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A plane electromagnetic wave with frequency `omega` falls upon an elastically bonded electron whose natural frequency equals `omega_(0)`. Neglecting the damping of oscillations, find the ratio of the mean enegry dissipated by the electron per unit time to the mean value of the enrgy flow density of the incident wave. |
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Answer» For the elastically bound electron `mddotoversetrarr(r) =m omega_(0)^(2) overset(r ) = eoverset(E_(0)) cos omegat` This equation has the particular intergal (i.e. neglecting the part which does not have the frequency of the impressed force ) `oversetrarr(r) = (eoversetrarr(E_(0)))/(m) (cos omegat)/(omega_(0)^(2) - omega^(2))` so and `ddotoversetrarr(p) =- (e^(2)oversetrarr(E_(0))omega^(2))/((omega_(0)^(2)-omega^(2))m)cos omegat` Hence `P =` mean radiated power `= (1)/(4piepsilon_(0)) (2)/(3c^(3)) ((e^(2)omega^(2))/(m(omega_(0)^(2)-omega2)))^(2)(1)/(2)E_(0)^(2)` The mean incident poynting flux is `lt S_("inx") gt = sqrt((epsilon_(0))/(mu_(0)) (1)/(2)E_(0)^(2)` Thus `(P)/( lt S_("inc") gt ) = (mu_(0)^(2))/(6pi) ((e^(2))/(m))^(2) (omega^(4))/((omega_(0)^(2) - omega^(2))^(2))`. |
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