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Time derivative of electric field? Electromagnetic radiation energy emitted

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by:

    [tex]\frac{dE}{dt}[/tex] = [tex]\frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}}[/tex]

    where c is the speed of light.
    (a). verify that this equation is dimensionally correct

    Relevant equations/ The attempt at a solution

    q -> Coulombs-> Amp*Seconds -> A*s
    a -> meters/second[tex]^{2}[/tex] --> m/s[tex]^{2}[/tex]
    [tex]\epsilon_{0}[/tex] --> [tex]\frac{C^{2}}{N*m^{2}}[/tex]--> [tex]\frac{A^{2}*s^{4}}{kg*m^{3}}[/tex]
    c-->[tex]\frac{m}{s}[/tex]


    now i have no problem doing all of that and boiling it down to [tex]\frac{dE}{dt}[/tex] = [tex]\frac{kg*m^{2}}{s^{3}}[/tex]

    but my problem comes in when it comes to proving that [tex]\frac{dE}{dt}[/tex] is supposed to have those units. The book asks to prove it. I took what they gave me and worked it down to [tex]\frac{kg*m^{2}}{s^{3}}[/tex]

    now what I seem to need is help with figuring out how to go the other way around the same circle by taking the derivative of the electric field with respect to time [tex]\frac{dE}{dt}[/tex] and showing that it has equivalent units to what i found above.

    I know the equation for electric field is:

    [tex]\vec{E}[/tex] = [tex]\hat{r} k \frac{q}{r^{2}}[/tex]

    but how do I go about taking the d/dt of that?!???


    Thanks for all the help!
    -Ben
     
  2. jcsd
  3. Feb 25, 2010 #2

    collinsmark

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    Hello GreenLantern,

    I believe E here represents energy, not electric field. Here, dE/dt is the change in energy per unit time.
     
  4. Feb 25, 2010 #3

    collinsmark

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    By the way, In case you haven't seen this:

    http://www.xkcd.com/687/

    :rofl:
     
  5. Feb 25, 2010 #4
    lol

    thanks for the input on getting me away from electric field
    okay so energy,

    energy of....??

    so that would energy flow per unit time which would be S (poynting vector magnitude) times area? (because S is the energy flow per unit time per unit area)

    I was figureing something along theses lines after i posted and then got lost in a world of confusing thoughts trying to figure it all out my self using S = (1/A) dU/dt = [tex]\epsilon_{0}[/tex] * c * E[tex]^{2}[/tex] (in vacuum)

    ugh still so confused. can anyone help me out some more with the solution to this???
     
  6. Feb 25, 2010 #5

    collinsmark

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    I'm not sure if this will help. But I'm pretty sure the units you are looking for is the Watt. Power is defined as

    [tex] P(t) = \frac{dE}{dt} [/tex]

    I'm pretty sure the relationship is more of a definition than a derivation. In the SI system, the unit of power is the Watt. I don't think you can prove/derive the relationship any further than that. The Watt unit can be broken down to J/s (the the Joule can be broken down further too, if you need to).

    Do you need to actually prove that

    [tex] \frac{dE}{dt} = \frac{q^{2}a^{2}}{6\pi\epsilon_{0}c^{3}} [/tex]

    or simply prove that the units are dimensionally correct? Because I think you've pretty much proven the units already (well, as long as you know that energy is measured in Joules).
     
  7. Feb 27, 2010 #6
    Ahhh okay i figured it out. Thanks so much for the help!!!!
    -GL
     
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