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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 cos θ 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 = (^) ( 5.00N 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 cos θ = (^) ( 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.
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.
(^)
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 CHAPTER 15
22. The magnitude of the electric force on the electron of charge e due to a uniform electric
field
G
is. 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.
E F = eE
6 CHAPTER 15
52. (a) downward (b) 3.43 μC
54. 2.51 × 10 −^10
56. (a) 0.307 s (b) Yes, the absence of gravity produces a 2.28% difference. 60. (d) E = 0 at x = +9.47 m, y =0.
JG
62. 2.0 μC
64. (a) 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
( )( )
2 9 9 (^1 2 9 ) 2 2 2
N m 4.5^10 C^ 2.8^10 C 8.99 10 1.1 10 N C (^) 3.2 m
F k^ eq q r
− − = = ^ × ⋅ ×^ × = × −
15.2 The electrical force would need to have the same magnitude as the current gravitational force, or
2 2 2 giving
E moon E moon e e
q M m GM m k G q r r k
This yields
( 11 2 2 )( 24 )( 22 ) 13 9 2 2
6.67 10 N m kg 5.98 10 kg 7.36 10 kg 5.71 10 C 8.99 10 N m C
q
× − ⋅ × ×
× ⋅
×
( )( ) 2
k (^) e 2 e 79 e F r
( )( )
( )
( )
2 19 2 9 2 14 2
N m^158 1.60^10 C 8.99 10 91 repulsion C (^) 2.0 10 m
−
−
⋅ ×
= × =
×
N
Electric Forces and Electric Fields 9
15.6 The attractive force between the charged ends tends to compress the molecule. Its magnitude is
( ) (^ ) ( )
2 2 19 2 9 1 2 2 6 2
(^1) N m 1.60^10 C 8.99 10 4.89 10 N C (^) 2.17 10 m
F k^ e e r
− − −
⋅ ×
= = × = ×
×
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
17 9 8
4.89 10 N
2.25 10 N m 2.17 10 m
F
x
− − −
×
= = = ×
×
k
15.7 1.00 g of hydrogen contains Avogadro’s number of atoms, each containing one proton
and one electron. Thus, each charge has magnitude q = NA e. The distance separating these charges is r = 2 RE , where is Earth’s radius. Thus,
( )
( )
R E
( )( )
( )
2
2
e A
E
e
= ×
23 19 2 2 5 2 6 2
N m^10 1.60^10 C 8.99 5.12 10 N C (^) 4 6.38 10 m
k N F R
−
⋅ ^ ×^ ×
= ×
×
15.8 The magnitude of the repulsive force between electrons must equal the weight of an
electron, Thus, k e e^2 r^2 = m ge
or
( )( ) ( )( )
2 9 2 2 19 2 31 2
8.99 10 N m C 1.60 10 C 5.08 m 9.11 10 kg 9.80 m s
e e
k e r m g
− −
× ⋅ ×
×
15.9 (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
( )( )
2 9 9 (^1 2 9 ) 2 2 2
N m^12 10 C^18 10 C 8.99 10 2.2 10 N C (^) 0.30 m
F k q^ e^ q r
− − = = ^ × ⋅ ×^ × = × −
10 CHAPTER 15
(b) When the spheres are connected by a conducting wire, the net charge will divide equally between the two identical spheres. Thus, the force is now
9 qnet q 1 (^) q 2 (^) 6.0 10 C = + = − × −
( ) ( )
2 2 9 2 9 2 2 2
(^2) N m 6.0 10 C 8.99 10 C (^) 4 0.30 m
F k^ e^ qnet r
−^ ×^ −
= = ×
or F = 9.0 × 10 −^7 N (repulsion)
15.10 The forces are as shown in the sketch at the right.
( )( )
( )
2 6 6 1 2 9 (^1 2 2) -2 2 12
N m 6.00^10 C^ 1.50^10 C 8.99 10 89.9 N C (^) 3.00 10 m
F k q q^ e r
× −^ × −
= = × =
×
( )( )
( )
2 6 6 (^1 3 ) (^2 2 2) -2 2 13
N m 6.00^10 C^ 2.00^10 C 8.99 10 43. C (^) 5.00 10 m
F k q^ e^ q r
× −^ × −
= = × =
×
N
( )( )
( )
2 6 6 (^2 3 ) (^3 2 2) -2 2 23
N m 1.50^10 C^ 2.00^10 C 8.99 10 67. C (^) 2.00 10 m
F k q^ e^ q r
× −^ × −
= = × =
×
6 C
N
The net force on the μ charge is 6 = 1 − F 2 = 46.7 N (to the left)
5 C
F F
The net force on the 1. μ charge is 1.5 = 1 + F 3 = 157 N (to the right)
2 C
F F
The net force on the− μ charge is F − 2 = F 2 + F 3 = 111 N (to the left)
(^)
12 CHAPTER 15
Thus, Σ =
and
The resultant force on the 6.00 nC charge is then
Fx (^) ( F 2 (^) + F 3 (^) ) cos 45.0 ° = 3.81 × 10 −^7 N
Σ F (^) y = (^) ( F 2 (^) − F 3 (^) ) sin 45.0 ° = −7.63 × 10 −^8 N
( ) (^) ( )
(^2 ) FR = Σ F (^) x + Σ F (^) y = 3.89 × 10 − Nat tan 1^ y 11. x
F
F
θ −^
or F R = 3.89 × 10 −^7 N at 11.3 °below + x ax
G
is
15.13 The forces on the 7.00 μC charge are shown at the right.
( )( )
( )(
2 6 6 9 1 2 2
2 6 6 9 (^2 )
N m 7.00^10 C^ 2.00^10 C 8.99 10 C (^) 0.500 m
0.503 N
N m 7.00^10 C^ 4.00^10 C 8.99 10 C (^) 0.500 m
1.01 N
F
F
− −
− −
⋅ ×^ ×
= ×
⋅ ×^ ×
= ×
Thus, Σ F x = ( F 1 + F 2 )cos 60.0 ° =0.755 N
Σ F y = ( F 1 − F 2 )sin 60.0 ° = −0.436 N
( ) (^) ( )
and
The resultant force on the 7.00 μC charge is
2 2 FR = Σ F (^) x + Σ F (^) y = 0.872 N at tan 1^ y 3 x
F
F
θ −^
or F R = 0.872 N at 30.0 °below the + x axis
G
)
^
Electric Forces and Electric Fields 13
15.14 Assume that the third bead has charge Q and is located at 0 < x < d. Then the forces exerted on it by the +3 q charge and by the +1 q charge have magnitudes
( ) (^3 )
k Qe 3 q F x
= and
( )
(^1 ) F k Q q^ e d x
respectively
These forces are in opposite directions, so charge Q is in equilibrium if F 3 = F 1. This gives
3 d − x^2 = x^2 , and solving for x , the equilibrium position is seen to be
d x = = d
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.
Σ F (^) y = 0 ⇒ T cos 5.0° = mg , or cos 5.
mg T = °
mg tan 5.0°
At equilibrium, the distance separating the two spheres is
Σ F (^) x = 0 ⇒ Fe = T sin 5.0° =
r = 2 L sin 5.0°.
Thus, Fe = mg tan 5.0°becomes ( )
2 2 2 sin 5.
k q e L
= mg tan 5.0° °
and yields
( )
( 3 )( 2 ) 9 2
0.20 10 kg 9. m s nC 8.99 10 m
−
tan 5.
C
× °
tan 5. 2 sin 5.
2 0.300 m sin 5.
e
mg q L k
× N⋅
Electric Forces and Electric Fields 15
15.19 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
( )
2 9 5 2 2 2
N m 40.0 C 8.99 10 3.60 10 N C downward e C (^) 1 000 m
q E k
- r
= = × = ×
The contribution due to the negative charge at 1 000 m altitude is
( )
2 9 5 2 2 2
N m 40.0 C 8.99 10 3.60 10 N C downward e C (^) 1 000 m
q E k − r
= = × = ×
The resultant field is then
7.20 105 N C downward E = E + (^) + E −= ×
G G G
15.20 (a) The magnitude of the force on the electron is F = q E = eE , and the acceleration is
( 19 )( ) 13 2 31
1.60 10 C 300 N C
5.27 10 m s e e 9.11^10 kg
F eE a m m
− −
×
= = = = ×
×
(b) v = v 0 + at = 0 + (^) ( 5.27 × 10 13 m s^2 )( 1.00 × 10 −^8 s (^) )= 5.27 × 105 m s
15.21 If the electric force counterbalances the weight of the ball, then
( 3 )( 2 ) 4 6
5.0 10 kg 9.8 m s or 1.2 10 N C 4.0 10 C
mg qE mg E q
− −
×
= = = = ×
×
15.22 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
net
The work-energy theorem, W = KE (^) f − KEi
, gives the magnitude of the field as
( ) ( )(^ )
15 4
3.25 10 J
0 1.63 10 N C
i 1.60 10 C 1.25 m qE x KE
×^ −
− ∆ = − = = ×
×
or i
KE
E
q x
16 CHAPTER 15
15.23 (a)
( 19 )( ) 10 2
1.60 10 C 640 N C
6.12 10 m s p 1.673^10 kg
F qE a m m
×^ −
= = = = ×
×
(b)
6 5 10 2
1.20 10 m s 1.96 10 s 19.6 s 6.12 10 m s
v a
t − μ
∆ ×
= × =
×
(c)
( ) ( )
2 2 6 2 0 10 2
1.20 10 m s 0 11.8 m 2 2 6.12 10 m s
x v^ f v a
− × −
×
(d) (^) ( )( )
1.673 10 kg 1.20 10 m s 1.20 10 J f (^) 2 p f 2 KE = m v = × −^ × = × −^15
15.24 The altitude of the triangle is
h = ( 0.500 m sin 60.0) ° =0.433 m
and the magnitudes of the fields due to each of the charges are
( )(
9 2 2 1 2 2
8.99 10 N m C 3. 0.433 m
N C
k q ×^ ⋅ = =
( )( )
9 2 2 9 2 3 2 2 2 2
8.99 10 N m C 8.00 10 C 1.15 10 N C 0.250 m
E k q^ e r
× ⋅ ×^ −
= = = ×
and
( )( )
9 2 2 9 3 (^3 ) 3
8.99 10 N m C 5.00 10 C 719 N C 0.250 m
E k^ eq r
× ⋅ ×^ −
(^9) )
(^) ^
(^) ^
^
^ ^
1
10 C
E e h
×^ −
18 CHAPTER 15
15.26 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.
Assume the point of interest is located on the y -axis at. Then, for equal magnitudes,
− 4.0 m < y <6.0 m
1 2 2 2 1 2
k e q ke q r r
= or ( ) ( )
2 2
C 8.0 C
m y y 4.0 m
Solving for y gives (^) ( )
6.0 m 9
y + 4.0 m = − y , or y = +0.85 m
^
15.27 If the resultant field is zero, the contributions from the two charges must be in opposite directions and also have equal magnitudes. Choose the line connecting the charges as the x -axis, with
the origin at the –2.5 μC charge. Then, the
two contributions will have opposite directions only in the regions x < 0 and
. 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 > 1.0 m x < 0.
Requiring equal magnitudes gives 2 1 22 1 2
k (^) e q k (^) eq r r
= or
2
2.5 6.0 C
d^2 1.0 m d
μC μ
Thus, ( )
m
- d = d
1.8 m
d =
Solving for d yields
, or 1.8 m to the left of the − 2.5 μCcharge
15.28 The magnitude of is three times the magnitude of because 3 times as many lines
emerge from as enter q.
q 2 q 1 q 2 (^) 1 q 2 (^) = 3 q 1
(a) Then, q 1 (^) q 2 = −1 3
Electric Forces and Electric Fields 19
(b) q 2 (^) > 0 because lines emerge from it,
and q 1 (^) < 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 charge in (a).
1 q
15.30 Rough sketches for these charge configurations are shown below.
15.31 (a) The sketch for (a) is shown at the right. Note that four times as many lines should leave as emerge from although, for clarity, this is not shown in this sketch.
q 1 q 2
(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.