The electrons in a particle beam each have a kinetic energy of 1.6×10−17J. What are the magnitude and direction of the electric field that stops these electrons in a distance of 10.0 cm
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a
103 v/m in the direction of velocity of electrons
b
103 v/m positive direction of velocity of electrons
c
103 v/m perpendicular to velocity of electrons
d
106 v/m perpendicular to velocity of electrons
answer is A.
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Detailed Solution
K.Ef-K.Ei=F×S0-1.6×10-17=-E×1.6×10-19×0.1E=103vm in the direction of velocity