (or 'dead sheep', as my friendly HF aerial Engineer used to say) and tuning for just one frequency is not a lot of use in antennae that carry signal frequencies covering a fractional bandwidth of ten percent.

The other statement that upset me a bit was the idea of power matching an AC load to the source. A really bad idea to expect the electrical load of a town to be matched to the alternator (or a 100W bulb to be 'matched' to the, eventually melting, battery source). Electrical supply philosophy involves as near a constant voltage supply (zero source impedance) as possible. The Max Power theorem is fine but definitely not always applicable.
 
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  • #26
hutchphd
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Honestly I didn't look at the entire video but was most interested in the graphic that represented the production of the radiation field. I thought it was particularly good. Points well taken...dead sheep...
 
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  • #27
anorlunda
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The other statement that upset me a bit was the idea of power matching an AC load to the source. A really bad idea to expect the electrical load of a town to be matched to the alternator (or a 100W bulb to be 'matched' to the, eventually melting, battery source). Electrical supply philosophy involves as near a constant voltage supply (zero source impedance) as possible. The Max Power theorem is fine but definitely not always applicable.
That's correct. It's amusing that your point needs to be repeated again and again, despite 140 years of experience with power grids.

(Source: Edison His Life and Inventions)
In 1882 Edison's Pearl Street Station became the first commercial electric power plant. The electrical engineering establishment of the day ridiculed the idea and said it could never work. They thought that Edison was an uneducated ruffian who didn't understand the mathematics of the maximum power transfer theorem. Edison eschewed mathematics. But using the maximum power theorem, 50% of the energy produced would be dissipated at the source. Edison said that the source resistance should be zero, and of course he was correct.

We can still fool EEs even today. At GE, we taught courses for utility power engineers, who already had EE degrees. In the first class, the teacher would ask "What is the optimum impedance of the generator source?" Usually, at least half of the students would answer, "Matching the load resistance."

What's the flaw? Why doesn't the maximum power transfer theorem apply to power grids? A battery has a fixed open circuit voltage and a source resistance that can not be adjusted. But a generator can adjust its source voltage to any reasonable value. There is no defined maximum power in the generator case. ##P=\frac{V^2}{R_{load}}## for the highest V we can tolerate.
 
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  • #28
sophiecentaur
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despite 140 years of experience with power grids.
That MP theorem is so often mis-applied. People say how good their Audio Power Amplifier is , with an efficiency of 'nearly' 100%. If the MPT was the driving consideration, it would be 50% at most - likewise for an 'efficient' transmitter amp.
 
  • #29
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The last couple of days I’ve been troubled with a specific part of electromagnetism. How will electric field lines be affected by an oscillating charge? More specific, what will happen with the “amplitude” of a wave in an electrical field line as the wave propagate away from the charge?

1. Will the amplitude be unchanged as the distance increase?
2. Will the amplitude increase as the distance increase?

When such wave propagation is described in figures / illustrations I actually find both cases (hence the confusion). Below you will find a short video that simulates case no.1 (no increase in amplitude with distance) using excel. I suspect, however that this is an incorrect description..


A time varying electric field will induce a time varying magnetic field. A description of the lines of electric field you are looking for lie in trying to represent electromagnetic energy propagation. As EM energy propagates away from a source, the magnetic and electric fields interact in complex ways and there is an 'inductive' near field where magnetic and electric field strength are in a complex ratio, and a 'far field' where the ratio of the electric and magnetic fields tend towards the propagation impedance of the medium they are moving through.
 
  • #30
tech99
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Since the matter of the Electric Induction Field has been raised, I would like to mention that this is the field that moves the charge in the first place.
 
  • #31
Costweist
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A time varying electric field will induce a time varying magnetic field. A description of the lines of electric field you are looking for lie in trying to represent electromagnetic energy propagation. As EM energy propagates away from a source, the magnetic and electric fields interact in complex ways and there is an 'inductive' near field where magnetic and electric field strength are in a complex ratio, and a 'far field' where the ratio of the electric and magnetic fields tend towards the propagation impedance of the medium they are moving through.

Yes I just realized that all the simulations visualized are really for an universe without the magnetic force haha..

If one wants to simulate vector fields
(Electric and magnetic) created by an accelarated point charge, is then Faraday’s law + the Lorentz force the sufficient?
 
  • #32
Ibix
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Yes I just realized that all the simulations visualized are really for an universe without the magnetic force haha..

If one wants to simulate vector fields
(Electric and magnetic) created by an accelarated point charge, is then Faraday’s law + the Lorentz force the sufficient?
Google the Lienard-Wiechert potentials.
 
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  • #33
sophiecentaur
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Since the matter of the Electric Induction Field has been raised, I would like to mention that this is the field that moves the charge in the first place.
In the case of a radio antenna, that's true but I would think the 'thing that moves the charge' could be gravitational, in the case of pulsars and other astronomical objects. (Just sayin')
 
  • #34
Costweist
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Google the Lienard-Wiechert potentials.

Thank you!
 
  • #35
tech99
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In the case of a radio antenna, that's true but I would think the 'thing that moves the charge' could be gravitational, in the case of pulsars and other astronomical objects. (Just sayin')
I don't remember this topic being discussed, but if the acceleration is caused mechanically, say, I suspect we do not have an electric induction field.
 

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