An electron has an initial velocity of 3.00 ? 10 6 m/s in a uniform 9.00 ? 10 5

Question

An electron has an initial velocity of 3.00 ? 106 m/s in a uniform 9.00 ? 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.

An electron has an initial velocity of 3.00 ? 106 m/s in a uniform 9.00 ? 105 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. What is the direction of the electric field? How far does the electron travel before coming to rest m How long does it take the electron to come to rest? s What is the electron's speed when it returns to its starting point? m/s

Explanation / Answer

a) same as initial velocity

b) Force = q*E

=>mass*acc = q*E

acc = q*E/m = (1.6*10^-19)*(9*10^5)/(9.1*10^-31) = -1.582*10^17 m/s^2

as

v^2 - u^2 = 2as

as,

V =0

=> s = -u^2/2*a = -(3*10^6)^2/(-2*1.582*10^17) = 2.843*10^-5 m

c) v = u + a*t

=> t = -u/a = -3*10^6/(-1.582*10^17) = 1.895*10^-11 s

d) when s =0

=>v^2 - u^2 = 2as

=> v = +-u

=>v = -3*10^6 m/s

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