(c) No, the fundamental charge could be any positive integer multiple of this charge. We can increase our certainty if we
had more experimental data.
(d) This charge is actually smaller than the actual fundamental charge and so we must conclude that there is some error in
the experiment. If it was a positive integer multiple of the actual fundamental charge then we might conclude that more
data is required.
Making Connections
8.
ε = 1.0 ¯ 102 N/C
(a) q = ?
ε=
kq
rE 2
q=
ε rE 2
k
(1.0 ×10
=
2
N/C )(6.4 ×10 6 m )
2
9.0 ×109 N ⋅ m 2 /C 2
q = 4.5 × 105 C
The charge on Earth is 4.5 ¯ 105 C. Since the field lines are directed towards Earth the charge is negative.
(b) (i) Both fields can be represented by straight field lines directed towards the centre of the earth.
(ii) Gravity always attracts objects while the electric field can either attract or repel objects.
(iii) Both fields are inverse square laws meaning they decrease with the square of the distance from the centre of Earth.
(c) m = ?
FE = Fg
eε = mg
eε
g
m=
(1.6 ×10
=
−19
C )(1.0 × 10 2 N/C )
9.8 N/kg
−18
m = 1.6 × 10 kg
The largest mass that could be suspended is 1.6 ¯ 10–18 kg.
(d) Both particles have small enough masses but the electric field points towards the centre of Earth meaning Earth has a
negative charge. It will attract the proton (which cannot be suspended) and repel the electron. The electron must be at a
higher altitude since its mass is so small.
9. The dust particles could have a negative charge and the electric field of the Earth would then suspend them above the
surface causing them to float. If this is true a negatively charged ebonite rod will repel them.
7.6 THE MOTION OF CHARGED PARTICLES IN ELECTRIC FIELDS
PRACTICE
(Page 368)
Understanding Concepts
1.
V = 1.5 ¯ 102 V
W = 0.24 J
q=?
W = Vq
q=
W
V
0.24 J
1.5 ×10 2 V
q = 1.6 × 10 −3 C
The magnitude of the charge is 1.6 ¯ 10–3 C.
=
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 453
2.
q = 2e
m = 6.6 ¯ 10–27 kg
V = 2.0 ¯ 103 V
v=?
(a) Using the equation:
W = qV =
v=
=
1 2
mv
2
2Vq
m
2 ( 2.0 × 103 V )(3.2 × 10−19 C )
6.6 × 10−27 kg
v = 4.4 × 105 m/s
The α particle would reach the plate at a velocity of 4.4 ¯ 105 m/s.
1
(b) V = ( 2.0 × 103 V ) = 1.0 × 103 V
2
2Vq
v=
m
=
2 (1.0 × 103 V )(3.2 × 10−19 C )
6.6 ×10 −27 kg
v = 3.1× 105 m/s
The α particle would reach the plate at a velocity of 3.1 ¯ 105 m/s.
3. m = 1.0 × 10–5 kg
q = 4.0 × 10–7 C
d = 50.0 cm = 0.50 m
V = 8.0 × 102 V
(a) W = ?
W = qV
= ( 4.0 × 10−7 C )(8.0 × 102 V )
W = 3.2 ×10 −4 J
The work done by the string is 3.2 ¯ 10–4 J.
(b) F = ?
W
F=
d
3.2 ×10 −4 J
=
0.50 m
F = 6.4 ×10 −4 N
The average force required is 6.4 ¯ 10–4 N.
(c) EK = ?
EK = W
EK = 3.2 × 10 −4 J
The kinetic energy of the pith ball is 3.2 ¯ 10–4 J.
454 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(d) v = ?
1 2
mv = EK
2
2 EK
v=
m
=
4.
2 (3.2 ×10 −4 J )
1.0 × 10 −5 kg
v = 8.0 m/s
The speed of the pith ball is 8.0 m/s.
r = 1.0 ¯ 10–12 m
EK = ?
v=?
kq q
EE = 1 2
r
(9.0 ×10 N ⋅ m /C )(1.6 ×10
=
9
2
2
−19
1.0 ×10 −12 m
C)
2
EE = 2.3 × 10 −16 J
Etotal = EK + EE = 2.3 × 10 −16 J
′ = EK′ + EE′
Etotal
′
Etotal
(since EK = 0)
1

= 2  mv 2  + 0
2


= mv 2
′ = 2.3 ×10 −16 J
But Etotal = Etotal
v=
=
5.
Etotal
m
2.3 × 10 −16 J
9.1× 10 −31 kg
v = 1.6 × 107 m/s
The kinetic energy of the each electron is 2.3 ¯ 10–16 J. The speed of each electron is 1.6 ¯ 107 m/s.
m = 3.3 ¯ 10–27 kg
q = 1.6 ¯ 10–19 C
v = 5.0 ¯ 106 m/s
V=?
1
W = qV = mv 2
2
2
mv
V=
2q
(3.3 ×10
=
−27
kg )(5.0 × 106 m/s )
2
2 (1.6 ×10 −19 C )
V = 2.6 × 105 V
The potential difference required is 2.6 ¯ 105 V.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 455
Section 7.6 Questions
(Page 371)
Understanding Concepts
1.
V = 1.2 ¯ 103 V
d = 0.12 m
(a) v = ?
W = qV =
v=
=
1 2
mv
2
2Vq
m
2 (1.2 × 103 V )(1.6 × 10−19 C )
9.11 ×10 −31 kg
v = 2.1 × 107 m/s
The electron will pass through the hole in the positive plate at a speed of 2.1 ¯ 107 m/s.
(b) The electron is not pulled back because there is no electric field outside of the parallel plate apparatus.
(c) Place a small hole in the negative plate instead. Some source of protons would be required. (A hot filament will not
suffice.)
(d) v = ?
2Vq
v=
m
=
2 (1.2 × 103 V )(1.6 × 10−19 C )
1.67 ×10 −27 kg
v = 4.8 × 105 m/s
The speed of the emerging proton is 4.8 ¯ 105 m/s.
2. ε = 2.5 ¯ 102 N/C
v = 1.2 ¯ 106 m/s
(a) r = 2.5 cm = 2.5 ¯ 10–2 m
W=?
W = qV
= qε r
= (1.6 × 10−19 C )( 2.5 × 10 2 N/C )( 2.5 × 10 −2 m )
W = 1.0 × 10−18 J
The work done is 1.0 ¯ 10–18 J.
(b) v΄ = ?
Knowing the relationship between work and energy, W = ∆EK = EK′ − EK , we can calculate the kinetic energy:
EK′ = W + EK
1
= W + mv 2
2
= 1.0 × 10 −18 J +
2
1
9.11× 10 −31 kg )(1.2 × 106 m/s )
(
2
EK′ = 1.7 × 10 −18 J
456 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
We can then calculate the speed:
EK′ = 12 mv′2
2 EK
m
v′ =
2 (1.7 ×10 −18 J )
=
9.11× 10−31 kg
v′ = 1.9 × 106 m/s
The speed of the electron at this point is 1.9 ¯ 106 m/s.
(c) r = ?
1
W = qV = mv 2
2
1 2
qε r = mv
2
1 2
mv
r= 2
qε
2
1
(9.11×10−31 kg )(1.2 ×106 m/s )
2
=
(1.6 ×10−19 C )( 2.5 ×102 N/C )
r = 1.6 × 10−2 m
The electron will move 1.6 ¯ 10–2 m, or1.6 cm into the field before it comes to rest.
3. v = 3.5 ¯ 106 m/s
(a) r = ?
ET = ET′
2 EK = EE′
mv 2 =
r=
=
ke 2
r2
ke 2
mv 2
k e
 
mv
9.0 × 109 N ⋅ m 2 /C 2  1.6 × 10−19 C 


9.11× 10 −31 kg  3.5 × 106 m/s 
r = 4.5 × 10 −6 m
The smallest possible distance between the two electrons is 4.5 ¯ 10–6 m.
(b) No, this would only occur during a head on collision, which is unlikely to occur for two electrons fired at each other
from far apart.
4. m = 6.64 ¯ 10–27 kg
q1 = +2e
r = 4.7 ¯ 10–15 m
q2 = +79e
EE = ?
kq q
EE = 1 2
r
=
(9.0 ×10 N ⋅ m /C ) (2 (1.6 ×10
9
=
2
2
4.7 ×10
−19
−15
)(
C ) 79 (1.6 × 10−19 C )
)
m
−12
EE = 7.7 × 10 J
The α particle must have an initial energy of 7.7 ¯ 10–12 J.
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 457
5.
v = 3.00 ¯ 106 m/s
ε = 120 N/C
∆l = 4.0 cm = 4.0 ¯ 10–2 m
(a) ∆dy = ?
We treat this like a projectile motion problem. The horizontal motion is uniform.
The time to cross the plates is:
∆l
∆t =
vx
4.0 ×10 −2 m
3.00 × 106 m/s
∆t = 1.3 × 10 −8 s
=
The acceleration of the electron is directed up with magnitude:
F
a= E
m
eε
=
m
(1.6 ×10−19 C ) (120 N/C )
=
9.11× 10−31 kg
a = 2.1× 1013 m/s 2
The initial velocity up is zero, therefore:
1
2
∆d y = a ( ∆t )
2
2
1
= ( 2.1 ×1013 m/s 2 )(1.3 ×10 −8 s )
2
∆d y = 1.8 × 10 −3 m
The vertical deflection of the electron is 1.8 ¯ 10–3 m.
(b) v′y = ?
v′y = 0 + a∆t
= ( 2.1 × 1013 m/s 2 )(1.3 × 10−8 s )
v′y = 2.7 ×105 m/s
The vertical component of the final velocity is 2.7 ¯ 105 m/s.
(c) θ = ?
v′y
tan θ =
vx
 2.7 × 105 m/s 
θ = tan −1 

6
 3.00 × 10 m/s 
θ = 5.1°
The angle at which the electron emerges is 5.1°.
Making Connections
6. (a) Have the beam of electrons pass through a pair of horizontal parallel plates. The electrons will accelerate towards the
positive plate. We can accelerate the electrons up or down by changing the amounts of charge on each plate and
changing which plate is negative.
(b) Have the beam pass through a pair of vertical parallel plates and deflect them in a similar fashion as explained above.
We can increase the deflection from left to right by gradually increasing the potential difference between the plates that
will increase the horizontal deflection. We can make the beam jump from right to left by reversing the charge on the
plates.
458 Unit 3 Electric, Gravitational, and Magnetic Fields
Copyright © 2003 Nelson
(c) The potential difference measured on the surface of the skin of the patient (which is used to indicate potential changes
from the heart) is used to control the potential changes on the horizontal plates causing the beam to deflect vertically.
The vertical plates change their potential independent of measurements to ensure the beam moves across the screen
horizontally at a regular rate.
(d) Answers may vary depending on the speed of the electrons, the length of the plates, and the electric field between the
plates. For example, in question 5 the deflection is 5.1º, which is close enough. If we double the electric field we will
double the acceleration of the electron meaning we will also double the final vertical and so the angle will be about 10º.
tan10°
This is true even though tangent is not a linear function since
= 2 (to two significant digits).
tan 5°
7. (a) The objects will accelerate towards each other and as they do so the acceleration and velocity will both increase, but
energy and momentum will be conserved. Electric potential energy and gravitational potential energy will be
transformed into kinetic energy.
(b) The motion will be similar but only gravitational potential energy will be available to convert into kinetic energy.
CHAPTER 7 LAB ACTIVITIES
Investigation 7.2.1: Factors Affecting Electric Force between Charges
(Pages 372–373)
Questions
(i) The force is larger when the distance between the charges is smaller and the charges are larger. The force is directly
proportional to the product of the charges, and inversely proportional to the distance between them squared.
kq q
(ii) FE = 12 2
r
Hypothesis/Prediction
(a) The force is directly proportional to the product of the charges and inversely proportional to the distance between them
kq q
squared: FE = 12 2 .
r
Analysis
(b)
T cos θ = mg
T sin θ = FE
F
T sin θ
= E
T cos θ mg
FE = mg tan θ
(c) Answers may vary. Students will plot a curve (an inverse square).
Copyright © 2003 Nelson
Chapter 7 Electric Charges and Electric Fields 459