Electric Fields: Exercise 2

An electron starts from rest from the bottom plate in Fig. 17.11. The potential difference across the plates is 1600 V and the separation of the plates is 15 mm. Calculate the time taken for the electron to reach the top plate. 

Solution:
F = Vq/d = (1600 x 1.6 x 10^-19) / 0.015 = 1.71 x 10^-14 N
a = force / mass = 1.71 x 10^-14 / 9.1 x 10^-31 = 1.88 x 10^16 ms^-2
s = ut + 0.5 at^2
0.015 = 0 + 0.5 (1.88 x 10^16) (t^2)
t^2 = 1.596 x 10^-18
t = 1.26 x 10^-9 s

For the electron in question 2  calculate:

a  the work done by the field on the electron,

b  the gain in kinetic energy,

c  the speed of the electron.

Solution:

a) work done = Fd = Eqd = Vq

W = 1.71 x 10^-14 x 0.015 = 2.565 x 10^-16 J

b) kinetic energy = Vq = 1600 x 1.6 x 10^-19 = 2.56 x 10^-16 J

c) v^2 = u^2 + 2as = 2 (1.88 x 10^16)(0.015) = 5.64 x 10^14

v = 2.37 x 10^7 ms^-1

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 × 103 N C−1.

Solution:

force = Eq = (3 × 103 × 1.6 x 10−19) = 4.8 × 10−16 N

acceleration = force/mass = 4.8 × 10−16 /9.1 × 10−31 = 5.27 × 1014 ms−2

v2 = 0 + 2as = 2 × 5.27 × 1014 ms−2 × 40 × 10−3 so speed v = 6.5 × 106 m s−1