Question

An electron (m = 9 x 10-31 kg) leaves the cathode of a radio tube with zero initial velocity and travels in a straight line to the anode, which is 1 cm away. It reaches the anode with a velocity of 6 x 106 m s-1
i)calculate the acceleration and the force ?
ii) the time taken for the electron to reach the anode ?

Answer 1

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Answer 2

F = m a
a = change in v/ change in t

v = Vi + a t
6*10^6 = 0 + a t = a t

x = Xi + Vi t + (1/2) a t^2

10^-2 meters = 0 + 0 +(1/2) a t^2
so

a t^2 = 2*10^-2
and
a = 6*10^6/t

so
6*10^6/t * t^2 = 2*10^-2
6 t = 2*10^-10 seconds
t = (1/3)10^-10 second
then
a = 18 *10^4 m/s^2

now F = m a

Answer 3

by the way, you could have used the average speed

Vav = 3*10^6
so
t = 10^-2/3*10^6 = (1/3)*10^-8
(not 10^-10)