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A particle with charge e and mass m flies with non-relativistic velocity v at a distance b past a stationary particle with charge q. Neglecting the bending of the trajectory of the moving particle, find the enegry lost by this particle due to radiation during the total flight time. |
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Answer» Solution :Most of the radiation occurs when the moving particle is closest of the stationary particle. In that ragion, we can write `R^(2) = b^(2) + v^(2) t^(2)` and apply the previous problem's formula THUS `DeltaW~~(1)/((4piepsilon_(0))^(3))(2)/(3c^(2)) int_(-OO)^(oo) ((qe^(2))/(m))^(2) (dt)/((b^(2)+v^(2)t^(2))^(2))` (the intergral can be taken between `+- oo` with littel error.) Now `int_(-oo)^(oo) (dt)/((b^(2)+bv^(2)t^(2))^(2)) = (1)/(v) int_(-oo)^(oo)(dx)/((b^(2)+X^(2))^(2)) = (PI)/(2vb^(3))`. Hence `Delta W ~~(1)/(4piepsilon_(0))^(3) (piq^(2)E^(4))/(3c^(3)m^(2)vb^(3))`. |
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