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Stuck on this dimensional analysis question.

  1. Nov 29, 2013 #1
    1. The problem statement, all variables and given/known data
    The total power radiated by an oscillating electric dipole is a function of the oscillation frequency ω, the dipole moment p(=Qd, where ±Q is the charge at each end of the dipole and d is the distance between charges), the speed of light c and the permittivity of free space ε0. Apply dimensional analysis to find the form of this function. Hence find the ratio of the power emitted by the same dipole at frequencies ω and 2ω

    2. Relevant equations



    3. The attempt at a solution
    I started off by writing out the units and trying to find the powers, this left me with:
    J/s ∝ (Hz)A (Qm)B (m/s)C (F/m)D
    My first thought was that C=1, so that there is s-1 on each side.
    Then I thought that D=2 so that the metres cancel out.
    Then I'm left with J ∝ (Hz)A Q F2 /s
    And now I'm completely stuck :(
    Any help appreciated, thanks.
     
    Last edited: Nov 29, 2013
  2. jcsd
  3. Nov 29, 2013 #2

    mfb

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    Hz = s-1.
    You cannot start with seconds like that. Actually, it is pointless to start with seconds, as you can always adjust A. Start with something different. It is also useful to express F in different units.
     
  4. Nov 30, 2013 #3
    I always seem to forget the simple things :(
    Thanks for the help

    So by converting everything to SI units I got A=-4, B=2, C=-3, D=-1
    I got this from:
    kgm2s-3 = (s-1)A (Asm)B (ms-1)C (s4A2m-3kg-1)D
    D is the only one dealing with kg, so that must equal -1.
    Then B must equal 2 so that A is cancelled out.
    Then C=-3 to bring m down to m2.
    A must equal -4 to achieve s-3.

    This leaves me with P = (p2)/(ω4c3ε)

    I'm not sure how to tackle the second bit about the ration of ω and 2ω, I don't really understand what it's asking me to do. Can you please help?
     
  5. Nov 30, 2013 #4

    mfb

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    Check the sign of the exponent A. Apart from that, it looks good (WolframAlpha can check this, for example).

    Concerning (b), imagine you have a function f(x)=x2. If you double x (if you replace x by 2x), what happens to the function value?
     
  6. Dec 1, 2013 #5
    Ah thank you very much.
    I probably wouldn't have noticed that I got the sign wrong :(

    If f(x)=x2 then f(2x)=4x2
    So the ratio of the power at frequencies ω and 2ω would be 4
     
  7. Dec 1, 2013 #6
    ah but as A=4, f(x)=x4, so f(2x)=16x4. Would that be correct? The ratio is 16?
     
  8. Dec 1, 2013 #7

    mfb

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    Right.
     
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