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

9702/Oct Nov/13/2013/Q31

A small charge q is placed in the electric field of a large charge Q.

Both charges experience a force F.

What is the electric field strength of the charge Q at the position of the charge q?

Solution:

Answer: D

A small charge q is placed in the electric field of a large charge Q. Both charges experience force F.

Electric force F at position of charge q = Eq

Note that F also account for the magnitudes of the charges. The electric force can also be given by another formula (Coulomb’s law: F = Qq / 4πϵ₀r²) but we don’t need to calculate this here. The electric force is already given to be F.

Electric field is the force per unit charge. The charge in the expression is that on which the force acts (not the charge creating the field), so q is involved rather than Q.

Electric field strength of charge Q at position of charge q, E = F/q

Reference: PYQ - Oct/Nov 2013 Paper 13 Q31