An electron falls through a distance of 1.5 cm in a uniform electric field of magnitude 2.0 × 10 4 N C –1 in the direction of the field . Then the direction of the field is reversed keeping its magnitude unchanged and a proton falls through the same distance . compute the time of fall in each case.
distance travelled by each particle will be given as
s = ut + (1/2)at2
here
initial velocity, u = 0 m/s
so,
s = (1/2)at2
or
time taken
t = (2s/a)1/2 ..................(1)
now,
the force exerted by the electric field causes acceleration in the particle, so
F = qE = ma
or
acceleration
a = qE/m .................(2)
thus,
equation (1) becomes
t = (2s / qE/m)1/2
or
t = (2ms / qE)1/2 ...................(3)
we shall use this equation to calculate the time taken by each particle
.
.
Times of fall
given
s = 1.5cm = 0.015m
E = 2.0 × 104 N/C
Now, for an electron
m = 9.1 x 10-31 kg
q = 1.6 x 10-19 C
thus, we get
telectron = [ (2 x 9.1x10-31 x 0.015) / (1.6x10-19 x 2.0 × 104)]1/2
or
telectron = 2.9 x 10-9 s
and
For a proton
m = 1.6 x 10-27 kg
q = 1.6 x 10-19 C
thus, we get
tproton = [ (2 x 1.6x10-27 x 0.015) / (1.6x10-19 x 2.0 × 104)]1/2
or
tproton = 1.3 x 10-7 s