ACADEMIC CENTRE OF EXCELLENCE (ACE) - PHYSICS 9702
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Saturday, August 3, 2019
Electric Fields: Exercise 1
Calculate the speed of an electron accelerated from rest through a distance of 40 mm by a uniform electric field of 3.0 × 10
3
N C
−1
.
Solution:
force =
Eq
= (3 ×
10
3
× 1.6 x 10
−19
)
=
4.8
×
10
−16
N
acceleration = force/mass = 4.8 × 10
−16
/9.1
× 10
−31
=
5.27
× 10
14
ms
−2
v
2
= 0 + 2
as
= 2 ×
5.27
× 10
14
ms
−2
× 40 × 10
−3
so speed
v
=
6.5 × 10
6
m s
−1
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