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A free electron is located in the field of a plane electromagnetic wave. Neglecting the magnetic component of the wave distrurbing its motion, find the ratio of the mean energy radiated by the oscillating electron per unit time to the mean value of the energy flow density of the incident wave. |
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Answer» Solution :If the electric field of the wave is `oversetrarr(E) = oversetrarr(E_(0)) cos omegat` then the induces a dipole MOMENT whose second derivative is `ddotoversetrarr(p) = (e^(2)oversetrarr(E_(0)))/(m) cos omegat` Hence radiated mean power `lt p gt = (1)/(4piepsilon_(0)) (2)/(3c^(3)) ((e^(2)E_(0))/(m))^(2) xx (1)/(2)` On the other HAND the mean Poynting flux of the INCIDENT radiation is `lt S_("inc") gt = sqrt((epsilon_(0))/(mu_(0))) xx (1)/(2)E_(0)^(2)` Thus `(P)/( lt S_("inc") gt ) = (1)/(4piepsilon_(0))(2)/(3) (epsilon_(0)mu_(0))^(3//1) ((e^(2))/(m))^(2) xx sqrt((mu_(0))/(epsilon_(0)))` `= (mu_(0)^(2))/(6pi) ((e^(2))/(m))^(2)` |
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