Le Fevre High School
SACE Stage 2 Physics
Motion of Charged Particles in Electric Fields 2
1.
An electron is accelerated from rest in the electron gun of a cathode ray
oscilloscope by a potential difference of 1.8 x 103 V. The electron then
enters the region between two parallel and horizontal deflecting plates
which are 4.00 cm apart and across which is applied a potential of 420.0 V.
The electric field between the plates deflects the electron from its straight
line path down the axis of the tube into a parabolic path.
(a) What is the horizontal velocity of the electron when it enters the deflecting
plates?
v1 = 0; accelerating voltage Va = 1800 Volt;
deflecting voltage Vd = 420 Volt
d = 4 cm = .04 m
Gain in kinetic energy in electron gun
K = Vaq
1
 m v2 = 1800 x 1.6 x 10 -19
2
v = 2.52 x 10 7 m s -1
 initial horizontal velocity
vH = 2.52 x 10 7 m s -1
(b) What is the horizontal velocity of the electron when it leaves the deflecting
plates?
vH is constant because the electric force on the electrons while they are
between the plates is constant in size and at 900 to the axis (field uniform).
Hence vH on leaving = 2.52 x 10 7 m s -1
(c) How long does the electron spend travelling between the plates if they are 6.60
cm long?
length of plate
Time =
horizontal velocity
0.066
=
2.52 x10 7
= 2.62 x 10-9 s
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(d) If the electric field accelerates the electron vertically downwards, what is its
vertical displacement as a result of travelling between the plates?
V
E = d
d
420
=
0.04
E = 10500 V m-1
F = Eq
= 10500 x 1.6 x 10-19
F = 1.68 x 10-15 N
F
a=
m
1.68 x 10 15
=
9.11 x 10 31
a = 1.85 x 1015 m s-2
1
s =
a t 2
2
(v1 = 0 in vertical direction)
1
= x 1.85 x 10 15 x (2.62 x 10 -9 ) 2
2
s = 6.35 x 10 -3 m (6.35 mm)
2.
A
C
1.5 m
B
The diagram shows three points A, B and C in a uniform electric field of strength
6.0 x 102 V m-1. The line joining the points A and B is parallel to the field and the
line joining A and C is perpendicular to the field.
The distance AB = distance AC = 1.5 m. A particle of mass 3.0 x 10-8 kg
carrying a positive charge of 2.0 x 10-10 C is initially at rest at A.
(a) Calculate the acceleration of the particle and hence find its displacement after
one second.
a = 4 m s-2
F
a=
displacement after 1 sec given by
m
1
Eq
s =
a t 2
=
2
m
= 0.5 x 4 x 12
6 x 10 2 x 2 x 10 10
=
= 2 m in direction of field
3 x 10  8
(b) What is the electric potential difference between the points A and B?
E = 600 V m-1
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AB = d = 1.5 m
V
 E=
d
V = Ed
= 600 x 1.5
V = 900 Volt
(c) What is the kinetic energy of the particle at the point B?
Gain in K = Vq
= 900.x 2.0 x 10-10
K = 1.8 x 10-7 J
(d) What is the electric potential difference between the points A and C?
Points A & C would lie on the same equipotential line, hence V = 0
(e) Describe how a uniform electric field can be produced in a laboratory.
Uniform electric fields can be produced by a pair of parallel metal plates
between which there is a potential difference. This is achieved by
connecting one plate to the positive terminal of a power supply and the
other to the negative terminal.
3.
In a cathode ray tube electrons are accelerated from rest through a
potential difference of 1000 Volt; they then pass between a pair of parallel
plates 3.0 cm long and 1.0 cm apart. A potential difference of 50 Volt is
applied to the plates. Assume the electric field between the plates to be
uniform and confined to the space between them. On emerging from the
region between the plates the electrons pass on to hit a screen some
distance away from the plates.
+
electron gun
d
1.0 cm
1000 V
3 cm
screen
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(a) Calculate the speed of the electrons emerging from the anode.
1
m v 2 = eV
2
2eV
v=
m
v=
2 x 1.6 x 10 19 x 1000
9.11 x 10 31
v = 1.88 x 10 7 m s -1
(b) Calculate the force on an individual electron when it is between the plates.
F = Ee
eV
d
50 x 1.6 x 10 19
=
0.01
F = 8.0 x 10-16 N vertically up
=
(c) How long will an electron take to pass between the plates? Explain your
working.
The horizontal component of velocity remains constant in the region between
the plates as there are no horizontal components of force in
that region. Hence
s
t = horizontal
vH
t =
3 x 10 2
1.88 x 10 7
= 1.6 x 10 -9 s
(d) What is the deflection (‘d’ as shown on the diagram) of the electrons at the
instant they leave the region between the plates?
1
 s = a t 2
2
1 F
s = . . t 2
2 m
1
8 x 10 16
 s = .
(1.6 x 10 -9)2
 31
2 9.11 x 10
= 1.1 x 10 -3 m
(e) Comment on the path of the electrons on leaving the region between the plates.
Path would be straight since no resultant force is acting - Newton's first Law
- velocity constant. Note that gravitational effects are negligible as the speed
of the electrons are so high.
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4.
Describe the main parts of a cyclotron and describe the purpose of each
part (Include a diagram).
Main Components:
Alternating
potential
difference
Evacuated outer
chamber
(a) Ion Source: heated filament produces
thermal electrons these collide with gas
molecules. The collisions release electrons
from the molecules thus causing ionisation.
(b) Dees: Hollow copper electrodes with an
alternating potential difference applied to
them. Results in an electric field in the gap
between the dees, but no electric field inside
interior of the dees.
(c) Evacuated outer chamber: Ensures no
energy losses by accelerated ions colliding with
air particles.
(d) Electromagnet: The magnetic field acts on
ions while inside the hollow dees resulting in a
semi-circular path.
Ion
Sour
ce
dee
dee
Target
Magnetic S
pole
dee
ion source dee
Magnetic N
pole
5.
Why is there no field insid of the interior of the dees?
No field exists inside a hollow charged conductor
6.
One of the original cyclotrons of Lawrence and Livingston had a dee
diameter of 24 cm and used an alternating potential difference of 4000 V
maximum between the dees. Protons were accelerated in the ,machine
through 150 complete revolutions before emerging.
(a) By how much did the kinetic energy (in eV) of a proton increase on
passing between the dees at maximum potential difference?
R = 12 cm = 0,12 m, V = 4000 V, n =150
(a) K = W = (qV) = (1.6 x 10-19)(4000)
= 6.4 x 10-16 J
or in eV = (1 e x 4000V) = 4000 eV
(b) What was the kinetic energy (in eV) of a proton emerging from the
cyclotron?
K = W = n (qV)
= 150{(1.6 x 10-19)(4000)x2}
(as cross the gap between the dees twice each revolution)
= 1.92 x 10-13 J or in eV = 150(4000)(2) = 1200 keV = 1.2 MeV
7.
The first prototype cyclotron built by Stanley Livingston as a student under
the supervision of Professor Ernest Lawrence had a dee diameter of just 9
cm and could be held in the hand. In one experiment with this cyclotron an
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alternating potential difference of 1000 V maximum was used to accelerate
singly charged ions of molecular hydrogen (H2+) to a kinetic energy of
80000 eV. How many revolutions did the ions com- plete in the prototype
to reach this energy?
W in accelerating the ions through one revolution = 2qV = 2(2e)(1000) = 4000 eV
Hence number revolutions to reach 80000 eV
= 80000 4000 = 20