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An electron moving with the speed 5 xx 10^6 m per second is shooted parallel to the electric field of intensity 1 xx 10^3 N/C. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant, (mass of e = 9.1 xx 10^(-31)kg) |
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Answer» 0.7 cm `qE = mA` `therefore a=(qE)/m =(1.6 XX 10^(-19) xx 10^(3))/(9.1 xx 10^(-31))` `therefore a=0.1758 xx 10^(15) m//s^(2)` Now equation of motion `v^(2) -v_(0)^(2) = 2ad` `therefore d =(v^(2)-v_(0)^(2))/(2A)`, where `v=0, v_(0)=5 xx 10^(6) m//s` `therefore =(25 xx 10^(12))/(3516 xx 10^(11)) = 0.00711 xx 10 = 0.0711` m `=7.11 cm = 7 cm` |
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