An electron is accelerated through a uniform electric field of strength, E= 20 [N/C]. Determine the speed after the the electron travels 0.5 m from rest.
E=F/Q. F= 20*(1.6910^-19)=3.3810^-18NF=ma m=9.1110-31 kgTherefore, a = 3.71 * 10^12 m/s^2Since v^2= u^2 + 2as and s= 0.5m,Then v=1.9310^6 m/s
LF
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