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An electron falls through a distance of 1.5cm in a uniform electric field of magnitude 2 times 10^4 NC^-1. The direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance. Compute the time of fall in each case. |
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Answer» Solution :Here, electric force, `F_(e) = qE = ma rArr a = (qE)/m` Linear distance travelledl at the end of time t is, `d = v_(0)t + 1/2 at^(2)` `therefore d = 1/2 (qE)/mt^(2), (therefore v_(0) =0 " and " a =(qE)/m)` `therefore t^(2) = (2md)/(qE)` `therefore t = sqrt((2md)/(eE))`......(1) (`therefore` For electron and proton magnitude of CHARGE is q=e) Linear distance travelled at the end of time t is, `d=v_(0)t + 1/2 at^(2)` `therefore d = 1/2((qE)/m)t^(2) (therefore v_(0)=0 " and " a =(qE)/m`) `therefore t^(2) = (2md)/(qE)` `therefore t = sqrt((2md)/(eE))`............(1) (`therefore` For electron and proton magnitude of charge is q = e) From equation (1), time taken by electron to pass through distance d (opposite to electric field) `t_(e) = sqrt((2m_(e)d)/(eE))` `= sqrt((2 xx 9.1 xx 10^(-31) xx 0.015)/(1.6 xx 10^(-19) xx 2 xx 10^(4))` `=2.92 xx 10^(-9)` s `therefore t_(e) = 2.92 ns` (Nano second)...........(2) From equation (1), time taken by proton to pass through distance d (PARALLEL to electric field) `t_(p) = sqrt((2m_(p)d)/(eE))` `=sqrt((2 xx 1.67 xx 10^(-27) xx 0.015)/(1.6 xx 10^(-19) xx 2 xx 10^(4))` `=1.251 xx 10^(-7)`s =`125.1 xx 10^(-9)`s `therefore t_(p) = 125.1`ns.............(3) From equation (2) and (3), `t_(p) GT t_(e)` (`therefore` Here, `t prop sqrt(m)` and `m_(p) gt m_(e)`) |
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