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Energy radiated by a parallel plate capacitor

  1. Nov 6, 2013 #1
    1. The problem statement, all variables and given/known data

    A parallel-plate capacitor C , with plate separation d , is given an initial
    charge ±Q[itex]_{0}[/itex]. It is then connected to a resistor R, and discharges, Q(t)=Q[itex]_{0}[/itex] e[itex]^{-t/RC}[/itex]
    2
    (a) What fraction of its energy Q[itex]_{0}[/itex][itex]^{2}[/itex]/2C does it radiate away?



    2. Relevant equations

    The equations for an electric dipole, involving the retarded time.

    3. The attempt at a solution

    I have been at this for hours and don't know what to do. I think I should treat the capacitor as a dipole at a very great distance away and integrate over a massive sphere as one does for an oscillating dipole. However this was taking me an age, and you get to the point where coshx is the only real solution so rather than being able to time average cos^2(x) as usual, you just can't do that.

    There must be something I'm missing. I know the energy loss from the capacitor Q[itex]_{0}[/itex][itex]^{2}[/itex]/2C , and I know the energy stored in the capacitor, but they appear to be the same amount, when logically they should be different, witht the difference in energy equalling the anergy radiated away. Any help would be great.
     
  2. jcsd
  3. Nov 7, 2013 #2

    rude man

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    Homework Helper
    Gold Member

    Not my field of expertise, so others may modify the following:

    The stored and dissipated energies do indeed compute to the same number. But these computations assume the quasi-stationary state. In reality, if you want to be really picky about it you'd have to start with the Maxwell relations modified for radiation. These include changing from del x E = 0 to del x E = -∂B/∂t and from del x H = i to del x H = i + ∂D/∂t. You would then have to compute the Poynting vector P = E x H and integrate it over a closed surface enveloping the entire circuit including the plates. Dn't ask me how that would be done. All I can tell you is that the answer is very, very close to zero for all practical applications..
     
  4. Nov 7, 2013 #3
    Thank you very much for the reply, I'll give it a shot!
     
  5. Mar 13, 2015 #4
    For those still seeking answers to this questoin (Griffiths Electrodynamics 11.8 in the 4th Edition), you should use the dipole equation $p=Qd$ then work from there.
     
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