Download Chapter 16 Solutions

ELECTRIC CHARGE AND ELECTRIC FIELD
16
Responses to Questions
1.
A plastic ruler is suspended by a thread and then rubbed with a cloth. As shown in Fig. 16–2, the ruler
is negatively charged. Now bring the charged comb close to the ruler. If the ruler is repelled by the
comb, then the comb is negatively charged. If the ruler is attracted by the comb, then the comb is
positively charged.
2.
The clothing becomes charged as a result of being tossed about in the dryer and rubbing against the
dryer sides and other clothes. When you put on the charged object (shirt), it causes charge separation
(polarization) within the molecules of your skin (see Fig. 16–9), which results in attraction between the
shirt and your skin.
3.
Fog or rain droplets tend to form around ions because water is a polar molecule (Fig. 16–4), with a
positive region and a negative region. The charge centers on the water molecule will be attracted to the
ions or electrons in the air.
4.
A plastic ruler that has been rubbed with a cloth is charged. When brought near small pieces of paper,
it will cause separation of charge (polarization) in the bits of paper, which will cause the paper to be
attracted to the ruler. A small amount of charge is able to create enough electric force to be stronger
than gravity. Thus the paper can be lifted.
On a humid day this is more difficult because the water molecules in the air are polar. Those polar
water molecules will be attracted to the ruler and to the separated charge on the bits of paper,
neutralizing the charges and thus reducing the attraction.
5.
See Fig. 16–9 in the text. The part of the paper near the
–+
charged rod becomes polarized—the negatively charged
electrons in the paper are attracted to the positively
–+
+++++++
charged rod and move toward it within their molecules.
–+
The attraction occurs because the negative charges in
the paper are closer to the positive rod than are the
–+
positive charges in the paper; therefore, the attraction
between the unlike charges is greater than the repulsion between the like charges.
6.
The net charge on a conductor is the sum of all of the positive and negative charges in the conductor.
If a neutral conductor has extra electrons added to it, then the net charge is negative. If a neutral
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16-1
16-2
Chapter 16
conductor has electrons removed from it, then the net charge is positive. If a neutral conductor has the
same amount of positive and negative charge, then the net charge is zero.
The “free charges” in a conductor are electrons that can move about freely within the material because
they are only loosely bound to their atoms. The “free electrons” are also referred to as “conduction
electrons.” A conductor may have a zero net charge but still have substantial free charges.
7.
For each atom in a conductor, only a small number of its electrons are free to move. For example,
every atom of copper has 29 electrons, but only 1 or 2 from each atom are free to move easily. Also,
not all of the free electrons move. As electrons move toward a region, causing an excess of negative
charge, that region then exerts a large repulsive force on other electrons, preventing them from all
moving to that same region.
8.
The electroscope leaves are connected together at the top. The horizontal
component of this tension force balances the electric force of repulsion.
The vertical component of the tension force balances the weight of the
leaves.
9.
The balloon has been charged. The excess charge on the balloon is able to polarize the water molecules
in the stream of water, similar to Fig. 16–9. This polarization results in a net attraction of the water
toward the balloon, so the water stream curves toward the balloon.
10.
Coulomb’s law and Newton’s law are very similar in form. When expressed in SI units, the magnitude
of the constant in Newton’s law is very small, while the magnitude of the constant in Coulomb’s law is
quite large. Newton’s law says the gravitational force is proportional to the product of the two masses,
while Coulomb’s law says the electrical force is proportional to the product of the two charges.
Newton’s law produces only attractive forces, since there is only one kind of gravitational mass.
Coulomb’s law produces both attractive and repulsive forces, since there are two kinds of electrical
charge.
11.
Assume that the charged plastic ruler has a negative charge residing on its surface. That charge
polarizes the charge in the neutral paper, producing a net attractive force. When the piece of paper then
touches the ruler, the paper can get charged by contact with the ruler, gaining a net negative charge.
Then, since like charges repel, the paper is repelled by the comb.
12.
For the gravitational force, we don’t notice it because the force is very weak, due to the very small
value of G, the gravitational constant, and the relatively small value of ordinary masses. For the
electric force, we don’t notice it because ordinary objects are electrically neutral to a very high degree.
We notice our weight (the force of gravity) due to the huge mass of the Earth, making a significant
gravity force. We notice the electric force when objects have a net static charge (like static cling from
the clothes dryer), creating a detectable electric force.
13.
The test charge creates its own electric field, The measured electric field is the sum of the original
electric field plus the field of the test charge. If the test charge is small, then the field that it causes is
small. Therefore, the actual measured electric field is not much different than the original field.
14.
A negative test charge could be used. For the purposes of defining directions, the electric field
might then be defined as the OPPOSITE of the force on the test charge, divided by the test charge.
G
G
Equation 16–3 might be changed to E = −F /q, q < 0.
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Electric Charge and Electric Field
16-3
15.
A
16.
The electric field is strongest to the right of the positive charge (on the line connecting the two
charges), because the individual fields from the positive charge and negative charge both are in the
same direction (to the right) at that point, so they add to make a stronger field. The electric field is
weakest to the left of the positive charge, because the individual fields from the positive charge and
negative charge are in opposite directions at that point, so they partially cancel each other. Another
indication is the spacing of the field lines. The field lines are closer to each other to the right of the
positive charge and farther apart to the left of the positive charge.
17.
At point A, the direction of the net force on a positive test
charge would be down and to the left, parallel to the nearby
electric field lines. At point B, the direction of the net force on
a positive test charge would be up and to the right, parallel to
the nearby electric field lines. At point C, the net force on a
positive test charge would be 0. In order of decreasing field
strength, the points would be ordered A, B, C.
18.
Electric field lines show the direction of the force on a test charge placed at a given location. The
electric force has a unique direction at each point. If two field lines cross, it would indicate that the
electric force is pointing in two directions at once, which is not possible.
19.
From rule 1: A test charge would be either attracted directly toward or repelled directly away from a
point charge, depending on the sign of the point charge. So the field lines must be
directed either radially toward or radially away from the point charge.
From rule 2: The magnitude of the field due to the point charge only depends on the distance from the
point charge. Thus the density of the field lines must be the same at any location around
the point charge for a given distance from the point charge.
From rule 3: If the point charge is positive, then the field lines will originate from the location of the
point charge. If the point charge is negative, then the field lines will end at the location
of the point charge.
Based on rules 1 and 2, the lines are radial and their density is constant for a given distance. This is
equivalent to saying that the lines must be symmetrically spaced around the point charge.
20.
The two charges are located as shown in the diagram.
2Q
Q
(a) If the signs of the charges are opposite, then the point
on the line where E = 0 will lie to the left of Q. In
ℓ
that region the electric fields from the two charges
will point in opposite directions, and the point will be
closer to the smaller charge.
(b) If the two charges have the same sign, then the point on the line where E = 0 will lie between
the two charges, closer to the smaller charge. In this region, the electric fields from the two
charges will point in opposite directions.
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16-4
Chapter 16
21.
We assume that there are no other forces (like gravity) acting on the test charge. The direction of the
electric field line shows the direction of the force on the test charge. The acceleration is always parallel
to the force by Newton’s second law, so the acceleration lies along the field line. If the particle is at
rest initially and then is released, the initial velocity will also point along the field line, and the particle
will start to move along the field line. However, once the particle has a velocity, it will not follow the
field line unless the line is straight. The field line gives the direction of the acceleration, or the
direction of the change in velocity.
22.
The electric flux depends only on the charge enclosed by the gaussian surface (Eq. 16–9), not on the
shape of the surface. ΦE will be the same for the cube as for the sphere.
Responses to MisConceptual Questions
1.
(a)
The two charges have opposite signs, so the force is attractive. Since Q2 is located on the
positive x axis relative to Q1 at the origin, the force on Q1 will be in the positive x direction.
2.
(d)
The signs of the charges are still opposite, so the force remains attractive. However, since Q2 is
now located at the origin with Q1 on the positive x axis, the force on Q1 will now be toward the
origin, or in the negative x direction.
3.
(e)
A common misconception is that the object with the greater charge and smaller mass has a
greater force of attraction. However, Newton’s third law applies here. The force of attraction is
the same for both lightning bugs.
4.
(d)
Students sometimes believe that electric fields only exist on electric field lines. This is incorrect.
The field lines represent the direction of the electric field in the region of the lines. The
magnitude of the field is proportional to the density of the field lines. At point 1 the field lines
are closer together than they are at point two. Therefore, the field at point 1 is larger than the
field at point 2.
5.
(d)
A common misconception students have is recognizing which object is creating the electric field
and which object is interacting with the field. In this question the positive point charge is
creating the field. The field from a positive charge always points away from the charge. When a
negative charge interacts with an electric field, the force on the negative charge is in the opposite
direction from the field. The negative charge experiences a force toward the positive charge.
6.
(a)
An object acquires a positive charge when electrons are removed from the object. Since
electrons have mass, as they are removed the mass of the object decreases.
7.
(a)
Students frequently think of the plates as acting like point charges with the electric field
increasing as the plates are brought closer together. However, unlike point charges, the electric
field lines from each plate are parallel with uniform density. Bringing the plates closer together
does not affect the electric field between them.
8.
(b)
The value measured will be slightly less than the electric field value at that point before the test
charge was introduced. The test charge will repel charges on the surface of the conductor, and
these charges will move along the surface to increase their distances from the test charge. Since
they will then be at greater distances from the point being tested, they will contribute a smaller
amount to the field.
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Electric Charge and Electric Field
16-5
9.
(d)
Students may equate the lack of the electric force between everyday objects as a sign that the
electric force is weaker than other forces in nature. Actually, the electric force between charged
particles is much greater than the gravitational force between them. The apparent lack of electric
force between everyday objects is because most objects are electrically neutral. That is, most
objects have the same number of positive and negative charges in them.
10.
(b)
In a lightning storm, charged particles in the clouds and ground create large electric fields that
ionize the air, creating lightning bolts. People are good conductors of electric charge and when
located in the electric field can serve as conduits of the lightning bolt. The inside of a metal car
acts like a cavity in a conductor, shielding the occupants from the external electric fields. In each
of the other options the person remains in the storm’s electric field and therefore may be struck
by the lightning. If the car is struck by lightning, then the electricity will pass along the exterior
of the car, leaving the people inside unharmed.
11.
(a, c, e) Lightning will generally travel from clouds to the tallest conductors in the area. If the person
is in the middle of a grassy field, near the tallest tree, or on a metal observation town, then he or
she is likely to be struck by lightning. A person inside a wooden building or inside a car is
somewhat shielded from the lightning.
12.
(d)
The electric field at the fourth corner is the vector sum of the electric fields from the charges at
the other three corners. The two positive charges create equal-magnitude electric fields pointing
in the positive x direction and in the negative y direction. These add to produce an electric field
in the direction of (d) with magnitude 2 times the magnitude of the electric field from one of
the charges. The electric field from the negative charge points in the direction of (b). However,
since the charge is 2 times farther away than the positive charges, the magnitude of the
electric field will be smaller than the field from the positive charges. The resulting field will then
be in the direction of (d).
13.
(e)
A common misconception is that the metal ball must be negatively charged. While a negatively
charged ball will be attracted to the positive rod, a neutral conductor will become polarized and
also be attracted to the positive rod.
Solutions to Problems
1.
Use Coulomb’s law (Eq. 16–1) to calculate the magnitude of the force.
F =k
2.
Q1Q2
r
2
= (8.988 × 109 N ⋅ m 2 /C 2 )
(1.602 × 10−19 C)(26 × 1.602 × 10−19 C)
(1.5 × 10
−12
m)
2
= 2.7 × 10−3 N
Use the charge per electron to find the number of electrons.
⎛
⎞
1 electron
= 2.996 × 1014 electrons ≈ 3.00 × 1014 electrons
(− 48.0 × 10−6 C) ⎜
⎜ −1.602 × 10−19 C ⎟⎟
⎝
⎠
3.
Use Coulomb’s law (Eq. 16–1) to calculate the magnitude of the force.
F =k
Q1Q2
r
2
= (8.988 × 109 N ⋅ m 2 /C2 )
(25 × 10−6 C)(2.5 × 10−3 C)
(0.16 m)
2
= 2.194 × 104 N ≈ 2.2 × 104 N
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16-6
4.
Chapter 16
Use Coulomb’s law (Eq. 16–1) to calculate the magnitude of the force.
F =k
5.
Q1Q2
r2
= (8.988 × 109 N ⋅ m 2 /C 2 )
(1.602 × 10−19 C) 2
(4.0 × 10−15 m) 2
= 14.42 N ≈ 14 N
The charge on the plastic comb is negative, so the comb has gained electrons and has a negative charge.
⎛
⎞ ⎛ 9.109 × 10−31 kg ⎞
1 e−
(−3.0 × 10−6 C) ⎜
⎟⎟ ⎜⎜
⎟⎟
−19
⎜
−1.602 × 10
C ⎠⎝
1 e−
Δm
⎝
⎠ = 1.90 × 10−15 ≈ (1.9 × 10−13 )%
=
m
0.0090 kg
6.
Since the magnitude of the force is inversely proportional to the square of the separation distance, F ∝
1
r2
,
if the distance is multiplied by a factor of 1/8, then the force will be multiplied by a factor of 64.
F = 64 F0 = 64(4.2 × 10−2 N) = 2.688 N ≈ 2.7 N
7.
Since the magnitude of the force is inversely proportional to the square of the separation distance,
1
F ∝ 2 , if the force is tripled, then the distance has been reduced by a factor of 3.
r
r=
8.
r0
3
=
6.52 cm
3
= 3.76 cm
Use the charge per electron and the mass per electron.
⎛
⎞
1 electron
(− 28 × 10−6 C) ⎜
= 1.748 × 1014 ≈ 1.7 × 1014 electrons
⎜ −1.602 × 10−19 C ⎟⎟
⎝
⎠
⎛ 9.109 × 10−31 kg ⎞
−16
(1.748 × 1014 e − ) ⎜
kg ≈ 1.6 × 10−16 kg
⎟⎟ = 1.592 × 10
−
⎜
1
e
⎝
⎠
9.
To find the number of electrons, convert the mass to moles and the moles to atoms and then multiply
by the number of electrons in an atom to find the total electrons. Then convert to charge.
⎛ 1 mole Au ⎞ ⎛ 6.022 × 1023 atoms ⎞ ⎛ 79 electrons ⎞ ⎛ −1.602 × 10−19 C ⎞
12 kg Au = (12 kg Au) ⎜
⎟⎟ ⎜
⎟⎟
⎟⎜
⎟ ⎜⎜
1 mole
electron
⎝ 0.197 kg ⎠ ⎜⎝
⎠ ⎝ 1 atom ⎠ ⎝
⎠
= −4.642 × 108 C ≈ −4.6 × 108 C
The net charge of the bar is 0 , since there are equal numbers of protons and electrons.
10.
Take the ratio of the electric force divided by the gravitational force. Note that the distance is not
needed for the calculation.
FE
=
FG
k
Q1Q2
(8.988 × 109 N ⋅ m 2 /C2 )(1.602 × 10−19 C) 2
r 2 = kQ1Q2 =
= 2.27 × 1039
m1m2 Gm1m2 (6.67 × 10−11 N ⋅ m 2 /kg 2 )(9.11 × 10−31 kg)(1.67 × 10−27 kg)
G 2
r
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Electric Charge and Electric Field
16-7
The electric force is about 2.27 × 1039 times stronger than the gravitational force for the given
scenario.
11.
Let the right be the positive direction on the line of charges. Use the fact that like charges repel and
unlike charges attract to determine the direction of the forces. In the following expressions,
k = 8.988 × 109 N ⋅ m 2 /C2 .
F+65 = − k
F+48 = k
(0.35 m) 2
(48 μ C)(65 μ C)
F−95 = − k
12.
(65 μ C)(48 μ C)
(0.35 m)
2
+k
(95 μ C)(65 μ C)
(0.70 m)
+k
2
(65 μ C)(95 μ C)
(48 μ C)(95 μ C)
= 563.49 N ≈ 560 N (to the right)
(0.35 m) 2
−k
= −115.65 N ≈ −120 N (to the left)
(0.70 m)2
(95 μ C)(48 μ C)
= −447.84 N ≈ −450 N (to the left)
(0.35 m) 2
The forces on each charge lie along a line connecting the charges. Let d represent
the length of a side of the triangle, and let Q represent the charge at each corner.
Since the triangle is equilateral, each angle is 60°.
F12 = k
F13 = k
Q2
d
2
Q
2
d2
→ F12 x = k
Q2
d
→ F13 x = −k
F1x = F12 x + F13 x = 0
F1 = F12x + F12y = 3k
cos 60°, F12 y = k
2
Q
2
d2
d2
d
cos 60°, F13 y = k
F1 y = F12 y + F13 y = 2k
Q2
Q2
2
Q2
d2
Q2
d
2
d
sin 60°
= 3(8.988 × 109 N ⋅ m 2 /C2 )
G
F12
Q1
Q2
sin 60°
sin 60° = 3k
G
F13
d
Q3
d
Q2
d2
(17.0 × 10−6 C) 2
(0.150 m) 2
= 199.96 N
≈ 2.00 × 102 N
G
The direction of F1 is in the y direction. Also notice that it lies along the bisector of the opposite side
of the triangle. Thus the force on the lower left charge is of magnitude 2.00 × 102 N and will point
30° below the −x axis . Finally, the force on the lower right charge is of magnitude 2.00 × 102 N and
will point 30° below the +x axis .
13.
The spheres can be treated as point charges since they are spherical. Thus Coulomb’s law may be used
to relate the amount of charge to the force of attraction. Each sphere will have a magnitude Q of
charge, since that amount was removed from one (initially neutral) sphere and added to the other.
F =k
Q1Q2
r2
=k
Q2
r2
→ Q=r
F
1.7 × 0−2 N
= (0.24 m)
k
8.988 × 109 N ⋅ m 2 /C2
⎛ 1 electron ⎞
= 3.3007 × 10−7 C ⎜
= 2.06 × 1012 ≈ 2.1 × 1012 electrons
⎜ 1.602 × 10−19 C ⎟⎟
⎝
⎠
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16-8
14.
Chapter 16
F41 = k
F42 = k
F43 = k
Q2
2
d
Q2
→ F41x = k
Q
d2
d
→ F42 x = k
2d 2
2
Q2
2
Q1
→ F43 x = 0, F43 y = k
Q
2Q 2
4d 2
, F42 y = k
G
F41
Q3
Q2
cos 45° = k
2d 2
Q4
d
, F41 y = 0
Q2
G
F42
G
F43
Determine the force on the upper right charge and then use the
symmetry of the configuration to determine the force on the other
three charges. The force at the upper right corner of the square is the
vector sum of the forces due to the other three charges. Let the
variable d represent the 0.100-m length of a side of the square, and let
the variable Q represent the 6.15-mC charge at each corner.
2Q 2
4d 2
2
d2
Add the x and y components together to find the total force, noting that F4 x = F4 y .
F4 x = F41x + F42 x + F43 x = k
F4 = F42x + F42y = k
Q2
d2
F4 y
F4 x
4d 2
+0 = k
Q2 ⎛
2⎞
1+
⎟ = F4 y
2 ⎜
⎜
4 ⎟⎠
d ⎝
Q2 ⎛
Q2 ⎛
2⎞
1⎞
k
1
2
+
=
⎜
⎟
⎜ 2+ ⎟
4 ⎟⎠
2⎠
d 2 ⎜⎝
d2 ⎝
= (8.988 × 109 N ⋅ m 2 /C2 )
θ = tan −1
2Q 2
+k
(6.15 × 10−3 C) 2 ⎛
1⎞
7
7
⎜ 2 + ⎟ = 6.507 × 10 N ≈ 6.51 × 10 N
2
2⎠
⎝
(0.100 m)
= 45° above the x direction
For each charge, the net force will be the magnitude determined above and will lie along the line from
the center of the square out toward the charge.
15.
Take the lower left-hand corner of the square to be the origin of coordinates. The 2Q will have a
rightward force on it due to Q, an upward force on it due to 3Q, and a diagonal force on it due to 4Q.
Find the components of each force, add the components, find the magnitude of the net force, and find
the direction of the net force. At the conclusion of the problem there is a diagram showing the net force
on the charge 2Q.
2Q: F2Qx = k
F2Qy = k
(2Q)Q
A
2
+k
(2Q)(3Q)
A2
(2Q)(4Q )
+k
2A
2
cos 45° = k
(2Q)(4Q)
2A 2
F2Q = F22Qx + F22Qy = 10.1
kQ 2
A
2
Q2
A
2
sin 45° = k
( 2 + 2 2 ) = 4.8284 kQA2
Q2
2
2
kQ
6 + 2 2 ) = 8.8284 2
2 (
A
A
θ 2Q = tan −1
F2 y
F2 x
= tan −1
8.8284
= 61°
4.8284
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Charge and Electric Field
16.
The wires form two sides of an isosceles triangle, so the two charges are separated
by a distance A = 2(78 cm) sin 26° = 68.4 cm and are directly horizontal from each
other. Thus the electric force on each charge is horizontal. From the free-body
diagram for one of the spheres, write the net force in both the horizontal and vertical
directions and solve for the electric force. Then write the electric force as given by
Coulomb’s law and equate the two expressions for the electric force to find the
charge.
∑ Fy = FT cos θ − mg = 0
∑ Fx = FT sin θ − FE = 0
FE = k
(Q /2) 2
A
2
(a)
→ Q = 2A
θ
G
mg
mg
sin θ = mg tan θ
cos θ
mg tan θ
k
(21 × 10−3 kg)(9.80 m/s 2 ) tan 26°
9
G
FT
mg
cos θ
→ FE = FT sin θ =
= mg tan θ
= 2(0.684 m)
17.
→ FT =
G
FE
16-9
2
2
(8.988 × 10 N ⋅ m /C )
= 4.572 × 10−6 C ≈ 4.6 × 10−6 C
If the force is repulsive, then both charges must be positive since the total charge is positive. Call
the total charge Q.
Q1 + Q2 = Q
Q1 =
F=
Q ± Q2 − 4
kQ1Q2
d2
=
kQ1 (Q − Q1 )
d2
→ Q12 − QQ1 +
Fd 2
=0
k
Fd 2
k
2
⎡
(12.0 N)(0.280 m) 2 ⎤
⎥
= 12 ⎢(90.0 × 10−6 C) ± (90.0 × 10−6 C) 2 − 4
⎢
(8.988 × 109 N ⋅ m 2 /C2 ) ⎥⎦
⎣
= 88.8 × 10−6 C, 1.2 × 10−6 C
(b)
If the force is attractive, then the charges are of opposite sign. The value used for F must then be
negative. Other than that, the solution method is the same as for part (a).
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
16-10
Chapter 16
Q1 =
Q ± Q2 − 4
Fd 2
k
2
⎡
(−12.0 N)(0.280 m) 2 ⎤
⎥
= 12 ⎢ (90.0 × 10−6 C) ± (90.0 × 10−6 C) 2 − 4
9
2 2
⎢
⎥
×
⋅
(8.988
10
N
m
/C
)
⎣
⎦
= 91.1 × 10−6 C, − 1.1 × 10−6 C
18.
The negative charges will repel each other, so the third charge must
q
−Q
−3Q
put an opposite force on each of the original charges. Consideration
x
l–x
of the various possible configurations leads to the conclusion that
the third charge must be positive and must be between the other two
l
charges. See the diagram for the definition of variables. For each
negative charge, equate the magnitudes of the two forces on the charge. Also note that 0 < x < A.
Left: k
k
k
Qq
x
2
Qq
x
2
Qq
x
=k
=k
2
=k
3Q 2
A
3Qq
(A − x)
3Q 2
A
2
2
2
Right: k
(A − x)
A
→ x=
→ q = 3Q
3Qq
3 +1
x2
A
2
2
=k
3Q 2
A2
→
= 0.366A
=Q
3
(
)
3 +1
2
= 0.402Q
Thus the charge should be of magnitude 0.40 Q , and a distance 0.37A from − Q toward − 3Q .
19.
Use Eq. 16–3 to calculate the force.
G
G F
E=
q
20.
G
G
→ F = qE = (−1.602 × 10−19 C)(2460 N/C east) = 3.94 × 10−16 N west
Use Eq. 16–3 to calculate the electric field.
G
G F 1.86 × 10−14 N south
= 1.16 × 105 N/C south
E= =
q
1.602 × 10−19 C
21.
Use Eq. 16–4a to calculate the electric field due to a point charge.
E=k
Q
r
2
= (8.988 × 109 N ⋅ m 2 /C2 )
33.0 × 10−6 C
(0.217 m)
2
= 6.30 × 106 N/C upward
Note that the electric field points away from the positive charge.
22.
Use Eq. 16–3 to calculate the electric field.
G
G F
6.4 N down
E= =
= 8.8 × 105 N/C up
q −7.3 × 10−6 C
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Electric Charge and Electric Field
23.
16-11
Assuming the electric force is the only force on the electron, then Newton’s second law may be used
with Eq. 16–5 to find the acceleration.
G
G
G
|q|
(1.602 × 10−19 C)
Fnet = ma = qE → a = E =
756 N/C = 1.33 × 1014 m/s 2
m
(9.11 × 10−31 kg)
Since the charge is negative, the direction of the acceleration is opposite to the field .
24.
E = E1 + E2 = k
=
Q1
r12
+k
Q2
r22
2
2
4(8.988 × 109 N ⋅ m /C )
(0.060 m) 2
=k
Q1
(A /2)
2
+k
Q2
(A /2)
2
G
E1
Q1 > 0
The electric field due to the positive charge (Q1 ) will
point away from it, and the electric field due to the
negative charge (Q2 ) will point toward it. Thus both
fields point in the same direction, toward the negative
charge, and the magnitudes can be added.
A 2
=
4k
A2
Q2 < 0
G
E2
( Q1 + Q2 )
(8.0 × 10−6 C + 5.8 × 10−6 C) = 1.3782 × 108 N/C ≈ 1.4 × 108 N/C
The direction is toward the negative charge .
25.
A
26.
Assuming the electric force is the only force on the electron, Newton’s second law may be used to find
the electric field strength.
Fnet = ma = qE →
E=
27.
ma (1.673 × 10−27 kg)(2.4 × 106 )(9.80 m/s 2 )
=
= 0.2456 N/C ≈ 0.25 N/C
q
(1.602 × 10−19 C)
Since the electron accelerates from rest toward the north, the net force on it must be to the north.
Assuming the electric force is the only force on the electron, Newton’s second law and Eq. 16–5 may
be used to find the electric field.
G
G
G mG
(9.11 × 10−31 kg)
G
(105 m/s 2 north) = 5.97 × 10−10 N/C south
Fnet = ma = qE → E = a =
q
(−1.602 × 10−19 C)
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16-12
28.
Chapter 16
The field due to the negative charge will point toward
the negative charge, and the field due to the positive
charge will point toward the negative charge. Thus the
magnitudes of the two fields can be added together to
find the charges.
Enet = 2 EQ = 2k
29.
Q
(A /2) 2
8kQ
=
→ Q=
A2
G
EQ
Q
E1 = k
E2 = k
E3 = k
A2
Q
2A 2
Q
A
2
→ E1x = k
Q
A2
→ E2 x = k
G
E −Q
l 2
EA2
(386 N/C)(0.160 m)2
=
= 1.37 × 10−10 C
8k
8(8.988 × 109 N ⋅ m 2 /C2 )
G
E3
The field at the upper right-hand corner of the square is the vector sum
of the fields due to the other three charges. Let A represent the 1.22-m
length of a side of the square, and let Q represent the charge at each
of the three occupied corners.
Q
−Q
G
E2
Q1
G
E1
l
, E1 y = 0
Q
2A 2
cos 45° = k
→ E3 x = 0, E1 y = k
2Q
4A 2
, E2 y = k
Q3
Q2
2Q
4A 2
Q
A2
Add the x and y components together to find the total electric field, noting that E x = E y .
E x = E1x + E2 x + E3 x = k
E = E x2 + E y2 = k
Q
A
2
+k
30.
Ey
Ex
4A
2
+0 = k
Q⎛
2⎞
⎜1 +
⎟ = Ey
2⎜
4 ⎟⎠
A ⎝
2⎞
1⎞
Q⎛
Q⎛
1+
⎟⎟ 2 = k 2 ⎜ 2 + ⎟
2⎜
⎜
4 ⎠
2⎠
A ⎝
A ⎝
= (8.988 × 109 N ⋅ m 2 /C2 )
θ = tan −1
2Q
(3.25 × 10−6 C) ⎛
1⎞
4
⎜ 2 + ⎟ = 3.76 × 10 N/C
2⎠
(1.22 m) 2 ⎝
= 45.0° from the x direction
The fields at the center due to the two −27.0 μ C negative
charges on opposite corners (lower right and upper left in the
diagram) will cancel each other, so only the other two charges
need to be considered. The field due to each of the other
charges will point directly toward its source charge.
Accordingly, the two fields are in opposite directions and can
be combined algebraically. The distance from each charge to
the center is A / 2.
Q2 = −27.0 μ C
Q2
l
G
E1
Q1 = −38.6 μ C
G
E2
Q2
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Electric Charge and Electric Field
E = E1 − E2 = k
31.
Q1
2
A /2
Q2
−k
2
A /2
=k
= (8.988 × 109 N ⋅ m 2 /C2 )
Q1 − Q2
(38.6 − 27.0) × 10−6 C
(0.425 m) 2 /2
2
A /2
16-13
= 1.15 × 106 N/C, toward the − 38.6 μ C charge
Choose the rightward direction to be positive. Then the field due to +Q will be positive, and the field
due to −Q will be negative.
E=k
Q
(x + a)
2
⎛ 1
1
= kQ ⎜
−
2
⎜
(x − a )
(x − a) 2
⎝ (x + a)
Q
−k
2
⎞
−4kQxa
⎟⎟ = 2
2 2
⎠ (x − a )
The negative sign means the field points to the left.
32.
For the net field to be zero at point P, the magnitudes of the fields created by Q1 and Q2 must be
equal. Also, the distance x will be taken as positive to the left of Q1. That is the only region where
the total field due to the two charges can be zero. Let the variable A represent the 12-cm distance.
G
G
E1 = E2
x=A
33.
(
→ k
Q1
Q2 − Q1
Q1
x
)
=k
2
Q2
→
(x + A) 2
= (12 cm)
32 μ C
(
45μ C − 32μ C
)
= 64.57 cm ≈ 65 cm
The field due to the charge at A will point straight downward, and the
field due to the charge at B will point along the line from the origin to
point B, 30° above the x axis. We first solve the problem symbolically
and then substitute in the values.
EA = k
EB = k
Q
2
A
Q
A2
→ EAx = 0, EAy = −k
→ EBx = k
EBy = k
E x = EAx + EBx = k
E = E x2 + E y2 =
=
Q
A2
Q
A2
3Q
2A
3k 2 Q 2
4A
4
+
3Q
2A 2
Q
4A
4
−Q
l
A2
sin 30° = k
=
l
Q
cos 30° = k
k 2Q 2
+Q
G
EB
B
l
G
EA
,
2A 2
E y = EAy + EBy = −k
2
A
4k 2 Q 2
4A
4
=
Q
2A 2
kQ
A2
(8.988 × 109 N ⋅ m 2 /C2 )(26 × 10−6 C)
θ = tan −1
= 3.651 × 107 N/C ≈ 3.7 × 107 N/C
(0.080 m) 2
Q
−k 2
Ey
2A = tan −1 −1 = 330°
= tan −1
Ex
3Q
3
k 2
2A
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16-14
34.
Chapter 16
The charges must be of the same sign so that the electric fields created by the two charges oppose each
other and add to zero. The magnitudes of the two electric fields must be equal.
E1 = E2
35.
→ k
Q1
(A /3)
2
=k
Q2
(2A /3)
2
→ 9Q1 =
9Q2
4
Q1
1
=
4
Q2
→
G
In each case, find the vector sum of the field caused by the charge on the left (Eleft ) and the field
G
G
G
caused by the charge on the right (Eright ).
Eright
E
left
Point A: From the symmetry of the geometry, in
calculating the electric field at point A, only the
vertical components of the fields need to be
considered. The horizontal components will cancel
each other.
θ = tan −1
d
d
θ
θ
+Q
+Q
5.0
= 26.6°
10.0
d = (0.050 m) 2 + (0.100 m) 2 = 0.1118 m
EA = 2
kQ
d
2
sinθ = 2(8.988 × 109 N ⋅ m 2 /C2 )
4.7 × 10−6 C
(0.1118 m)
2
sin 26.6° = 3.0 × 106 N/C
Point B: Now the point is not symmetrically placed.
The horizontal and vertical components of each
individual field need to be calculated to find the
resultant electric field.
θleft = tan −1
5.0
= 45°
5.0
θleft = tan −1
5.0
= 18.4°
15.0
G
Eright
dleft
+Q
θ left
θ A = 90°
G
Eleft
d right
θ right
+Q
dleft = (0.050 m) 2 + (0.050 m) 2 = 0.0707 m
d right = (0.050 m) 2 + (0.150 m) 2 = 0.1581 m
G
G
Q
Q
E x = (Eleft ) x + (Eright ) x = k 2 cos θ left − k 2 cos θ right
d left
d right
⎡ cos 45°
cos 18.4° ⎤
= (8.988 × 109 N ⋅ m 2 /C2 )(4.7 × 10−6 C) ⎢
−
= 4.372 × 106 N/C
2
2⎥
(0.1581 m) ⎦⎥
⎣⎢ (0.0707 m)
G
G
Q
Q
E y = (Eleft ) y + (Eright ) y = k 2 sin θ left + k 2 sin θ right
dleft
d right
⎡ sin 45°
sin 18.4° ⎤
6
= (8.988 × 109 N ⋅ m 2 /C2 )(4.7 × 10−6 C) ⎢
+
⎥ = 6.509 × 10 N/C
2
(0.1581 m) 2 ⎥⎦
⎢⎣ (0.0707 m)
Ey
θ B = tan −1
EB = E x2 + E y2 = 7.841× 106 N/C ≈ 7.8 × 106 N/C
= 56.11° ≈ 56°
Ex
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Electric Charge and Electric Field
16-15
The results are consistent with Fig. 16–32b. In the figure, the field at point A points straight up,
matching the calculations. The field at point B should be to the right and vertical, matching the
calculations. Finally, the field lines are closer together at point B than at point A, indicating that the
field is stronger there, again matching the calculations.
36.
We assume that gravity can be ignored, which is proven in part (b).
(a) The electron will accelerate to the right. The magnitude of the acceleration can be found from
setting the net force equal to the electric force on the electron. The acceleration is constant, so
constant-acceleration relationships can be used.
Fnet = ma = |q| E → a =
|q| E
m
υ 2 = υ02 + 2aΔx → υ = 2aΔx = 2
= 2
(b)
|q| E
Δx
m
(1.602 × 10−19 C)(1.45 × 104 N/C)
(9.11 × 10
−31
kg)
(1.60 × 10−2 m) = 9.03 × 106 m/s
The value of the acceleration caused by the electric field is compared to g.
a=
|q| E (1.602 × 10−19 C)(1.45 × 104 N/C)
=
= 2.55 × 1015 m/s 2
31
−
m
(9.11 × 10 kg)
a /g = (2.55 × 1015 m/s 2 )/(9.80 m/s 2 ) = 2.60 × 1014
The acceleration due to gravity can be ignored compared to the acceleration caused by the electric
field.
37.
(a)
The net force between the thymine and adenine is due to the following forces.
O–H attraction:
FOH = k
O–N repulsion:
FON = k
N–N repulsion:
FNN = k
H–N attraction:
FHN = k
(0.4e)(0.2e)
(1.80 × 10−10 m) 2
(0.4e)(0.2e)
(2.80 × 10−10 m) 2
(0.2e)(0.2e)
(3.00 × 10−10 m) 2
(0.2e)(0.2e)
(2.00 × 10−10 m) 2
FA −T = FOH − FON − FNN + FHN
= (0.02004)
=
=
=
=
0.08ke2
(1.80 × 10−10 m) 2
0.08ke 2
(2.80 × 10−10 m) 2
0.04ke2
(3.00 × 10−10 m) 2
0.04ke 2
(2.00 × 10−10 m) 2
0.04
0.04
⎛ 0.08 0.08
=⎜
−
−
+
2
2
2
2.80
3.00
2.002
⎝ 1.80
(8.988 × 109 N ⋅ m 2 /C2 )(1.602 × 10−19 C) 2
(1.0 × 10
−10
m)
2
2
⎞ ke 2
1
⎞⎛
⎜
⎟
⎟⎜
⎠ ⎝ 1.0 × 10−10 m ⎟⎠ d 2
= 4.623 × 10−10 N ≈ 5 × 10−10 N
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16-16
Chapter 16
(b)
The net force between the cytosine and guanine is due to the following forces.
O–H attraction:
FOH = k
O–N repulsion:
FON = k
H–N attraction:
FHN = k
N–N repulsion:
FNN = k
(0.4e)(0.2e)
(1.90 × 10−10 m) 2
(0.4e)(0.2e)
(2.90 × 10−10 m) 2
(0.2e)(0.2e)
(2.00 × 10
(c)
m)
2
(0.2e)(0.2e)
(3.00 × 10−10 m) 2
FC −G = 2 FOH − 2 FON − FNN + FHN
= (0.03085)
−10
=
=
=
=
0.08ke2
(1.90 × 10−10 m) 2
(two of these)
0.08ke2
(2.90 × 10−10 m) 2
(two of these)
0.04ke2
(2.00 × 10−10 m) 2
0.04ke2
(3.00 × 10−10 m) 2
0.08
0.04
0.04
⎛ 0.08
= ⎜2
−2
−
+
2
2
2
2.90
3.00
2.002
⎝ 1.90
(8.988 × 109 N ⋅ m 2 /C 2 )(1.602 × 10−19 C) 2
(1.0 × 10
−10
m)
2
2
⎞ ke2
1
⎞⎛
⎟
⎟ ⎜⎜
⎠ ⎝ 1.0 × 10−10 m ⎟⎠ d 2
= 7.116 × 10−10 N ≈ 7 × 10−10 N
For the 105 pairs of molecules, we assume that half are A–T pairs and half are C–G pairs. We
average the above results and multiply by 105.
Fnet = 12 105 (FA −T + FC −G ) = 105 (4.623 × 10−10 N + 7.116 × 10−10 N)
= 5.850 × 10−5 N ≈ 6 × 10−5 N
38.
Use Gauss’s law to determine the enclosed charge. Note that the size of the box does not enter into the
calculation.
ΦE =
39.
(a)
Qencl
ε0
Use Gauss’s law to determine the electric flux.
ΦE =
(b)
→ Qencl = ΦE ε 0 = (1850 N ⋅ m 2 /C)(8.85 × 10−12 C2 /N ⋅ m 2 ) = 1.64 × 10−8 C
Qencl
ε0
=
−1.0 × 10−6 C
8.85 × 10
−12
2
C /N ⋅ m
2
= −1.1 × 105 N ⋅ m 2 /C
Since there is no charge enclosed by surface A 2 , Φ E = 0 .
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Electric Charge and Electric Field
40.
(a)
(b)
Assuming that there is no charge contained within the cube,
the net flux through the cube is 0. All of the field lines that
enter the cube also leave the cube.
Four faces have no flux through them, because no field lines
pass through them. In the diagram, the left face has a
positive flux and the right face has the opposite amount of
negative flux.
16-17
l
Φleft = EA = E A 2 = (7.50 × 103 N/C)(8.50 × 10−2 m) 2
Φleft = 54.2 N ⋅ m 2 /C ; Φright = −54.2 N ⋅ m 2 /C ; Φother = 0
41.
Equation 16–10 applies. Note that the separation distance does not enter into the calculation.
E=
42.
Q /A
ε0
→ Q = ε 0 EA = (8.85 × 10−12 C 2 /N ⋅ m 2 )(130 N/C)(0.85 m) 2 = 8.3 × 10−10 C
The electric field can be calculated by Eq. 16–4a, and that can be solved for the charge.
E=k
Q
r2
→ Q=
Er 2 (3.75 × 102 N/C)(3.50 × 10−2 m) 2
=
= 5.11 × 10−11 C
9
2
2
k
8.988 × 10 N ⋅ m /C
This is about 3 × 108 electrons. Since the field points toward the ball, the charge must be negative.
43.
See Fig. 16–34 in the text for additional insight into this problem.
Q
(a)
Inside the inner radius of the shell, the field is that of the point charge, E = k
(b)
There is no field inside the conducting material: E = 0 .
(c)
Outside the shell, the field is that of the point charge, E = k
(d)
The shell does not affect the field due to Q alone, except in the shell material, where the field is 0.
Q
r2
r2
.
.
The charge Q does affect the shell—it polarizes it. There will be an induced charge of −Q
uniformly distributed over the inside surface of the shell and an induced charge of +Q uniformly
distributed over the outside surface of the shell.
44.
Set the magnitude of the electric force equal to the magnitude of the force of gravity and solve for the
distance.
Felectric = Fgravitational
r=e
45.
→ k
e2
r2
= mg
→
k
(8.988 × 109 N ⋅ m 2 /C2 )
= (1.602 × 10−19 C)
= 5.08 m
mg
(9.11 × 10−31 kg)(9.80 m/s 2 )
Water has an atomic mass of 18, so 1 mole of water molecules has a mass of 18 grams. Each water
molecule contains 10 protons.
⎛ 6.02 × 1023 H 2 O molecules ⎞ ⎛ 10 protons ⎞ ⎛ 1.60 × 10−19 C ⎞
9
75 kg ⎜
⎟⎟ ⎜
⎟⎟ = 4.0 × 10 C
⎟ ⎜⎜
⎜
0.018
kg
1
molecule
proton
⎝
⎠
⎝
⎠
⎝
⎠
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16-18
46.
Chapter 16
Calculate the total charge on all electrons in 3.0 g of copper and compare 32 μ C to that value. The
molecular weight of copper is 63.5 g, and its atomic number is 29.
⎛ 1 mole ⎞ ⎛ 6.02 × 1023 atoms ⎞ ⎛ 29 e ⎞ ⎛ 1.602 × 10−19 C ⎞
5
Total electron charge = 3.0 g ⎜
⎟⎟ ⎜
⎟⎟ = 1.32 × 10 C
⎟ ⎜⎜
⎟ ⎜⎜
mole
1e
⎝ 63.5 g ⎠ ⎝
⎠ ⎝ atoms ⎠ ⎝
⎠
Fraction lost =
47.
32 × 10−6 C
1.32 × 105 C
= 2.4 × 10−10
Use Eq. 16–4a to calculate the magnitude of the electric charge on the Earth. We use the radius of the
Earth as the distance.
E=k
Q
r2
→ Q=
Er 2 (150 N/C)(6.38 × 106 m) 2
=
= 6.8 × 105 C
k
8.988 × 109 N ⋅ m 2 /C2
Since the electric field is pointing toward the Earth’s center, the charge must be negative.
48.
(a)
From Problem 47, we know that the electric field is pointed toward the Earth’s center. Thus an
electron in such a field would experience an upward force of magnitude FE = eE. The force of
gravity on the electron will be negligible compared to the electric force.
FE = eE = ma →
a=
(b)
eE (1.602 × 10−19 C)(150 N/C)
=
= 2.638 × 1013 m/s 2 ≈ 2.6 × 1013 m/s 2 , up
−
31
m
(9.11 × 10 kg)
A proton in the field would experience a downward force of magnitude FE = eE. The force of
gravity on the proton will be negligible compared to the electric force.
FE = eE = ma →
a=
(c)
49.
eE (1.602 × 10−19 C)(150 N/C)
=
= 1.439 × 1010 m/s 2 ≈ 1.4 × 1010 m/s 2 , down
−27
m
(1.67 × 10
kg)
For the electron:
a 2.638 × 1013 m/s 2
=
≈ 2.7 × 1012
2
g
9.80 m/s
For the proton:
a 1.439 × 1010 m/s 2
=
≈ 1.5 × 109
2
g
9.80 m/s
For the droplet to remain stationary, the magnitude of the electric force on the droplet must be the
same as the weight of the droplet. The mass of the droplet is found from its volume times the density
of water. Let n be the number of excess electrons on the water droplet.
FE = |q|E = mg
n=
→ neE = 34 π r 3 ρ g
→
4π r 3 ρ g 4π (1.8 × 10−5 m)3 (1.00 × 103 kg/m3 )(9.80 m/s 2 )
=
= 9.96 × 106 ≈ 1.0 × 107 electrons
3eE
3(1.602 × 10−19 C)(150 N/C)
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Electric Charge and Electric Field
50.
16-19
There are four forces to calculate. Call the rightward direction the positive direction. The value of k is
8.988 × 109 N ⋅ m 2 /C2 and the value of e is 1.602 × 10−19 C.
Fnet = FCH + FCN + FOH + FON =
k (0.40e)(0.20e) ⎡
1
1
1
1 ⎤
−
+
+
−
⎥
2 ⎢
2
2
2
−9
(1 × 10 m) ⎢⎣ (0.30)
(0.40)
(0.18)
(0.28) 2 ⎥⎦
= 2.445 × 10−10 N ≈ 2.4 × 10−10 N
51.
The electric force must be a radial force in order for the electron to move in a circular orbit.
FE = Fradial
rorbit = k
52.
→ k
Q2
mυ
2
Q2
2
rorbit
=
mυ 2
rorbit
→
= (8.988 × 109 N ⋅ m 2 /C2 )
(1.602 × 10−19 C) 2
(9.109 × 10
−31
6
kg)(2.2 × 10 m/s)
2
= 5.2 × 10−11 m
If all of the angles to the vertical (in both cases) are assumed
G
G
to be small, then the spheres only have horizontal
FT1
FT2
θ1
displacement, and the electric force of repulsion is always
θ2
horizontal. Likewise, the small-angle condition leads to
G
G
tan θ ≈ sin θ ≈ θ for all small angles. See the free-body
G
FE1 m gG
FE2
m
g
diagram for each sphere, showing the three forces of gravity,
1
2
tension, and the electrostatic force on each charge. Take the
right to be the positive horizontal direction and up to be the
positive vertical direction. Since the spheres are in equilibrium, the net force in each direction is zero.
(a)
∑ F1x = FT1 sin θ1 − FE1 = 0
∑ F1y = FT1 cos θ1 − m1 g
→ FE1 = FT1 sin θ1
m1 g
cos θ1
→ FT1 =
→ FE1 =
m1 g
sin θ1 = m1 g tan θ1 = m1 gθ1
cos θ1
A completely parallel analysis would give FE2 = m2 gθ 2 . Since the electric forces are a Newton’s third
law pair, they can be set equal to each other in magnitude. Note that the explicit form of Coulomb’s
law was not necessary for this analysis.
FE1 = FE2
(b)
→ m1 gθ1 = m2 gθ 2
The horizontal distance from one sphere to the other, represented by d,
is found by the small-angle approximation. See the diagram. Use the
relationship derived above that FE = mgθ to solve for the distance.
d = A(θ1 + θ 2 ) = 2Aθ1 → θ1 =
m1 gθ1 = FE1 =
53.
→ θ1 /θ 2 = m2 /m1 = 1
kQ(2Q)
d
2
= mg
d
2A
d
2A
1/3
→
⎛ 4AkQ 2 ⎞
d =⎜
⎜ mg ⎟⎟
⎝
⎠
l θ1 θ2 l
l sin θ 1
l sin θ 2
Because of the inverse square nature of the electric
Q1
Q2
field, any location where the field is zero must be closer
to the weaker charge (Q2 ). Also, in between the two
l
d
charges, the fields due to the two charges are in the same direction and cannot cancel. Thus the only
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16-20
Chapter 16
place where the field can be zero is closer to the weaker charge, but not between them. In the diagram,
this means that A must be positive. Evaluate the net field at that location.
E = −k
Q1 −
Q2
Q1
(A + d ) 2
d=
=0 →
Q2 (A + d ) 2 = Q1A 2
5.0 × 10−6 C
2.5 × 10
−5
C − 5.0 × 10
−6
C
→
(2.4 m) =
1.9 m from Q2 ,
4.3 m from Q1
The electric field at the surface of the pea is given by Eq. 16–4a. Solve that equation for the charge.
E=k
55.
A
+k
2
Q2
A=
54.
Q2
Q
→ Q=
r2
Er 2 (3 × 106 N/C)(3.75 × 10−3 m) 2
=
= 5 × 10−9 C (≈ 3 × 109 electrons)
k
8.988 × 109 N ⋅ m 2 /C 2
There will be a rightward force on Q1 due to Q2 , given by Coulomb’s law. There will be a leftward
force on Q1 due to the electric field created by the parallel plates. Let right be the positive direction.
∑F = k
Q1Q2
x2
− Q1 E
= (8.988 × 109 N ⋅ m 2 /C2 )
(6.7 × 10−6 C)(1.8 × 10−6 C)
(0.47 m) 2
− (6.7 × 10−6 C)(5.3 × 104 N/C)
= 0.14 N, right
56.
The weight of the sphere is the density times the volume. The electric force is given by Eq. 16–1, with
both spheres having the same charge, and the separation distance equal to their diameter.
mg = k
Q=
57.
Q2
(d ) 2
→ ρ 34 π r 3 g =
kQ 2
(2r ) 2
→
16 ρπ gr 5
16(35 kg/m3 )π (9.80 m/s 2 )(1.0 × 10−2 m)5
=
= 8.0 × 10−9 C
3k
3(8.99 × 109 N ⋅ m 2 /C2 )
Since the electric field exerts a force on the charge in the
same direction as the electric field, the charge is positive.
Use the free-body diagram to write the equilibrium equations
for both the horizontal and vertical directions and use those
equations to find the magnitude of the charge.
θ
G
FT
G
mg
G
FE
43cm
θ
l = 55cm
43
= 38.6°
55
∑ Fx = FE − FT sin θ = 0 → FE = FT sin θ = QE
θ = cos −1
∑ Fy = FT cos θ − mg = 0
Q=
→ FT =
mg
cos θ
→ QE = mg tan θ
mg tan θ (1.0 × 10−3 kg)(9.80 m/s 2 ) tan 38.6°
=
= 8.2 × 10−7 C
E
(9500 N/C)
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Electric Charge and Electric Field
58.
(a)
The force of sphere B on sphere A is given by Coulomb’s law.
kQ 2
FAB =
(b)
R2
(a)
R2
=
kQ 2
2R2
, away from B
(3Q /4)(Q /2)
R
2
=
3kQ 2
8R 2
, away from B
Outside of a sphere, the electric field of a uniformly charged sphere is the same as if all the
charge were at the center of the sphere. Thus we can use Eq. 16–4a. According to Example 16–6,
each toner particle has the same charge as 20 electrons. Let N represent the number of toner
particles.
E=k
N=
(b)
QQ /2
The result of touching sphere A to sphere C is that the charge on the two spheres is shared, so the
charge on A is reduced to 3Q /4. Again use Coulomb’s law.
FAB = k
59.
, away from B
The result of touching sphere B to uncharged sphere C is that the charge on B is shared between
the two spheres, so the charge on B is reduced to Q /2. Again use Coulomb’s law.
FAB = k
(c)
16-21
Q
r
2
=k
NQ0
r2
→
Er 2
(5000N/C)(0.50 m) 2
=
= 4.346 × 1010 ≈ 4 × 1010 particles
−
9
2
2
19
kQ0 (8.988 × 10 N ⋅ m /C )(20)(1.6 × 10
C)
According to Example 16–6, each toner particle has a mass of 9.0 × 10−16 kg.
m = Nm0 = (4.346 × 1010 )(9.0 × 10−16 kg) = 3.911 × 10−5 ≈ 4 × 10−5 kg
60.
Since the gravity force is downward, the electric force must be upward. Since the charge is positive,
the electric field must also be upward. Equate the magnitudes of the two forces and solve for the
electric field.
Felectric = Fgravitational
61.
→ qE = mg
→ E=
mg (1.67 × 10−27 kg)(9.80 m/s 2 )
=
= 1.02 × 10−7 N/C, up
q
(1.602 × 10−19 C)
The weight of the mass is only about 2 N. Since the tension in the string is more than
that, there must be a downward electric force on the positive charge, which means that
the electric field must be pointed down. Use the free-body diagram to write an
expression for the magnitude of the electric field.
∑ F = FT − mg − FE = 0 → FE = QE = FT − mg
E=
→
G
FT
G
mg
G
FE
FT − mg 5.18 N − (0.185 kg)(9.80 m/s 2 )
=
= 9.90 × 106 N/C
Q
3.40 × 10−7 C
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16-22
62.
Chapter 16
(a)
Since the field is uniform, the electron will experience a constant force in the direction opposite to
its velocity, so the acceleration is constant and negative. Use Eq. 2–11c with a final velocity of 0.
F = ma = qE = −eE → a = −
Δx = −
(b)
υ 02
2a
=−
υ02
⎛ eE ⎞
2⎜ − ⎟
⎝ m⎠
=
eE 2
; υ = υ02 + 2aΔx = 0 →
m
mυ02 (9.11 × 10−31 kg)(5.32 × 106 m/s) 2
=
= 8.53 × 10−3 m
2eE 2(1.60 × 10−19 C)(9.45 × 103 N/C)
Find the elapsed time from constant-acceleration relationships. Upon returning to the original
position, the final velocity will be the opposite of the initial velocity. Use Eq. 2–11a.
υ = υ0 + at →
t=
63.
υ − υ0
a
=
2mυ0 2(9.11 × 10−31 kg)(5.32 × 106 m/s)
−2υ0
=
=
= 6.41 × 10−9 s
3
−19
eE
⎛ eE ⎞
×
×
(1.60
10
C)(9.45
10
N/C)
⎜− ⎟
⎝ m⎠
On the x axis, the electric field can only be zero at a location closer to the smaller magnitude charge.
The field will never be zero to the left of the midpoint between the two charges. In between the two
charges, the field due to both charges will point to the left. Thus the total field cannot be zero there. So
the only place on the x axis where the field can be zero is to the right of the negative charge, and x
must be positive. Calculate the field at point P and set it equal to zero.
E=k
(−Q /2)
x2
+k
Q
(x + d ) 2
= 0 → 2 x 2 = (x + d ) 2
→ x=
d
2 −1
=d
(
)
2 + 1 ≈ 2.41 d
The field cannot be zero at any points off the x axis. For any point off the x axis, the electric fields due
to the two charges will not be along the same line, so they can never combine to give 0.
64.
To find the number of electrons, convert the mass to moles and the moles to atoms and then multiply
by the number of electrons in an atom to find the total electrons. Then convert to charge.
⎛ 1 mole Al ⎞ ⎛ 6.02 × 1023 atoms ⎞ ⎛ 13 electrons ⎞ ⎛ −1.602 × 10−19 C ⎞
25 kg Al = (25 kg Al) ⎜
⎜ 2.7 × 10−2 kg ⎟⎟ ⎜⎜
⎟⎟ ⎜⎝ 1 molecule ⎟⎠ ⎜⎜
⎟⎟
1 mole
electron
⎝
⎠⎝
⎠
⎝
⎠
= −1.2 × 109 C
65.
The electric field will put a force of magnitude FE = QE on
each charge. The distance of each charge from the pivot point is
A /2, so the torque caused by each force is τ = FE r⊥ = QE
( )
1A
2
.
Both torques will tend to make the rod rotate counterclockwise
⎛ QE A ⎞
in the diagram. The net torque is τ net = 2 ⎜
⎟ = QE A , and it
⎝ 2 ⎠
is counterclockwise. However, if the charges were switched, the
torque would be clockwise.
G
FE
−Q
G
FE
+Q
G
E
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Electric Charge and Electric Field
66.
Enet = 2 E sin θ = 2
=
=
67.
+Q
From the diagram, we see that the x components
of the two fields will cancel each other at the
point P. Thus the net electric field will be in the
negative y direction and will be twice the y
component of either electric field vector.
2kQ
kQ
x2 + a2
a
a
x
sin θ
a
x + a 2 (x 2 + a 2 )1/2
2kQa
G
E− Q
θ
G
EQ
−Q
2
(x 2 + a 2 )3/2
16-23
in the negative y direction
One sphere will have a positive charge, and the other sphere will have the same amount of negative
charge. First solve for that charge by equating the electric force to the gravitational force. Then
compare that charge to the total charge. N represents the number of electrons moved from one sphere
to the other.
Felectric = Fgravitational
N=
→ k
(Ne) 2
( 0.12 kg )
m G
=
e k
1.602 × 10219 C
fraction =
(
r2
)
6.453 × 106 electrons
24
7.22 × 10 electrons
=G
m2
r2
6.67 × 10211
8.988 × 109
→
= 6.453 × 106 electrons
= 8.94 × 10219
Solutions to Search and Learn Problems
1.
(a)
(b)
(c)
(d)
When the leaves are charged by induction, no additional charge is added to the leaves. As the
charged rod is near the top of the electroscope, it repels the charge onto the leaves, causing them
to separate. When the rod is removed, the charge returns to its initial equilibrium position and the
leaves come back together.
When the leaves are charged by conduction, positive charge is placed onto the electroscope from
the rod, causing the leaves to separate. When the rod is removed, the charge remains on the
electroscope and the leaves remain separated.
Yes. The electroscope has a negative charge on the top sphere and on the leaves. Therefore, the
electroscope has a total net negative charge, so it must have been charged by conduction.
In Fig. 16–11a the electroscope is charged by induction, so the net charge is zero. Electric field lines
leaving the positive end of the electroscope end on the negative end. In Fig. 16–11b the electroscope
has a net positive charge. Electric field lines originate on the electroscope and point away from it.
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16-24
2.
Chapter 16
A negative charge must be placed at the center of the square. Let
Q = 6.4 μ C be the charge at each corner, let −q be the magnitude of
negative charge in the center, and let d = 9.2 cm be the side length of
the square. By the symmetry of the problem, if we make the net force
on one of the corner charges be zero, then the net force on each other
corner charge will also be zero. The force on the charge in the center
is also zero, from the symmetry of the problem.
F41 = k
F42 = k
F43 = k
F4q = k
Q2
→ F41x = k
d2
Q2
Q2
d2
→ F42 x = k
2d 2
Q2
qQ
d 2 /2
G
F4q
G
F41
−q
Q3
, F41 y = 0
Q2
2d 2
→ F4 qx = −k
d
Q2
cos 45° = k
→ F43 x = 0, F43 y = k
d2
Q1
G
F42
G
F43
Q4
2qQ
d2
2Q 2
4d 2
, F42 y = k
2Q 2
4d 2
Q2
d2
cos 45° = −k
2qQ
d2
= F4 qy
The net force in each direction should be zero.
Q2
∑ Fx = k d 2 + k
2Q 2
4d
2
+0−k
⎛ 1 1⎞
q = Q⎜
+ ⎟ = (6.4 × 10−6
⎝ 2 4⎠
2qQ
=0 →
d2
⎛ 1 1⎞
C) ⎜
+ ⎟ = 6.1 × 10−6 C
⎝ 2 4⎠
So the charge to be placed is −q = −6.1 × 10−6 C . Note that the length of the sides of the square does
not enter into the calculation.
This is an unstable equilibrium. If the center charge were slightly displaced, say toward the right, then
it would be closer to the right charges than the left and would be attracted more to the right. Likewise,
the positive charges on the right side of the square would be closer to it and would be attracted more to
it, moving from their corner positions. The system would not have a tendency to return to the
symmetric shape; rather, it would have a tendency to move away from it if disturbed.
3.
This is a constant-acceleration situation, similar to projectile motion in a uniform gravitational field.
Let the width of the plates be A, the vertical gap between the plates be h, and the initial velocity be
υ0 . Notice that the vertical motion has a maximum displacement of h/2. Let upward be the positive
vertical direction. We calculate the vertical acceleration produced by the electric field as the electron
crosses the region.
Fy = ma y = qE = −eE → a y = −
eE
m
Since the electron has constant horizontal velocity in the region, we use the horizontal component of
the velocity and the length of the region to solve for the time of flight across the region.
A = υ0 cos θ 0 (t ) → t =
A
υ0 cos θ0
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Electric Charge and Electric Field
16-25
In half this time, the electron will travel vertically from the middle of the region to the highest point.
At the highest point the vertical velocity must be zero. Using Eq. 2–11a with the known acceleration
and time to reach the highest point, we can solve for the square of the initial speed of the electron.
υ y = υ0 y + a y ttop
top
⎞
⎞
A
eE ⎛
A
⎛ eE ⎞ ⎛
2
→ 0 = υ0 sin θ 0 + ⎜ − ⎟ ⎜ 12
⎟ → υ0 =
⎜
⎟
2m ⎝ sin θ 0 cos θ 0 ⎠
⎝ m ⎠ ⎝ υ0 cos θ 0 ⎠
Finally, using Eq. 2–11a we solve for the initial angle.
ytop = y0 + υ0 y ttop + 12 a y t 2
h = A tan θ 0 −
eE A 2
→
1
2
1
2
⎛
A
h = υ0 sin θ 0 ⎜ 12
⎝ υ0 cos θ 0
= A tan θ 0 −
eE A 2
⎞ 1 eE ⎛ 1
⎞
A
⎟− 2 ⎜2
⎟
m ⎝ υ0 cos θ 0 ⎠
⎠
1
⎞
A
4m cos θ 0 υ
4m cos θ 0 eE ⎛
⎜
⎟
2m ⎝ sin θ 0 cos θ 0 ⎠
2h
2(1.0 cm)
h = 12 A tan θ 0 → θ 0 = tan −1
= tan −1
= 15.52° ≈ 16°
A
7.2 cm
2
2
0
2
2
→
= A tan θ 0 − 12 A tan θ 0
4.
Coulomb observed experimentally that the force between two charged objects is directly proportional
to the charge on each one. For example, if the charge on either object is tripled, then the force is
tripled. This is not in agreement with a force that is proportional to the sum of the charges instead of to
the product of the charges. Also, a charged object is not attracted to or repelled from a neutral
insulating object. But if the numerator in Coulomb’s law were proportional to the sum of the charges,
then there would be a force between a neutral object and a charged object, because their sum would not
be 0.
5.
Two forces act on the mass: gravity and the electric force. Since the acceleration is smaller than the
acceleration of gravity, the electric force must be upward, opposite the direction of the electric field.
The charge on the object is therefore negative. Using Newton’s second law, with down as the positive
direction, we can calculate the magnitude of the charge on the object.
G
G
∑ F = ma
q=
6.
→ mg + qE = ma →
m(a − g ) (1.0 kg)(8.0 m/s 2 − 9.8 m/s 2 )
=
= −1.2 × 10−2 C
150 N/C
E
The magnitude of the electric field at the third vertex is the vector sum of the
electric fields from each of the two charges. If we place the two charges at the
lower two vertices, then the electric field at the top vertex will be vertically
downward. By symmetry, the horizontal components of the electric field will
cancel, leaving only the vertical components for the net field.
E = Ey =
ke
A
2
sin 30° +
ke
A
2
sin 30° =
l
ke
A2
The electric force on a third negative charge would be upward (opposite the direction of the electric
field) and equal to the product of the electric field magnitude and the magnitude of the charge.
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16-26
Chapter 16
F = eE =
7.
ke2
A2
, upward
Set the Coulomb electrical force equal to the Newtonian gravitational force on the Moon.
FE = FG
Q=
→ k
Q2
2
rorbit
=G
M Moon M Earth
2
rorbit
→
GM Moon M Earth
(6.67 × 10−11 N ⋅ m 2 /kg 2 )(7.35 × 1022 kg)(5.98 × 1024 kg)
=
= 5.71 × 1013 C
9
2 2
k
(8.988 × 10 N ⋅ m /C )
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