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Energy transfer in EM field

  1. Jul 22, 2011 #1

    gbz

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    When an accelerated charged particle is shot into an electromagnetic field such that the accelerated particle comes to a complete stop in the field, we naturally have loss of energy from the charged particle. But what is this energy converted to? And where? Does the particle release radiation (photons) as it slows down and comes to a stop or does the equipment generating the electric field acquire the energy of the particle and heat up/radiate?
     
  2. jcsd
  3. Jul 22, 2011 #2

    Dale

    Staff: Mentor

    Potential energy. Just like rolling a rock up a hill.
     
  4. Jul 22, 2011 #3

    Drakkith

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    Staff: Mentor

    It releases photons and also exerts a force on whatever is generating the force against it. Some of the energy would be lost as the object flexes and such.
     
  5. Jul 23, 2011 #4

    gbz

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    Thanks
     
  6. Jul 23, 2011 #5
    As the particle decelerates (or for that matter when it was accelerated), it sends out a "ripple" in the electromagnetic field that radiates away the energy.

    If you shake a charge particle (accelerates it, then decelerates it, then accelerates it, then decelerates it...), the particle will also send out "ripple" in the form of oscillating electromagnetic field. We used that everyday for wireless communication. It is the same principle.
     
  7. Jul 23, 2011 #6
    I'm not sure the situation has been covered adequately.

    Consider a conducting plate connected to the negative terminal of an ideal constant voltage source, with a negatively charged particle heading towards the plate. The postitive terminal would be grounded. As the particle approaches the plate, wouldn't electrons be driven from the plate towards the negative terminal? This would be a situation where energy is absorbed by the voltage source.

    Of course the potential energy of the charged particle would be increased as well.
     
    Last edited: Jul 23, 2011
  8. Jul 23, 2011 #7

    Drakkith

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    Staff: Mentor

    There would be a force against the electrons in the plate sure, but there is also a force holding them there. They negative terminal would not absorb any energy, it would expend it trying to hold all those like charges on the plate.
     
  9. Jul 23, 2011 #8
    That is not correct. Power absorbed by an element is equal to voltage times current. If no current flows through the source, it does not absorb or expend any energy.

    Here's a way I thought of it. As the negatively charged particle approaches the plate, the negative charge in the plate has more potential energy per charge (voltage), due to the electric field of the charged particle. Thus, the plate discharges through the source in order for constant voltage on the plate to be maintained.
     
  10. Jul 23, 2011 #9

    Drakkith

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    Staff: Mentor

    Is the plate negatively charged by the voltage source? If so it is using energy to keep it charged. I don't know a setup where doing this would result in the production or storage of energy, other than in the approaching particle.
     
  11. Jul 23, 2011 #10
    I explained to you why you are wrong. Now you've re-asserted your incorrect idea without any counter-argument. The only way the source expends energy is if it moves negative charge out of the negative terminal (towards the plate) in response to the incoming negative particle.


    Think of the negative charge headed to the negative plate as current. What if the charge hit the plate? Then, would you agree, that negative electricity would flow into the negative terminal, and thus the source would absorb energy? Even if the particle doesn't hit the plate, the negative charges in the plate would be still be repulsed away from the particle as it approaches, and thus negative electricity would flow towards the negative terminal.
     
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