Where Does the Energy Go in an EM Field When a Charged Particle Stops?

AI Thread Summary
When a charged particle decelerates in an electromagnetic field, it loses energy, which can be converted into radiation (photons) or potential energy. The particle's deceleration generates ripples in the electromagnetic field, similar to how wireless communication works. As the particle approaches a conducting plate connected to a voltage source, it influences the distribution of charges, potentially causing energy absorption by the source if current flows. The discussion highlights the importance of current in energy transfer, emphasizing that without current, no energy is absorbed or expended by the source. Overall, the energy dynamics involve complex interactions between the charged particle, the electromagnetic field, and the voltage source.
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?
 
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gbz said:
But what is this energy converted to?
Potential energy. Just like rolling a rock up a hill.
 
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.
 
Thanks
 
gbz said:
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?

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.
 
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.
 
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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.
 
Drakkith said:
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.

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.
 
MisterX said:
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.

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.
 
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Drakkith said:
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.

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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