Avichal said:
When two such particles, in this case when two electrons are present the total mass will be the sum of the individual masses? Why is this?
Suppose here's a charged particle, for example, an still electron.
We know that the charged particle will emit radiation (photon) when it's accelerating.
So, let's push this charged particle and make its velocity be non-zero velocity from zero.
Assume its mass is
m.
The change of its kinetic energy will be ΔE
k = [itex]\frac{1}{2}[/itex]mv
final2 - [itex]\frac{1}{2}[/itex]m0
2
--
However, it's not the energy you lose. Don't forget the emitting radiation. Let's be some energy,
Ephoton.
--
So the total energy you lose (or, the work done by you) during this process is
( [itex]\frac{1}{2}[/itex]mv
final2 - [itex]\frac{1}{2}[/itex]m0
2 ) +
Ephoton = [itex]\frac{1}{2}[/itex]mv
final2 +
Ephoton
So, it's like you push a new-mass charged particle:
[itex]\frac{1}{2}[/itex]mv
final2 + E
photon = [itex]\frac{1}{2}[/itex]
m'v
final2 - [itex]\frac{1}{2}[/itex]
m'0
2
Obviously,
m' > m
That's why you'll feel its mass greater than its still mass.
Same argument can explain the case you mentioned, the scenario that two electrons are present.
ps. actually, I prefer to think the electric field has some mass.