Chapter 15
Electric Forces and
Electric Fields
Quick Quizzes
1.
(b). Object A must have a net charge because two neutral objects do not attract each other.
Since object A is attracted to positively-charged object B, the net charge on A must be
negative.
2.
(b). By Newton’s third law, the two objects will exert forces having equal magnitudes but
opposite directions on each other.
3.
(c). The electric field at point P is due to charges other than the test charge. Thus, it is
unchanged when the test charge is altered. However, the direction of the force this field
exerts on the test change is reversed when the sign of the test charge is changed.
4.
(a). If a test charge is at the center of the ring, the force exerted on the test charge by charge
on any small segment of the ring will be balanced by the force exerted by charge on the
diametrically opposite segment of the ring. The net force on the test charge, and hence the
electric field at this location, must then be zero.
5.
(c) and (d). The electron and the proton have equal magnitude charges of opposite signs.
The forces exerted on these particles by the electric field have equal magnitude and
opposite directions. The electron experiences an acceleration of greater magnitude than
does the proton because the electron’s mass is much smaller than that of the proton.
6.
(a). The field is greatest at point A because this is where the field lines are closest together.
The absence of lines at point C indicates that the electric field there is zero.
7.
(c). When a plane area A is in a uniform electric field E, the flux through that area is
Φ E = EA cosθ where θ is the angle the electric field makes with the line normal to the
plane of A. If A lies in the xy-plane and E is in the z-direction, then θ = 0° and
Φ E = EA = ( 5.00 N C ) ( 4.00 m 2 ) = 20.0 N ⋅ m 2 C .
8.
(b). If θ = 60° in Quick Quiz 15.7 above, then
Φ E = EA cosθ = ( 5.00 N C ) ( 4.00 m 2 ) cos ( 60° ) = 10.0 N ⋅ m 2 C
9.
(d). Gauss’s law states that the electric flux through any closed surface is equal to the net
enclosed charge divided by the permittivity of free space. For the surface shown in Figure
15.28, the net enclosed charge is Q = −6 C which gives Φ E = Q ∈0 = − ( 6 C ) ∈0 .
1
2
10.
CHAPTER 15
(b) and (d). Since the net flux through the surface is zero, Gauss’s law says that the net
change enclosed by that surface must be zero as stated in (b). Statement (d) must be true
because there would be a net flux through the surface if more lines entered the surface
than left it (or vise-versa).
Electric Forces and Electric Fields
3
Answers to Even Numbered Conceptual Questions
2.
Conducting shoes are worn to avoid the build up of a static charge on them as the wearer
walks. Rubber-soled shoes acquire a charge by friction with the floor and could discharge
with a spark, possibly causing an explosive burning situation, where the burning is
enhanced by the oxygen.
4.
Electrons are more mobile than protons and are more easily freed from atoms than are
protons.
6.
No. Object A might have a charge opposite in sign to that of B,
but it also might be neutral. In this latter case, object B causes
object A to be polarized, pulling charge of one sign to the near
face of A and pushing an equal amount of charge of the
opposite sign to the far face. Then the force of attraction
exerted by B on the induced charge on the near side of A is
slightly larger than the force of repulsion exerted by B on the
induced charge on the far side of A. Therefore, the net force
on A is toward B.
+++
+
+
B
+
+
+++
–+
– +A
–+
8.
If the test charge was large, its presence would tend to move the charges creating the field
you are investigating and, thus, alter the field you wish to investigate.
10.
She is not shocked. She becomes part of the dome of the Van de Graaff, and charges flow
onto her body. They do not jump to her body via a spark, however, so she is not shocked.
12.
An electric field once established by a positive or negative charge extends in all directions
from the charge. Thus, it can exist in empty space if that is what surrounds the charge.
14.
No. Life would be no different if electrons were positively charged and protons were
negatively charged. Opposite charges would still attract, and like charges would still
repel. The designation of charges as positive and negative is merely a definition.
16.
The antenna is similar to a lightning rod and can induce a bolt to strike it. A wire from the
antenna to the ground provides a pathway for the charges to move away from the house
in case a lightning strike does occur.
18.
(a) If the charge is tripled, the flux through the surface is also tripled, because the net flux
is proportional to the charge inside the surface. (b) The flux remains constant when the
volume changes because the surface surrounds the same amount of charge, regardless of
its volume. (c) The flux does not change when the shape of the closed surface changes. (d)
The flux through the closed surface remains unchanged as the charge inside the surface is
moved to another location inside that surface. (e) The flux is zero because the charge
inside the surface is zero. All of these conclusions are arrived at through an understanding
of Gauss’s law.
20.
(a) –Q (b) +Q (c) 0 (d) 0 (e) +Q (See the discussion of Faraday’s ice-pail experiment in the
textbook.)
4
22.
CHAPTER 15
The magnitude of the electric force on the electron of charge e due to a uniform electric
G
field E is F = eE . Thus, the force is constant. Compare this to the force on a projectile of
mass m moving in the gravitational field of the Earth. The magnitude of the gravitational
force is mg. In both cases, the particle is subject to a constant force in the vertical direction
and has an initial velocity in the horizontal direction. Thus, the path will be the same in
each case—the electron will move as a projectile with an acceleration in the vertical
direction and constant velocity in the horizontal direction. Once the electron leaves the
region between the plates, the electric field disappears, and the electron continues moving
in a straight line according to Newton’s first law.
Electric Forces and Electric Fields
Answers to Even Numbered Problems
2.
5.71 × 1013 C
4.
F = 1.91( ke q2 a2 ) along the diagonal toward the negative charge
6.
2.25 × 10 −9 N m
8.
5.08 m
10.
F6 = 46.7 N ( left ) , F1.5 = 157 N ( right ) , F−2 = 111 N ( left )
12.
3.89 × 10 −7 N at 11.3° below + x axis
14.
x = 0.634 d , stable if third bead has positive charge
16.
1.45 m beyond the –3.00 nC charge
18.
(a)
2.00 × 107 N C to the right
20.
(a)
5.27 × 1013 m s 2
22.
1.63 × 10 4 N C directed opposite to the proton’s velocity
24.
1.88 × 10 3 N C at 4.40° below + x axis
26.
at y = + 0.85 m
28.
(a)
34.
(a) zero
36.
~ 1 µm
38.
(a)
40.
(a) –55.7 nC
(b) negative, with a spherically symmetric distribution
42.
5.65 × 10 5 N ⋅ m 2 C
48.
(a) 8.2 × 10 −8 N
50.
5.25 µ C
q1 q2 = − 1 3
2.0 × 10 6 N ⋅ m 2 C
(b)
(b) 40.0 N to the left
5.27 × 10 5 m s
(b)
q2 > 0, q1 < 0
(b)
1.8 × 10 6 N C
(b) 0
(b) 2.2 × 10 6 m s
(c)
1.1 × 10 5 N C
5
6
CHAPTER 15
3.43 µ C
52.
(a) downward
54.
2.51 × 10 −10
56.
(a) 0.307 s
(b) Yes, the absence of gravity produces a 2.28% difference.
60.
(d)
62.
2.0 µ C
64.
(a)
(b)
JG
E = 0 at x = +9.47 m, y = 0.
37.0° or 53.0°
(b)
1.66 × 10 −7 s and 2.21 × 10 −7 s
Electric Forces and Electric Fields
7
Problem Solutions
15.1
Since the charges have opposite signs, the force is one of attraction .
Its magnitude is
F=
15.2
k e q1 q2
r2
−9
−9

N ⋅ m 2  ( 4.5 × 10 C )( 2.8 × 10 C )
=  8.99 × 109
= 1.1 × 10 −8 N

2
2
C 
( 3.2 m )

The electrical force would need to have the same magnitude as the current gravitational
force, or
ke
q2
M m
= G E 2 moon
2
r
r
q=
giving
GME mmoon
ke
This yields
q=
15.3
F=
( 6.67 × 10
−11
N ⋅ m 2 kg 2 )( 5.98 × 10 24 kg )( 7.36 × 10 22 kg )
8.99 × 10 N ⋅ m C
9
2
2
k e ( 2 e )( 79 e )
r2
2
( 158 ) ( 1.60 × 10 −19 C )

9 N⋅m 
=  8.99 × 10
= 91 N

2
C2 

( 2.0 × 10−14 m )
2
( repulsion )
= 5.71 × 1013 C
8
15.4
CHAPTER 15
F1 = F2 =
and
and
F3 =
ke q
2
(a 2 )
2
=
ur
F1
2
ke q
2a2
k q 
ke q
kq
+ e 2 ( 0.707 ) = 1.35  e 2 
2
2a
a
 a 
ΣFy = F1 + F3 sin 45° =
 ke q2 
ke q2 ke q2
+
=
0.707
1.35
(
)
 2 
a2
2a2
 a 
( ΣFx )
2
( )
+ ΣFy
2
2
= 1.91
2
2
+q
a 2
ur
F3
45°
a
ur
F2
 ΣFy 
ke q2
−1
and θ = tan −1 
 = tan ( 1) = 45°
2
a
F
Σ
 x
G
 k q2 
F R = 1.91  e 2  along the diagonal toward the negative charge
 a 
so
(a)
ke q
a2
2
–q
a
r
ΣFx = F2 + F3 cos 45° =
FR =
15.5
–q
The attractive forces exerted on the positive charge by
the negative charges are shown in the sketch and have
magnitudes
F=
k
2e 2 
e( )
r2
N ⋅ m2
=  8.99 × 10 9
C2

−19 2 

  4 ( 1.60 × 10 ) 
= 36.8 N

2
 ( 5.00 × 10 −15 m )
(b) The mass of an alpha particle is m = 4.002 6 u , where u = 1.66 × 10 −27 kg is the
unified mass unit. The acceleration of either alpha particle is then
a=
F
36.8 N
=
= 5.54 × 10 27 m s 2
m 4.002 6 ( 1.66 × 10 −27 kg )
–q
Electric Forces and Electric Fields
15.6
The attractive force between the charged ends tends to compress the molecule. Its
magnitude is
−19
2
k ( 1e ) 
N ⋅ m 2  ( 1.60 × 10 C )
F = e 2 =  8.99 × 109
= 4.89 × 10 −17 N .

2
2
−
6
r
C  ( 2.17 × 10 m )

2
The compression of the “spring” is
x = ( 0.010 0 ) r = ( 0.010 0 ) ( 2.17 × 10 −6 m ) = 2.17 × 10 −8 m ,
so the spring constant is k =
15.7
F 4.89 × 10 −17 N
=
= 2.25 × 10 −9 N m
x 2.17 × 10 −8 m
1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton
and one electron. Thus, each charge has magnitude q = N A e . The distance separating
these charges is r = 2 RE , where RE is Earth’s radius. Thus,
F=
(
ke N A e
( 2 RE )
)
2
2

N ⋅ m2
=  8.99 × 10 9
C2

15.8
2
The magnitude of the repulsive force between electrons must equal the weight of an
electron, Thus, k e e 2 r 2 = me g
or
15.9
−19
23
 ( 6.02 × 10 )( 1.60 × 10 C ) 
= 5.12 × 10 5 N

2
6

4 ( 6.38 × 10 m )
k e2
r= e =
me g
( 8.99 × 10 N ⋅ m C )(1.60 × 10 C )
( 9.11 × 10 kg )( 9.80 m s )
9
2
−31
2
−19
2
2
= 5.08 m
(a) The spherically symmetric charge distributions behave as if all charge was located
at the centers of the spheres. Therefore, the magnitude of the attractive force is
F=
k e q1 q2
r2
−9
−9

N ⋅ m 2  ( 12 × 10 C )( 18 × 10 C )
=  8.99 × 109
= 2.2 × 10 −5 N

2
2
C
( 0.30 m )


9
10
CHAPTER 15
(b) When the spheres are connected by a conducting wire, the net charge
qnet = q1 + q2 = −6.0 × 10 −9 C will divide equally between the two identical spheres.
Thus, the force is now
F=
k e ( qnet 2 )
r2
2
2
−6.0 × 10 −9 C )

9 N⋅m  (
=  8.99 × 10

C 2  4 ( 0.30 m )2

2
or F = 9.0 × 10 −7 N (repulsion)
15.10
The forces are as shown in the sketch
at the right.
+6.00 mC
ur
ur
F2
F1
ur ur
+1.50 mC F1 F2 –2.00 mC
ur ur
F3 F3
3.00 cm
F1 =
F2 =
F3 =
2.00 cm
2
6.00 × 10 −6 C )( 1.50 × 10 −6 C )
k e q1 q2 
9 N⋅m  (
=
8.99
×
10
= 89.9 N


2
r122
C2 

( 3.00 × 10-2 m )
ke q1 q3
r132
ke q2 q3
r232
2
6.00 × 10 −6 C )( 2.00 × 10 −6 C )

9 N⋅m  (
=  8.99 × 10
= 43.2 N

2
C2 

( 5.00 × 10-2 m )
−6
−6

N ⋅ m 2  ( 1.50 × 10 C )( 2.00 × 10 C )
=  8.99 × 109
= 67.4 N

2
C2 

( 2.00 × 10-2 m )
The net force on the 6 µ C charge is F6 = F1 − F2 = 46.7 N (to the left)
The net force on the 1.5 µ C charge is F1.5 = F1 + F3 = 157 N (to the right)
The net force on the −2 µ C charge is F−2 = F2 + F3 = 111 N (to the left)
11
Electric Forces and Electric Fields
15.11
ur 5.00 nC
6.00 nC
F6
m
q
ur 0.100 0.300
m
ur
F
3
FR
–3.00 nC
In the sketch at the right, FR is the resultant of the
forces F6 and F3 that are exerted on the charge at
the origin by the 6.00 nC and the –3.00 nC charges
respectively.
2
6.00 × 10 −9 C )( 5.00 × 10 −9 C )

9 N⋅m  (
F6 =  8.99 × 10

2
C2 
( 0.300 m )

= 3.00 × 10 −6 N
−9
−9

N ⋅ m 2  ( 3.00 × 10 C )( 5.00 × 10 C )
= 1.35 × 10 −5 N
F3 =  8.99 × 109

2
2
C
0.100
m
(
)


The resultant is FR =
F
2
+ ( F3 ) = 1.38 × 10 −5 N at θ = tan −1  3
 F6

 = 77.5°

G
FR = 1.38 × 10 −5 N at 77.5° below − x axis
Consider the arrangement of charges shown in the sketch at
the right. The distance r is
r=
2
2
( 0.500 m ) + ( 0.500 m ) = 0.707 m
The forces exerted on the 6.00 nC charge are

N ⋅ m  ( 6.00 × 10 C )( 2.00 × 10
F2 =  8.99 × 109

2
C2 
( 0.707 m )

2
−9
−9
C)
3.00 nC
0.500 m
15.12
2
0.500 m
or
( F6 )
ur
F2
r
45.0°
0.500 m
r
45.0°
6.00 nC
2.00 nC
= 2.16 × 10 −7 N
and
2
6.00 × 10 −9 C )( 3.00 × 10 −9 C )

9 N⋅m  (
F3 =  8.99 × 10
= 3.24 × 10 −7 N

2
2
C
0.707
m
(
)


ur
F3
12
CHAPTER 15
Thus, ΣFx = ( F2 + F3 ) cos 45.0° = 3.81 × 10 −7 N
and
ΣFy = ( F2 − F3 ) sin 45.0° = −7.63 × 10 −8 N
The resultant force on the 6.00 nC charge is then
FR =
or
15.13
( ΣFx )
2
( )
+ ΣFy
2
 ΣFy
= 3.89 × 10 −7 N at θ = tan −1 
 ΣFx

 = − 11.3°

G
FR = 3.89 × 10 −7 N at 11.3° below +x axis
The forces on the 7.00 µC charge are shown at the right.
y
2
7.00 × 10 −6 C )( 2.00 × 10 −6 C )

9 N⋅m  (
F1 =  8.99 × 10

2
C2 
( 0.500 m )

= 0.503 N
−6
−6

N ⋅ m 2  ( 7.00 × 10 C )( 4.00 × 10 C )
F2 =  8.99 × 109

2
C2 
( 0.500 m )

+ 7.00 mC
ur
F2
60.0°
x
–
+
0.500
m
2.00 mC
–4.00 mC
= 1.01 N
Thus, ΣFx = ( F1 + F2 ) cos 60.0° = 0.755 N
and
ΣFy = ( F1 − F2 ) sin 60.0° = −0.436 N
The resultant force on the 7.00 µC charge is
FR =
or
( ΣFx )
2
( )
+ ΣFy
2
 ΣFy 
= 0.872 N at θ = tan −1 
 = −30.0°
 ΣFx 
G
FR = 0.872 N at 30.0° below the +x axis
ur
F1
Electric Forces and Electric Fields
15.14
13
Assume that the third bead has charge Q and is located at 0 < x < d . Then the forces
exerted on it by the +3q charge and by the +1q charge have magnitudes
F3 =
k e Q ( 3q )
x
2
and F1 =
keQ ( q )
(d − x)
2
respectively
These forces are in opposite directions, so charge Q is in equilibrium if F3 = F1 . This gives
3 ( d − x ) = x 2 , and solving for x, the equilibrium position is seen to be
2
x=
d
= 0.634 d
1+1 3
This is a position of stable equilibrium if Q > 0 . In that case, a small displacement from
the equilibrium position produces a net force directed so as to move Q back toward the
equilibrium position.
15.15
Consider the free-body diagram of one of the spheres given
at the right. Here, T is the tension in the string and Fe is the
repulsive electrical force exerted by the other sphere.
mg
ΣFy = 0 ⇒ T cos 5.0° = mg , or T =
cos 5.0°
ΣFx
+y
5.0°
ur
Fe
= 0 ⇒ Fe = T sin 5.0° = mg tan 5.0°
q = ( 2 L sin 5.0° )
ke q2
( 2 L sin 5.0° )
2
= mg tan 5.0° and yields
mg tan 5.0°
ke
=  2 ( 0.300 m ) sin 5.0°
( 0.20 × 10
−3
+x
ur
mg
At equilibrium, the distance separating the two spheres is r = 2 L sin 5.0° .
Thus, Fe = mg tan 5.0° becomes
ur
T
kg )( 9.80 m s 2 ) tan 5.0°
8.99 × 109 N ⋅ m 2 C 2
= 7.2 nC
14
15.16
CHAPTER 15
6.00 nC –3.00 nC
The required position is shown in the
sketch at the right. Note that this places q
closer to the smaller charge, which will
allow the two forces to cancel. Requiring
that
ur
F3
q
ur
F6
x
0.600 m
F6 = F3 gives
k e ( 6.00 nC ) q
( x + 0.600 m )
=
2
k e ( 3.00 nC ) q
2
, or 2 x 2 = ( x + 0.600 m )
x2
Solving for x gives the equilibrium position as
x=
15.17
0.600 m
= 1.45 m beyond the − 3.00 nC charge
2 −1
For the object to “float” it is necessary that the electrical force support the weight, or
qE = mg
15.18
−6
qE ( 24 × 10 C ) ( 610 N C )
m=
=
= 1.5 × 10 −3 kg
2
g
9.8 m s
or
+6.00 mC
q1
(a) Taking to the right as
positive, the resultant
electric field at point P is
given by
ur
E2
P
2.00 cm
ur
E3 +1.50 mC
ur
q2
E1
1.00 cm
3.00 cm
ER = E1 + E3 − E2
=
k e q1
2
1
r
+
k e q3
2
3
r
−
k e q2
r22

N ⋅ m2
=  8.99 × 109
C2

  6.00 × 10 −6 C 2.00 × 10 −6 C 1.50 × 10 −6 C 

+
−

2
2
2
  ( 0.020 0 m )
( 0.030 0 m ) ( 0.010 0 m ) 
This gives ER = + 2.00 × 107 N C
G
or E R = 2.00 × 107 N C to the right
G
G
(b) F = qE R = ( −2.00 × 10 −6 C )( 2.00 × 107 N C ) = −40.0 N
G
or F = 40.0 N to the left
–2.00 mC
q3
Electric Forces and Electric Fields
15.19
15
We shall treat the concentrations as point charges. Then, the resultant field consists of
two contributions, one due to each concentration.
The contribution due to the positive charge at 3 000 m altitude is
E+ = k e
2

9 N ⋅ m  ( 40.0 C )
=
×
= 3.60 × 10 5 N C
8.99
10


C 2  ( 1 000 m )2
r2 
q
( downward )
The contribution due to the negative charge at 1 000 m altitude is
E− = k e
2

9 N ⋅ m  ( 40.0 C )
=
×
= 3.60 × 10 5 N C
8.99
10


C 2  ( 1 000 m )2
r2 
q
( downward )
The resultant field is then
G G
G
E = E + + E − = 7.20 × 10 5 N C ( downward )
15.20
(a) The magnitude of the force on the electron is F = q E = eE , and the acceleration is
−19
F
eE ( 1.60 × 10 C ) ( 300 N C )
=
=
= 5.27 × 1013 m s 2
a=
−31
9.11 × 10 kg
me me
(b) v = v0 + at = 0 + ( 5.27 × 1013 m s 2 )( 1.00 × 10 −8 s ) = 5.27 × 10 5 m s
15.21
If the electric force counterbalances the weight of the ball, then
qE = mg
15.22
or
−3
2
mg ( 5.0 × 10 kg )( 9.8 m s )
E=
=
= 1.2 × 10 4 N C
−6
q
4.0 × 10 C
The force an electric field exerts on a positive change is in the direction of the field. Since
this force must serve as a retarding force and bring the proton to rest, the force and
hence the field must be in the direction opposite to the proton’s velocity .
The work-energy theorem, Wnet = KE f − KEi , gives the magnitude of the field as
− ( qE ) ∆x = 0 − KEi
or
E=
KEi
3.25 × 10 −15 J
=
= 1.63 × 10 4 N C
q ( ∆x ) ( 1.60 × 10 -19 C ) ( 1.25 m )
−19
F qE ( 1.60 × 10 C ) ( 640 N C )
a= =
=
= 6.12 × 1010 m s 2
m mp
1.673 × 10 -27 kg
(a)
(b) t =
1.20 × 10 6 m s
∆v
=
= 1.96 × 10 −5 s = 19.6 µ s
a 6.12 × 1010 m s 2
∆x =
(c)
v 2f − v02
(d) KE f =
2
6
10
2
11.8 m
2
1
1
mp v 2f = ( 1.673 × 10 −27 kg )( 1.20 × 106 m s ) = 1.20 × 10 −15 J
2
2
q1 = 3.00 nC
The altitude of the triangle is
h = ( 0.500 m ) sin 60.0° = 0.433 m
00
15.24
2a
(1.20 × 10 m s ) − 0 =
=
2 ( 6.12 × 10 m s )
m
15.23
CHAPTER 15
0.5
16
and the magnitudes of the fields due to
each of the charges are
q2 = 8.00 nC
E1 =
9
2
2
−9
k e q1 ( 8.99 × 10 N ⋅ m C )( 3.00 × 10 C )
=
2
h2
( 0.433 m )
60.0°
0.250 m
= 144 N C
9
2
2
−9
k e q2 ( 8.99 × 10 N ⋅ m C )( 8.00 × 10 C )
E2 = 2 =
= 1.15 × 10 3 N C
2
r2
0.250
m
(
)
and
E3 =
k e q3
r32
( 8.99 × 10
=
9
N ⋅ m 2 C 2 )( 5.00 × 10 −9 C )
( 0.250 m )
2
= 719 N C
h
ur
E2
ur
ur E q2 = –5.00 nC
E1 3
Electric Forces and Electric Fields
17
Thus, ΣEx = E2 + E3 = 1.87 × 10 3 N C and ΣEy = −E1 = −144 N C
giving
ER =
( ΣEx )
2
(
+ ΣEy
)
2
= 1.88 × 10 3 N C
and
θ = tan −1 ( ΣEy ΣEx ) = tan −1 ( −0.0769 ) = −4.40°
JG
Hence E R = 1.88 × 10 3 N C at 4.40° below the +x axis
15.25
From the symmetry of the charge distribution, students
should recognize that the resultant electric field at the
center is
q
ur
E2
ur
E1
30°
G
ER = 0
30°
ur
E3
If one does not recognize this intuitively, consider:
G
G G
G
E R = E1 + E 2 + E 3 , so
Ex = E1x − E2 x =
ke q
r
2
q
cos 30° −
ke q
r2
cos 30° = 0
and
Ey = E1 y + E2 y − E3 =
Thus, ER = Ex2 + Ey2 = 0
ke q
r
2
sin 30° +
ke q
r
2
sin 30° −
ke q
r2
=0
q
18
15.26
CHAPTER 15
If the resultant field is to be zero, the contributions
of the two charges must be equal in magnitude
and must have opposite directions. This is only
possible at a point on the line between the two
negative charges.
q1 = –9.0 mC
r1
6.0 m
Assume the point of interest is located on the
y-axis at − 4.0 m < y < 6.0 m . Then, for equal
magnitudes,
k e q1
r12
=
k e q2
r22
or
9.0 µ C
( 6.0 m − y )
Solving for y gives y + 4.0 m =
15.27
2
=
ur
E1
ur
E2
y
+x
4.0 m
8.0 µ C
( y + 4.0 m )
r2
2
q2 = –8.0 mC
8
( 6.0 m − y ) , or y = + 0.85 m
9
If the resultant field is zero, the
+y
contributions from the two charges must
r1 = d
1.0 m
be in opposite directions and also have
+x
ur
ur
equal magnitudes. Choose the line
q
q
E2
E1
1 = –2.5 mC
2 = 6.0 mC
connecting the charges as the x-axis, with
r2 = 1.0 m + d
the origin at the –2.5 µC charge. Then, the
two contributions will have opposite
directions only in the regions x < 0 and
x > 1.0 m . For the magnitudes to be equal, the point must be nearer the smaller charge.
Thus, the point of zero resultant field is on the x-axis at x < 0 .
Requiring equal magnitudes gives
Thus,
(1.0 m + d )
k e q1
2
1
r
=
k e q2
2
2
r
or
2.5 µ C
6.0 µ C
=
2
2
d
( 1.0 m + d )
2.5
=d
6.0
Solving for d yields
d = 1.8 m ,
15.28
or
1.8 m to the left of the − 2.5 µ C charge
The magnitude of q2 is three times the magnitude of q1 because 3 times as many lines
emerge from q2 as enter q1 .
(a) Then,
q1 q2 = − 1 3
q2 = 3 q1
Electric Forces and Electric Fields
(b)
19
q2 > 0 because lines emerge from it,
and q1 < 0 because lines terminate on it.
15.29
Note in the sketches at the right that
electric field lines originate on
positive charges and terminate on
negative charges. The density of
lines is twice as great for the −2 q
charge in (b) as it is for the 1q
charge in (a).
–2q
q>0
(a)
15.30
15.31
(b)
Rough sketches for these charge configurations are shown below.
+1 mC
–2 mC
(a)
(b)
(a) The sketch for (a) is shown at
the right. Note that four times
as many lines should leave q1
as emerge from q2 although, for
clarity, this is not shown in this
sketch.
(b) The field pattern looks the same
here as that shown for (a) with
the exception that the arrows
are reversed on the field lines.
–2 mC
+1 mC
(c)
q1 = 4q2
q2
20
15.32
CHAPTER 15
(a) In the sketch for (a) at
the right, note that there
are no lines inside the
sphere. On the outside
of the sphere, the field
lines are uniformly
spaced and radially
outward.
(b) In the sketch for (b) above,
note that the lines are
(a)
(b)
perpendicular to the surface
at the points where they emerge. They should also be symmetrical about the
symmetry axes of the cube. The field is zero inside the cube.
15.33
(a)
Zero net charge on each surface of the sphere.
(b) The negative charge lowered into the sphere repels − 5 µ C to the outside surface,
and leaves + 5 µ C on the inside surface of the sphere.
(c) The negative charge lowered inside the sphere neutralizes the inner surface, leaving
zero charge on the inside . This leaves − 5µ C on the outside surface of the
sphere.
(d) When the object is removed, the sphere is left with − 5.00 µ C on the outside
surface and zero charge on the inside .
15.34
(a) The dome is a closed conducting surface. Therefore, the electric field is zero
everywhere inside it.
At the surface and outside of this spherically symmetric charge distribution, the
field is as if all the charge were concentrated at the center of the sphere.
(b) At the surface,
E=
9
2
2
−4
k e q ( 8.99 × 10 N ⋅ m C )( 2.0 × 10 C )
=
= 1.8 × 10 6 N C
2
R2
( 1.0 m )
Electric Forces and Electric Fields
(c) Outside the spherical dome, E =
( 8.99 × 10
E=
15.35
9
ke q
. Thus, at r = 4.0 m ,
r2
N ⋅ m 2 C2 )( 2.0 × 10 −4 C )
( 4.0 m )
21
2
= 1.1 × 10 5 N C
For a uniformly charged sphere, the field is strongest at the surface.
Thus, Emax =
k e qmax
,
R2
6
R2 Emax ( 2.0 m ) ( 3.0 × 10 N C )
=
=
= 1.3 × 10 −3 C
ke
8.99 × 109 N ⋅ m 2 C 2
2
qmax
or
15.36
If the weight of the drop is balanced by the electric force, then mg = q E = eE or the
mass of the drop must be
m=
−19
4
eE ( 1.6 × 10 C )( 3 × 10 N C )
=
≈ 5 × 10 −16 kg
2
g
9.8 m s
 3m 
4

But, m = ρV = ρ  π r 3  and the radius of the drop is r = 

3

 4πρ 
 3 ( 5 × 10 −16 kg ) 

r=
3
 4π ( 858 kg m ) 
15.37
(a)
13
= 5.2 × 10 −7 m
or
r~ 1 µ m
F = qE = ( 1.60 × 10 −19 C )( 3.0 × 10 4 N C ) = 4.8 × 10 −15 N
(b) a =
15.38
13
F
4.8 × 10 −15 N
=
= 2.9 × 1012 m s 2
−27
mp 1.673 × 10 kg
The flux through an area is Φ E = EA cosθ , where θ is the angle between the direction of
the field E and the line perpendicular to the area A.
(a)
Φ E = EA cosθ = ( 6.2 × 10 5 N C )( 3.2 m 2 ) cos 0° = 2.0 × 106 N ⋅ m 2 C
(b) In this case, θ = 90° and Φ E = 0
22
15.39
CHAPTER 15
The area of the rectangular plane is A = ( 0.350 m )( 0.700 m ) = 0.245 m 2 .
(a) When the plane is parallel to the yz plane, θ = 0° , and the flux is
Φ E = EA cosθ = ( 3.50 × 10 3 N C )( 0.245 m 2 ) cos 0° = 858 N ⋅ m 2 C
(b) When the plane is parallel to the x-axis, θ = 90° and Φ E = 0
(c)
15.40
Φ E = EA cosθ = ( 3.50 × 10 3 N C )( 0.245 m 2 ) cos 40.0° = 657 N ⋅ m 2 C
In this problem, we consider part (b) first.
(b) Since the field is radial everywhere, the charge distribution generating it must be
spherically symmetric . Also, since the field is radially inward, the net charge
inside the sphere is negative charge .
keQ
. Thus, just
r2
outside the surface where r = R , the magnitude of the field is E = k e Q R2 , so
(a) Outside a spherically symmetric charge distribution, the field is E =
R2 E ( 0.750 m ) ( 890 N C )
Q=
=
= 5.57 × 10 −8 C = 55.7 nC
9
2
2
ke
8.99 × 10 N ⋅ m C
2
Since we have determined that Q < 0 , we now have Q = − 55.7 nC
15.41
Φ E = EA cosθ and Φ E = Φ E , max when θ = 0°
Thus, E =
15.42
Φ E , max
A
=
Φ E , max
πd 4
2
=
4 ( 5.2 × 10 5 N ⋅ m 2 C )
π ( 0.40 m )
2
= 4.1 × 106 N C
k q
Φ E = EA cosθ =  e 2 ( 4π R2 ) cos 0° = 4π k e q
R 
Φ E = 4π ( 8.99 × 10 9 N ⋅ m 2 C 2 )( 5.00 × 10 −6 C ) = 5.65 × 10 5 N ⋅ m 2 C
Electric Forces and Electric Fields
15.43
23
We choose a spherical gaussian surface, concentric with the charged spherical shell and
of radius r. Then, ΣEA cosθ = E ( 4π r 2 ) cos 0° = 4π r 2 E .
(a) For r > a (that is, outside the shell), the total charge enclosed by the gaussian
surface is Q = + q − q = 0 . Thus, Gauss’s law gives 4π r 2 E = 0, or E = 0 .
(b) Inside the shell, r < a , and the enclosed charge is Q = + q .
Therefore, from Gauss’s law, 4π r 2 E =
q
q
kq
, or E =
= e2
2
r
4π ∈0 r
∈0
G kq
The field for r < a is E = e2 directed radially outward .
r
15.44
Construct a gaussian surface just barely inside the surface of the conductor, where E = 0 .
Q
= 0 inside. Thus, any excess charge residing on
Since E = 0 inside, Gauss’ law says
∈0
the conductor must be outside our gaussian surface (that is, on the surface of the
conductor).
15.45
E = 0 at all points inside the conductor, and cosθ = cos 90° = 0 on the cylindrical surface.
Thus, the only flux through the gaussian surface is on the outside end cap and Gauss’s
Q
.
law reduces to ΣEA cosθ = EAcap =
∈o
The charge enclosed by the gaussian surface is Q = σ A , where A is the cross-sectional
area of the cylinder and also the area of the end cap, so Gauss’s law becomes
EA =
15.46
σA
∈o
, or E =
σ
∈o
Choose a very small cylindrical gaussian surface with one end inside the conductor.
Position the other end parallel to and just outside the surface of the conductor.
Since, in static conditions, E = 0 at all points inside a conductor, there is no flux through
the inside end cap of the gaussian surface. At all points outside, but very close to, a
conductor the electric field is perpendicular to the conducting surface. Thus, it is parallel
to the cylindrical side of the gaussian surface and no flux passes through this cylindrical
side. The total flux through the gaussian surface is then Φ = EA , where A is the crosssectional area of the cylinder as well as the area of the end cap.
24
CHAPTER 15
The total charge enclosed by the cylindrical gaussian surface is Q = σ A , where σ is the
charge density on the conducting surface. Hence, Gauss’s law gives
EA =
15.47
F=
15.48
(a)
k e q1 q2
r2
F=
σA
or E =
∈0
σ
∈o
9
2
−19
k e 2 ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C )
= e2 =
= 57.5 N
2
r
( 2.00 × 10-15 m )
2
k e q1 q2
r2
=
( 8.99 × 10
=
9
ke e 2
r2
N ⋅ m 2 C 2 )( 1.60 × 10 −19 C )
( 0.53 × 10
−10
m)
2
= 8.2 × 10 −8 N
2
(b) F = me ac = me ( v 2 r ) , so
r⋅F
=
v=
me
15.49
( 0.53 × 10
m )( 8.2 × 10 −8 N )
−10
9.11 × 10
-31
kg
The three contributions to the resultant
electric field at the point of interest are
shown in the sketch at the right.
= 2.2 × 10 6 m s
y
– 4.0 nC 5.0 nC
The magnitude of the resultant field is
ER = −E1 + E2 + E3
ER = −
k e q1
r12
+
k e q2
r22
+
k e q3

N ⋅ m2
ER =  8.99 × 109
C2

r32
ur
E1
3.0 nC
r3 = 1.2 m
r2 = 2.0 m
r1 = 2.5 m
 q1
q2
q3 
= ke  − 2 + 2 + 2 
r2
r3 
 r1
  4.0 × 10 −9 C 5.0 × 10 −9 C 3.0 × 10 −9 C 

+
+
 −
2
2
2
( 1.2 m ) 
  ( 2.5 m )
( 2.0 m )
G
ER = + 24 N C , or E R = 24 N C in the +x direction
ur
E2
ur
E3
Electric Forces and Electric Fields
15.50
ur
y
T q
Consider the free-body diagram shown at the right.
ΣFy = 0 ⇒ T cosθ = mg or T =
mg
cosθ
ur
Fe
ΣFx = 0 ⇒ Fe = T sin θ = mg tan θ
qE = mg tan θ , or q =
15.51
( 2.00 × 10
−3
mg tan θ
E
kg )( 9.80 m s 2 ) tan 15.0°
1.00 × 10 N C
3
= 5.25 × 10 −6 C = 5.25 µ C
(a) At a point on the x-axis, the contributions by the two
charges to the resultant field have equal magnitudes
kq
given by E 1 = E2 = e2 .
r
The components of the resultant field are
k q
k q
Ey = E1 y − E2 y =  e2  sin θ −  e2  sin θ = 0
 r 
 r 
Since
y
q
r
a
a
q
b
q
 k e ( 2q ) 
k q
k q
Ex = E1x + E2 x =  e2  cosθ +  e2  cosθ = 
 cosθ
2
 r 
 r 
 r

and
cosθ b r b
b
= 2 = 3 =
, the resultant field is
32
2
r
r
r
( a2 + b 2 )
G
ER =
k e ( 2q ) b
(a
2
+ b2 )
32
x
ur
mg
Since Fe = qE , we have
q=
25
in the +x direction
r
ur
E1
q
q
ur
E2
x
26
CHAPTER 15
(b) Note that the result of part (a) may be written as ER =
ke (Q ) b
(a
2
+ b2 )
32
where Q = 2q is
the total charge in the charge distribution generating the field.
In the case of a uniformly charged circular ring, consider the ring to consist of a
very large number of pairs of charges uniformly spaced around the ring. Each pair
consists of two identical charges located diametrically opposite each other on the
ring. The total charge of pair number i is Qi . At a point on the axis of the ring, this
pair of charges generates an electric field contribution that is parallel to the axis and
ke bQi
.
has magnitude Ei =
32
( a2 + b2 )
The resultant electric field of the ring is the summation of the contributions by all
pairs of charges, or


ke b
 ΣQi = ke bQ 3 2
ER = ΣEi = 
32
2
2
(a + b ) 
( a2 + b 2 )


where Q = ΣQi is the total charge on the ring.
G
ER =
15.52
(a)
ay =
(a
keQ b
2
vy2 − v02 y
2 ( ∆y )
+ b2 )
32
( 21.0
=
in the +x direction
m s) − 0
2
2 ( 5.00 m )
= 44.1 m s 2
( downward )
Since ay > g , the electrical force must be directed downward, aiding the
gravitational force in accelerating the bead. Because the bead is positively charged,
the electrical force acting on it is in the direction of the electric field. Thus, the field
is directed downward .
(b) Taking downward as positive, ΣFy = qE + mg = may .
Therefore,
q=
(
m ay − g
)
E
(1.00 × 10
=
kg ) ( 44.1 − 9.80 ) m s 2 
= 3.43 × 10 −6 C = 3.43 µ C
1.00 × 10 4 N C
−3
Electric Forces and Electric Fields
15.53
+8.00 mC
Because of the spherical symmetry of the
charge distribution, any electric field
present will be radial in direction. If a field
does exist at distance R from the center, it is
the same as if the net charge located within
r ≤ R were concentrated as a point charge
at the center of the inner sphere. Charge
located at r > R does not contribute to the
field at r = R .
2.00 cm
5.00 cm
4.00 cm
(a) At r = 1.00 cm , E = 0 since static
–4.00 mC
electric fields cannot exist within
conducting materials.
(b) The net charge located at r ≤ 3.00 cm is Q = +8.00 µ C .
Thus, at r = 3.00 cm ,
E=
keQ
r2
( 8.99 × 10
=
9
N ⋅ m 2 C 2 ) ( 8.00 × 10 −6 C )
( 3.00 × 10
−2
m)
2
= 7.99 × 107 N C ( outward )
(c) At r = 4.50 cm , E = 0 since this is located within conducting materials.
(d) The net charge located at r ≤ 7.00 cm is Q = + 4.00 µ C .
Thus, at r = 7.00 cm ,
E=
=
keQ
r2
( 8.99 × 10
9
N ⋅ m 2 C2 ) ( 4.00 × 10 −6 C )
( 7.00 × 10
−2
m)
2
27
= 7.34 × 10 6 N C ( outward )
28
15.54
CHAPTER 15
The charges on the spheres will be equal in magnitude and opposite in sign. From
F = k e q2 r 2 , this charge must be
q=
F ⋅ r2
=
ke
(1.00 × 10
4
N ) ( 1.00 m )
8.99 × 109 N ⋅ m 2 C 2
2
= 1.05 × 10 −3 C
The number of electrons transferred is
n=
q 1.05 × 10 −3 C
=
= 6.59 × 1015
e 1.60 × 10 −19 C
The total number of electrons in 100-g of silver is
electrons 

23 atoms   1 mole 
25
N =  47
 ( 100 g ) = 2.62 × 10
 6.02 × 10

atom
mole
107.87
g




Thus, the fraction transferred is
n 6.59 × 1015
=
= 2.51 × 10 −10 (that is, 2.51 out of every 10 billion).
25
N 2.62 × 10
15.55
Φ E = EA cosθ
= ( 2.00 × 10 4 N C ) ( 6.00 m )( 3.00 m )  cos10.0° = 3.55 × 10 5 N ⋅ m 2 C
15.56
(a) The downward electrical force acting on the ball is
Fe = qE = ( 2.00 × 10 −6 C )( 1.00 × 10 5 N C ) = 0.200 N
The total downward force acting on the ball is then
F = Fe + mg = 0.200 N+ ( 1.00 × 10 -3 kg )( 9.80 m s 2 ) = 0.210 N
Thus, the ball will behave as if it was in a modified gravitational field where the
effective free-fall acceleration is
“ g” =
F
0.210 N
=
= 210 m s 2
-3
m 1.00 × 10 kg
Electric Forces and Electric Fields
29
The period of the pendulum will be
T = 2π
(b)
L
0.500 m
= 2π
= 0.307 s
" g"
210 m s 2
Yes . The force of gravity is a significant portion of the total downward force
acting on the ball. Without gravity, the effective acceleration would be
“ g” =
Fe
0.200 N
=
= 200 m s 2
m 1.00 × 10 -3 kg
0.500 m
= 0.314 s
200 m s 2
giving T = 2π
a 2.28% difference from the correct value with gravity included.
15.57
The sketch at the right gives a free-body diagram of the
positively charged sphere. Here, F1 = k e q
2
10°
ur
T
2
r is the attractive
force exerted by the negatively charged sphere and F2 = qE
is exerted by the electric field.
ΣFy = 0 ⇒ T cos10° = mg or T =
ur
F1
mg
cos10°
ΣFx = 0 ⇒ F2 = F1 +T sin10° or qE =
ke q
r2
y
ur
F2
ur
mg
2
+ mg tan10°
At equilibrium, the distance between the two spheres is r = 2 ( L sin10° ) . Thus,
E=
=
ke q
4 ( L sin10° )
( 8.99 × 10
9
2
+
mg tan10°
q
N ⋅ m 2 C2 )( 5.0 × 10 −8 C )
4 ( 0.100 m ) sin10° 
or the needed electric field strength is
2
+
( 2.0 × 10
−3
kg )( 9.80 m s 2 ) tan10°
( 5.0 × 10
E = 4.4 × 10 5 N C
−8
C)
x
30
15.58
CHAPTER 15
y
As shown in the sketch, the electric field
at any point on the x-axis consists of two
parts, one due to each of the charges in the
dipole.
r–
+q
–q
E = E+ − E− =
E=
ke q
( x − a)
2
−
ke q
r+2
−
a
ur
E–
a
r+
ke q
( x + a)
x
r−2
ke q
2
ur
E+


 ( x + a )2 − ( x − a )2 
4 ax 

= ke q 
=
k
q

e
2
2
 ( x 2 − a 2 )2 
 ( x − a ) ( x + a ) 


4 k e qa
 4 ax 
Thus, if x 2 >> a 2 , this gives E ≈ k e q  4  =
x3
 x 
15.59
y
(a) Consider the free-body diagram for the ball given in the sketch.
ΣFx = 0 ⇒ T sin 37.0° = qEx or T =
qEx
sin 37.0°
and
ΣFy = 0 ⇒ qEy + T cos 37.0° = mg or qEy + qEx cot 37.0° = mg
ur 37.0°
T
ur
qEy
ur
mg
ur
qEx
x
1.00 × 10 −3 kg )( 9.80 m s 2 )
(
mg
Thus, q =
=
Ey + Ex cot 37.0°  5.00 + ( 3.00 ) cot 37.0°  × 10 5 N C
= 1.09 × 10 −8 C = 10.9 nC
(b) From ΣFx = 0 , we found that T =
Hence,
15.60
(1.09 × 10
T=
−8
qEx
.
sin 37.0°
C )( 3.00 × 10 5 N C )
sin 37.0°
= 5.44 × 10 −3 N
(a) At any point on the x-axis in the range 0 < x < 1.00 m , the contributions made to the
resultant electric field by the two charges are both in the positive x direction. Thus,
it is not possible for these contributions to cancel each other and yield a zero field.
Electric Forces and Electric Fields
31
(b) Any point on the x-axis in the range x < 0 is located closer to the larger magnitude
charge ( q = 5.00 µ C ) than the smaller magnitude charge ( q = 4.00 µ C ) . Thus, the
contribution to the resultant electric field by the larger charge will always have a
greater magnitude than the contribution made by the smaller charge. It is not
possible for these contributions to cancel to give a zero resultant field.
(c) If a point is on the x-axis in the region x > 1.00 m , the contributions made by the
two charges are in opposite directions. Also, a point in this region is closer to the
smaller magnitude charge than it is to the larger charge. Thus, there is a location in
this region where the contributions of these charges to the total field will have equal
magnitudes and cancel each other.
(d) When the contributions by the two charges cancel each other, their magnitudes
must be equal. That is,
ke
( 5.00 µ C )
x
2
= ke
( 4.00 µ C )
4
or x − 1.00 m = +
x
2
5
( x − 1.00 m )
Thus, the resultant field is zero at
15.61
x=
1.00 m
= + 9.47 m
1− 4 5
We assume that the two spheres have equal charges, so the
repulsive force that one exerts on the other has magnitude
Fe = k e q2 r 2 .
ur
T 10°
ur
Fe
From Figure P15.61 in the textbook, observe that the distance
separating the two spheres is
r = 3.0 cm + 2 ( 5.0 cm ) sin10°  = 4.7 cm = 0.047 m
From the free-body diagram of one sphere given above, observe
that
ΣFy = 0 ⇒ T cos10° = mg or T = mg cos10°
and
 mg 
ΣFx = 0 ⇒ Fe = T sin10° = 
 sin10° = mg tan10°
 cos10° 
ur
mg
32
CHAPTER 15
Thus, k e q2 r 2 = mg tan10°
( 0.015 kg ) ( 9.8
mgr 2 tan10°
=
q=
ke
or
m s 2 ) ( 0.047 m ) tan10°
2
8.99 × 109 N ⋅ m 2 C 2
giving q = 8.0 × 10 −8 C or q ~ 10 −7 C
15.62
Consider the free-body diagram of the rightmost charge given below.
ΣFy = 0 ⇒ T cosθ = mg
ur
T
T = mg cosθ
or
and
ΣFx = 0 ⇒ Fe =T sin θ = ( mg cosθ ) sin θ = mg tan θ
But,
Fe =
Thus,
5k e q 2
= mg tan θ or q =
4 L2 sin 2 θ
q
ur
Fe
ur
mg
ke q2 ke q2
ke q2
ke q2
5k e q 2
+
=
+
=
2
2
r12
r22
( L sin θ ) ( 2L sin θ ) 4L2 sin 2 θ
4 L2 mg sin 2 θ tan θ
5k e
If θ = 45°, m = 0.10 kg, and L = 0.300 m then
4 ( 0.300 m ) ( 0.10 kg ) ( 9.80 m s 2 ) sin 2 ( 45° ) tan ( 45° )
2
q=
or
15.63
5 ( 8.99 × 10 9 N ⋅ m 2 C 2 )
q = 2.0 × 10 −6 C = 2.0 µ C
(a) When an electron (negative charge) moves distance ∆x in the direction of an
electric field, the work done on it is
W = Fe ( ∆x ) cosθ = eE ( ∆x ) cos180° = − eE ( ∆x )
(
)
From the work-energy theorem Wnet = KE f − KEi with KE f = 0 , we have
− eE ( ∆x ) = − KEi , or E =
KEi
1.60 × 10 −17 J
=
= 1.00 × 10 3 N C
e ( ∆x ) ( 1.60 × 10 −19 C ) ( 0.100 m )
33
Electric Forces and Electric Fields
(b) The magnitude of the retarding force acting on the electron is Fe = eE , and Newton’s
second law gives the acceleration as a = − Fe m = − eE m . Thus, the time required to
bring the electron to rest is
t=
2m ( KEi )
v − v0 0 − 2 ( KEi ) m
=
=
− eE m
a
eE
or
t=
2 ( 9.11 × 10 −31 kg )( 1.60 × 10 −17 J )
(1.60 × 10
−19
C )( 1.00 × 10 N C )
3
= 3.37 × 10 −8 s = 33.7 ns
(c) After bringing the electron to rest, the electric force continues to act on it causing the
electron to accelerate in the direction opposite to the field at a rate of
a=
15.64
−19
3
eE ( 1.60 × 10 C )( 1.00 × 10 N C )
=
= 1.76 × 1014 m s 2
9.11 × 10 -31 kg
m
(a) The acceleration of the protons is downward (in the direction of the field) and
−19
Fe eE ( 1.60 × 10 C ) ( 720 N C )
=
= 6.90 × 1010 m s 2
ay = =
1.67 × 10 −27 kg
m m
The time of flight for the proton is twice the time required to reach the peak of the
arc, or
v
0y
t = 2tpeak = 2 
 ay

 2v sin θ
= 0

ay

ur
v0
The horizontal distance traveled in this time is
 2v sin θ
R = v0 x t = ( v0 cosθ )  0

ay

 v 2 sin 2θ
= 0

ay

ur
a
q
R
34
CHAPTER 15
Thus, if R = 1.27 × 10 −3 m , we must have
sin 2θ =
ay R
v02
=
( 6.90 × 10
10
m s 2 )( 1.27 × 10 −3 m )
( 9 550
m s)
2
= 0.961
giving 2θ = 73.9° or 2θ = 180° − 73.9° = 106.1° . Hence, θ = 37.0° or 53.0°
(b) The time of flight for each possible angle of projection is:
For θ = 37.0° :
t=
For θ = 53.0° :
t=
2v0 sin θ
ay
2v0 sin θ
ay
=
=
2 ( 9 550 m s ) sin 37.0°
6.90 × 1010 m s 2
2 ( 9 550 m s ) sin 53.0°
6.90 × 10
10
m s
2
= 1.66 × 10 −7 s
= 2.21 × 10 −7 s