Download MODULE: FROM IDEAS TO IMPLEMENTATION Chapter

MODULE: FROM IDEAS TO IMPLEMENTATION
Chapter 10: Cathode rays and the development of television
(Questions, pages 185-186)
Note: the charge on a single electron is taken as –1.6x10-19C.
1.
B = 6.0 x 10-2 T
v = 2.5 x 107 m s-1
q = 1.6 x 10-19 C
F = Bqv
= 6.0 x 10-2 x 1.6 x 10-19 x 2.5 x 107
= 2.4 x 10-13 N
2.
v = 3.0 x 104 m s-1
B = 9.0 x 10-2 T
q = 2 x 1.6 x 10-19 C
F = Bqv
= 9.0 x 10-2 x 2 x 1.6 x 10-19 x 3.0 x 104
= 8.6 x 10-16 N
3.
v = 1.8 x 106 m s-1
B = 0.60 T
q = 1.6 x 10-19 C
F = Bqv
= 0.60 x 1.6 x 10-19 x 1.8 x 106
= 1.7 x 10-13 N
4.
F = ma
m = 9.1 x 10-31 kg
1.7 x 10-13 = 9.1 x 10-31a
a = 1.9 x 1017 m s-2
5.
d = 5 x 10-2 m
V = 200 V
(a) E =
V
d
200
= 5 x 10 -2
= 4.0 x 103 V m-1 (to the left)
Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 10
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(b) q = 1.6 x 10-19 C
F = qE
= 1.6 x 10-19 x 4.0 x 103
= 6.4 x 10-16 N to the right
(c) F on proton is 6.4 x 10-16 N to left as same charge magnitude
(d) Opposite direction as opposite charges on electron and proton.
(e) W = qV
= 1.6 x 10-19 x 200
= 3.2 x 10-17 J each
6. Conservation of charge means the total charge remains constant.
The positive charge remained on the cathode and the negative charge moved around
the circuit back towards the cathode.
7. <take in figure 10A> The electric field is strongest near each charge where the
electric field lines are closest together.
8.
q = 1.0 x 10-6 C
E = 20 N C-1
F = qE
= 1.0 x 10-6 x 20
= 2.0 x 10-5 N
9. The drawing should closely resemble figure 10.7(b) on page 174 of the textbook.
10. A ‘uniform’ electric field means a constant strength field in which a unit charge
experiences the same force at every point. At the edges, the field gradually becomes
weaker.
11.
m = 2.4 x 10-12 kg
E = 4.9 x 107 N C-1
upward force = downward force
qE = mg
q x 4.9 x 107 = 2.4 x 10-12 x 9.8
q = 4.8 x 10-19 C
<take in figure 10B>
12.
d = 10.0 cm = 0.1 m
V = 20.0 V
V
(a) E =
d
=
20
0 .1
= 2.0 x 102 V m-1 or N C-1
Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 10
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(b) q = 2.0 x 10-3 C
F = qE
= 2.0 x 10-3 x 2.0 x 102
= 0.40 N to negative plate
13.
q = 5.25 mC = 5.25 x 10-3 C
v = 300 m s-1 NE
B = 0.310 T
F = Bqv as v ⊥ B
= 0.310 x 5.25 x 10-3 x 300
= 4.9 x 10-1 N
14. In air, cathode rays collide with air molecules which become ionised and the
cathode rays lose energy or become part of these molecules. The rays do not continue
in straight line motion.
In a vacuum tube, the rays are attracted to the cathode and move in a straight
line. The rays are observed when they hit a phosphorescent material (such as zinc
sulphide) coated on a screen or cause a green glow on glass.
15. If charged particles enter a uniform magnetic field at right angles to their velocity,
they experience a force perpendicular both to their velocity and to the magnetic field
resulting in uniform circular motion.
16. If charged particles enter a uniform magnetic field at an angle other than at right
angles to their velocity, they experience a force perpendicular to both the magnetic
field and to the component of their velocity perpendicular to the magnetic field
resulting in circular motion. The other velocity component will be at right angles to
the plane of the circular motion so that the charges move in a helical path.
17.
(a) Cathode rays behave as waves when they:
ƒ move in straight lines (evidenced by sharp shadows of opaque objects)
ƒ pass through thin metal foils
ƒ reflect at equal angles
ƒ produce chemical changes, e.g. change the colour of silver salts
ƒ produce fluorescence.
(b) Cathode rays behave as particles when they:
ƒ move in straight lines (evidenced by sharp shadows of opaque objects)
ƒ reflect at equal angles
ƒ are deflected by a magnetic field
ƒ are deflected by an electric field
ƒ exert momentum (evidenced by turning the paddlewheel)
ƒ move at less than the speed of light or other electromagnetic waves
ƒ move away from the cathode at right angles to cathode.
Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 10
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18. There was a sharp shadow of the metal cross, formed on the end of the glass tube,
evidenced by the areas of green glow around the shape of the cross. This showed that
the rays travelled in straight lines.
The paddle wheel must be pushed by a particle with momentum if it is to start rolling.
19. <take in figure 10C> An electron entering an electric field will experience a force
in the opposite direction to the electric field. This force will cause acceleration in this
direction. If the electron enters at right angles to the electric field, then this velocity
will be constant while the velocity perpendicular to its original velocity will increase
from zero. The combined motion is in a parabolic path while it is in the electric field.
20. The electron must enter at right angles to the magnetic field. The path will be
semicircular and the electron leaves the magnetic field in the opposite direction to
entering.
<take in figure 10D> If the magnetic field is switched on or increased while the
electron is in the middle of the magnetic field region, then uniform circular motion
can result if the radius of the motion is such that the electron remains in the magnetic
field.
21.
B = 1.00 x 10-2 T
d = 10 x 10-3 m
V = 300 V
V
(a) E =
d
300
=
10x10 −3
= 3.00 x 104 V m-1 or N C-1
(b) qE = Bqv
E = Bv
3.00 x 104 = 1.00 x 10-2 v
v = 3.00 x 106 m s-1
(c) qE = qvB
qvB = qE
= 1.6 x 10-19 x 3 x 104
magnetic force = 4.8 x 10-15 T
22. J. J. Thomson developed the ‘plum pudding’ model of the atom, consisting of a
sphere of positive material embedded with the electrons in regular rings. Note that he
did not see the atom as a haphazard mixture of positive and negative charges. Only
some of the electrons were able to be released from the atom.
23. The basic difference is that cathode-ray beams are controlled by electric fields
produced by charged plates (capacitors) in the CRO and by magnetic fields produced
by Helmholtz coils in television sets.
Physics 2: HSC Course, 2nd edition (Andriessen et al, 2003), Chapter 10
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