find the velocity acquired by an electron that has flown through an electric field between two points at potential of 100v and 300v if its initial velocity is 5"10^6 m/s
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Work done in moving a charge in potential difference is
W= qV
W = 1.6x 10^-19x (300-100)
W = 3.2 x10^-17
By work energy theorem
change in KE=W
1/2m(vf^2 -vi^2) = 3.2x10^-17
vf2 - vi^2 = 7.03 x 10^13
vf^2 = 9.53 x 10^13
vf = 9.7 x 10^6
Hope this helps
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