An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of
magnitude 2 x 10' N/C. The time taken by the electron to fall to this distance is
1) 1.3 x 102 g 2)2.1 x 10-12 g 3)1.6 x 10-10 g 4)2.9 x 100 g
Answers
S=1.5cm=1.5×10
−2
m
u=0m/s
m
e
: mass of electron =9.10×10
−31
Kg
m
p
: mass of proton =1.672×10
−27
Kg
from coloumb's force,
F
=q
E
and given
E
=2×10
4
N/c
force proton
F
(+)
=1.6×10
−19
×2×10
4
=3.2×10
−19
N
force on electron,
F
(−)
=−1.6×10
−19
×2×10
4
=−3.2×10
−15
N
From Newton's Second law,
F
=m
a
∴ for proton, a
(+)
=
m
p
F
(+)
=1.913×10
12
m/s
2
for electron, a
(−)
=
m
e
F
(−)
=−0.35×10
16
m/s
2
from equ of motion, S=ut+
2
1
at
2
⟶(1)
ATQ, The direction of the field was reversed, hence we will ignore the sign convention for acceleration.
∴ Substituting values in equ (1) for proton,
1.5×10
−2
=
2
1
×1.913×10
12
×t
2
1.568×10
−14
=t
2
⇒t=1.252×10
−7
seconds
Substituting values in equ(1) for electron,
1.5×10
−2
=
2
1
×3.5×10
15
×t
2
0.857×10
−17
=t
2
⇒t
0
=2.927×10
−19
seconds
JAY HIND