Destruction of e-p pair:energy?

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SUMMARY

The discussion centers on the annihilation process of particle-antiparticle pairs, specifically the interaction between charges a+q and a-q. When these charges coalesce, they neutralize, collapsing the charge configuration and eliminating the electric field. The energy released during this process is determined by the relativistic equation E = γmc², where m represents the rest mass of the particles. If the particles are at rest, the total energy released is 2mc², reflecting the conservation of energy in particle interactions.

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Kolahal Bhattacharya
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When a+q & a-q charges colesce together, the neutralization collapses the charge configuration & the electric field is gone.what is the amount of energy released?
Please note that it is a personal query-not homework
 
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Kolahal Bhattacharya said:
When a+q & a-q charges colesce together, the neutralization collapses the charge configuration & the electric field is gone.what is the amount of energy released?
Please note that it is a personal query-not homework
It depends on how far apart they end up and how big (diameter) they are! What is the amount of work done by the charges in moving from infinity separation to the final state?

AM
 
Kolahal Bhattacharya said:
When a+q & a-q charges colesce together, the neutralization collapses the charge configuration & the electric field is gone.what is the amount of energy released?
Please note that it is a personal query-not homework
The language you are using is not standard but I am assuming you are taking about an *annihilation* process (liek an electron-positron annihilation, say).

Well, energy is conserved. So the total energy released is simply the sum of the energies of the two particle annihilating (using the relativistic expression [itex]E = \gamma mc^2[/itex]). *If* the two particles are initially at rest, then the energy released is 2mc^2 (notice that a particle and its antiparticle always have the same rest mass).
 

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