# 16 Instructor Solutions Manual

## Full text

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### 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

LECTRIC

HARGE AND

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IELD

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+ + + + + + +

–+

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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

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15.

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. (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 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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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.

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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.

19 19

9 2 2 3

1 2

2 12 2

(1.602 10 C)(26 1.602 10 C)

(8.988 10 N m /C ) 2.7 10 N

(1.5 10 m)

Q Q F k

r

− −

− −

× × ×

= = × ⋅ = ×

×

2. Use the charge per electron to find the number of electrons.

( 48.0 10 6C) 1 electron19 2.996 10 electrons14 3.00 10 electrons14

1.602 10 C

⎛ ⎞

− × ⎜= × ≈ ×

− ×

⎝ ⎠

3. Use Coulomb’s law (Eq. 16–1) to calculate the magnitude of the force.

6 3

9 2 2 4 4

1 2

2 2

(25 10 C)(2.5 10 C)

(8.988 10 N m /C ) 2.194 10 N 2.2 10 N

(0.16 m)

Q Q F k

r

− −

× ×

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4. Use Coulomb’s law (Eq. 16–1) to calculate the magnitude of the force.

19 2

9 2 2

1 2

2 15 2

(1.602 10 C)

(8.988 10 N m /C ) 14.42 N 14 N

(4.0 10 m)

Q Q F k

r

− −

×

= = × ⋅ = ≈

×

5. The charge on the plastic comb is negative, so the comb has gained electrons and has a negative charge.

31 6

19

15 13

1 9.109 10 kg

( 3.0 10 C)

1.602 10 C 1

1.90 10 (1.9 10 )% 0.0090 kg

e

e m

m

− −

− −

− −

⎛ ⎞⎛ ×

− × ⎜ ⎟⎜⎟⎜

− ×

Δ = ⎝ ⎠⎝ ⎠= × ×

6. Since the magnitude of the force is inversely proportional to the square of the separation distance,F 12,

r

if the distance is multiplied by a factor of 1/8, then the force will be multiplied by a factor of 64. F=64F0=64(4.2 10× −2N) 2.688 N= ≈ 2.7 N

7. Since the magnitude of the force is inversely proportional to the square of the separation distance, 2

1 ,

F r

∝ if the force is tripled, then the distance has been reduced by a factor of 3.

0 6.52 cm 3.76 cm

3 3

r

r= = =

8. Use the charge per electron and the mass per electron.

6 14 14

19 31

14 16 16

1 electron

( 28 10 C) 1.748 10 1.7 10 electrons

1.602 10 C

9.109 10 kg

(1.748 10 ) 1.592 10 kg 1.6 10 kg

1

e

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.

23 19

8 8

1 mole Au 6.022 10 atoms 79 electrons 1.602 10 C

12 kg Au (12 kg Au)

0.197 kg 1 mole 1 atom electron

4.642 10 C 4.6 10 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.

1 2

9 2 2 19 2

2 39

E 1 2

11 2 2 31 27

1 2

G 1 2

2

(8.988 10 N m /C )(1.602 10 C) 2.27 10 (6.67 10 N m /kg )(9.11 10 kg)(1.67 10 kg)

Q Q k

F r kQ Q

m m

F G Gm m

r

− × ⋅ − × −

= = = = ×

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The electric force is about 2.27 10× 39 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,

9 2 2

8.988 10 N m /C .

k= × ⋅

65 2 2

48 2 2

95 2

(65 C)(48 C) (65 C)(95 C) 115.65 N 120 N (to the left)

(0.35 m) (0.70 m)

(48 C)(65 C) (48 C)(95 C) 563.49 N 560 N (to the right)

(0.35 m) (0.35 m)

(95 C)(65 C) (95 C)(48 C)

(0.70 m) (0.35

F k k

F k k

F k k

μ μ μ μ

μ μ μ μ

μ μ μ μ

+

+

= − + = − ≈ −

= + = ≈

= − − 2 447.84 N 450 N (to the left)

m) = − ≈ −

12. 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°.

2 2 2

12 2 12 2 12 2

2 2 2

13 2 13 2 13 2

2 2

1 12 13 1 12 13 2 2

cos 60 , sin 60

cos 60 , sin 60

0 2 sin 60 3

x y

x y

x x x y y y

Q Q Q

F k F k F k

d d d

Q Q Q

F k F k F k

d d d

Q Q

F F F F F F k k

d d

= → = ° = °

= → = − ° = °

= + = = + = ° =

2 6 2

2 2 9 2 2

1 1 1 2 2

2

(17.0 10 C)

3 3(8.988 10 N m /C ) 199.96 N (0.150 m)

2.00 10 N

x y Q

F F F k

d

×

= + = = × ⋅ =

≈ ×

The direction of F1 G

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 10 N× 2 and will point

30 below the° −x axis . Finally, the force on the lower right charge is of magnitude 2.00 10 N× 2 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.

2 2

1 2

2 2 9 2 2

7 12 12

19 1.7 0 N (0.24 m)

8.988 10 N m /C 1 electron

3.3007 10 C 2.06 10 2.1 10 electrons 1.602 10 C

Q Q Q F

F k k Q r

k

r r

×

= = → = =

× ⋅

⎛ ⎞

= × ⎜= × ≈ ×

×

⎝ ⎠

1 Q

2

Q Q3

12 FG

13 FG

### d

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14. 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.

2 2

41 Q2 41x Q2, 041y

F k F k F

d d

= → = =

2 2 2 2

42 2 42 2 2 42 2

2 2

43 2 43 43 2

2 2

cos 45 ,

2 2 4 4

0,

x y

x y

Q Q Q Q

F k F k k F k

d d d d

Q Q

F k F F k

d d

= → = ° = =

= → = =

Add the x and y components together to find the total force, noting that F4x=F4y.

2 2 2

4 41 42 43 2 2 2 4

2 2

2 2

4 4 4 2 2

3 2

9 2 2 7 7

2 4

1 4

abov

2 2

0 1

4 4

2 1

1 2 2

4 2

(6 15 10 C) 1

(8.988 10 N m /C ) 2 6.507 10 N 6.51 10 N

2 (0.100 m)

tan 45 e t

x x x x y

x y

y

x

Q Q Q

F F F F k k k F

d d d

Q Q

F F F k k

d d

F F θ

⎛ ⎞

= + + = + + = ⎜ + ⎟=

⎝ ⎠

⎛ ⎞

= + = ⎜ + ⎟ = ⎜ + ⎟

⎝ ⎠

. × ⎛ ⎞

= × ⋅ ⎜ + ⎟= × ≈ ×

⎝ ⎠

= = ° he directionx

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.

2 :Q

### )

2 2

2 (2 )2 (2 )(4 )2 cos 45 2 2 2 2 4.8284 2 2

Qx Q Q Q Q Q kQ

F =k +k ° =k + =

A A A A

### )

2 2

2 2 2 2 2

2

2

2 2 1 1

2 2 2 2 2

2

(2 )(3 ) (2 )(4 )sin 45 6 2 2 8.8284

2

8.8284

10.1 tan tan 61

4.8284

Qy

y

Q Qx Qy Q

x

Q Q Q Q Q kQ

F k k k

F kQ

F F F

F

θ − −

= + ° = + =

= + = = = = °

A A A A

A

41

FG

1

Q

2

Q Q3

d

4

Q

43

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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.

T T

T E E T

cos 0

cos

sin 0 sin sin tan

cos

y

x

mg

F F mg F

mg

F F F F F mg

θ

θ

θ θ θ θ θ

= − = → =

= − = → = = =

### ∑

2

E 2

3 2

6 6

9 2 2

( /2) tan 2 tan

(21 10 kg)(9.80 m/s ) tan 26

2(0.684 m) 4.572 10 C 4.6 10 C

(8 988 10 N m /C )

Q mg

F k mg Q

k

θ θ

− −

= = → =

× °

= = × ≈ ×

. × ⋅

A A

17. (a) If the force is repulsive, then both charges must be positive since the total charge is positive. Call the total charge Q.

2 2

1 2 1 1

1 2 kQ Q2 kQ Q Q( 2 ) 1 1 Fd 0

Q Q Q F Q QQ

k

d d

+ = = = → − + =

2 2 1

2

6 6 2

1

2 9 2 2

6 6

4 2

(12.0 N)(0.280 m) (90.0 10 C) (90.0 10 C) 4

(8.988 10 N m /C ) 88.8 10 C, 1.2 10 C

Fd

Q Q

k Q

− −

− −

± −

=

⎡ ⎤

⎢ ⎥

= × ± × −

⎢ × ⋅ ⎥

⎣ ⎦

= × ×

(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).

E

T

### G

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2 2 1

2

6 6 2

1

2 9 2 2

6 6

4 2

( 12.0 N)(0.280 m) (90.0 10 C) (90.0 10 C) 4

(8.988 10 N m /C ) 91.1 10 C, 1.1 10 C

Fd

Q Q

k Q

− −

− −

± −

=

⎢ ⎥

= × ± × −

×

⎣ ⎦

= × − ×

18. The negative charges will repel each other, so the third charge must put an opposite force on each of the original charges. Consideration of the various possible configurations leads to the conclusion that the third charge must be positive and must be between the other two 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.

### )

2 2

2 2 2 2

2 2

2 2

2 2 2 2

3 3 3

Left: Right:

( )

3

0.366 3 1

( )

3 3 3 0.402

3 1

Qq Q Qq Q

k k k k

x x

Qq Qq

k k x

x x

Qq Q x

k k q Q Q Q

x

= = →

= → = =

+ −

= → = = =

+

A A A

A

A A

A A

Thus the charge should be of magnitude 0.40Q, and a distance 0.37 fromA −Q toward 3 .− Q

19. Use Eq. 16–3 to calculate the force.

q ( 1.602 10 19 C)(2460 N/C east) 3.94 10 16 N west

q

− −

=F → = = − × = ×

E F E

G

G G G

20. Use Eq. 16–3 to calculate the electric field.

14

5 19

1.86 10 N south

1.16 10 N/C south

1.602 10 C

q

− −

×

= = = ×

× F

E

G G

21. Use Eq. 16–4a to calculate the electric field due to a point charge.

6

9 2 2 6

2 2

33.0 10 C

(8.988 10 N m /C ) 6.30 10 N/C upward

(0.217 m)

Q E k

r

×

= = × ⋅ = ×

Note that the electric field points away from the positive charge. 22. Use Eq. 16–3 to calculate the electric field.

6.4 N down6 8.8 10 N/C up5

7.3 10 C

q

= = = ×

− ×

F E

G G

Q

l

l

### Q

(11)

23. 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.

19

14 2

net | | (1.602 1031 C)756 N/C 1.33 10 m/s

(9.11 10 kg)

q

m q a E

m

− −

×

= = → = = = ×

×

FG aG EG

Since the charge is negative, the direction of the acceleration is opposite to the field .

24. 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.

1 2 1 2

1 2 2 2 2 2 2 1 2

1 2

9 2 2

6 6 8 8

2

4

( )

( /2) ( /2)

4(8.988 10 N m /C )

(8.0 10 C 5.8 10 C) 1.3782 10 N/C 1.4 10 N/C

(0.060 m)

Q Q Q Q k

E E E k k k k Q Q

r r

− −

= + = + = + = +

× ⋅

= × + × = × ≈ ×

A A A

The direction is toward the negative charge .

25.

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.

net

27 6 2

19

(1.673 10 kg)(2.4 10 )(9.80 m/s )

0.2456 N/C 0.25 N/C (1.602 10 C)

F ma qE ma E

q

= = →

× ×

= = = ≈

×

27. 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.

31

2 10

net (9.11 10 19kg) (105 m/s north) 5.97 10 N/C south ( 1.602 10 C)

m m q

q

− −

×

= = → = = = ×

− ×

FG aG EG EG aG

### A

1 0

Q > EG1 Q2<0

2

A E2

(12)

28. 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.

2 2

10 net 2 2 2 82 (386 N/C)(0.160 m)9 2 2 1.37 10 C

8

( /2) 8(8.988 10 N m /C )

Q Q kQ E

E E k Q

k

= = = → = = = ×

× ⋅

A

A A

29. 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.

1 2 1 2 1

2 2 2 2 2 2 2

3 2 3 1 2

, 0

2 2

cos 45 ,

2 2 4 4

0,

x y

x y

x y

Q Q

E k E k E

Q Q Q Q

E k E k k E k

Q Q

E k E E k

= → = =

= → = ° = =

= → = =

A A

A A A A

A A

Add the x and y components together to find the total electric field, noting that Ex =Ey.

1 2 3 2 2 2

2 2

2 2

6

9 2 2 4

2 1

2 2

0 1

4 4

2 1

1 2 2

4 2

(3.25 10 C) 1

(8.988 10 N m /C ) 2 3.76 10 N/C 2

(1.22 m) tan 45.0 from the direction

x x x x y

x y

y

x

Q Q Q

E E E E k k k E

Q Q

E E E k k

E

x E

θ

⎛ ⎞

= + + = + + = ⎜ + ⎟=

⎝ ⎠

⎛ ⎞

= + = ⎜ + ⎟ = +

⎝ ⎠

⎝ ⎠

× ⎛ ⎞

= × ⋅ ⎜ + ⎟= ×

= = °

A A A

A A

30. The fields at the center due to the two −27.0 Cμ 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 / 2.A

Q

EG −Q

2

l

Q

EG Q

1

EG

3

EG

2

EG

1

Q

2

Q Q3

l

2

Q

1 38.6 C

Q = − μ

### l

2

Q 1

EG 2 EG

2 27.0 C

(13)

A EG

Q

+

l

l

l

Q

B EG

A

B

1 2 21 22 12 2

/2 /2 /2

Q Q Q Q

E=EE =kk =k

A A A

6

9 2 2

2 6

(38.6 27.0) 10 C (8.988 10 N m /C )

(0.425 m) /2 1.15 10 N/C, toward the 38.6 C chargeμ

− ×

= × ⋅

= × −

31. 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.

2 2 1 2 1 2 24 2 2

( ) ( ) ( ) ( ) ( )

Q Q kQxa

E k k kQ

x a x a x a x a 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.

### )

1 2

1 2 2 2

1

2 1

( )

32 C

(12 cm) 64.57 cm 65 cm

45 C 32 C

Q Q

k k

x x

Q x

Q Q

μ

μ μ

= → = →

+

= = = ≈

− −

EG EG

A

A

33. 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.

A 2 A A 2

B 2 B 2 2

B 2 2

A B 2 A B 2

0,

3

cos30 ,

2 sin 30

2 3

2 2

x y

x

y

x x x y y y

Q Q

E k E E k

Q Q Q

E k E k k

Q Q

E k k

Q Q

E E E k E E E k

= → = = −

= → = ° =

= ° =

= + = = + = −

A A

A A A

A A

A A

2 2 2 2 2 2

2 2

4 4 4 2

9 2 2 6

7 7

2

2

1 1 1

2

3 4

4 4 4

(8.988 10 N m /C )(26 10 C) 3.651 10 N/C 3.7 10 N/C (0.080 m)

1 2

tan tan tan 330

3 3

2 x y

y

x

k Q k Q k Q kQ

E E E

Q k E

E k Q

θ

− − −

= + = + = =

× ⋅ ×

= = × ≈ ×

= = = = °

A A A A

A

(14)

34. 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.

1 2 12 2 2 1 2 1

2

9 1

9

4 4

( /3) (2 /3)

Q Q Q Q

E E k k Q

Q

= → = → = → =

A A

35. In each case, find the vector sum of the field caused by the charge on the left (Eleft) G

and the field caused by the charge on the right (Eright).

G

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.

1

2 2

5.0

tan 26.6

10.0

(0.050 m) (0.100 m) 0.1118 m

d

θ== °

= + =

6

9 2 2 6

A 2 2sin 2(8.988 10 N m /C )4.7 10 C2sin 26.6 3.0 10 N/C A 90

(0.1118 m)

kQ E

d θ θ

×

= = × ⋅ ° = × = °

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.

1 1

left left

2 2

left

2 2

right

5.0 5.0

tan 45 tan 18.4

5.0 15.0

(0.050 m) (0.050 m) 0.0707 m

(0.050 m) (0.150 m) 0.1581 m

d

d

θ == ° θ == °

= + =

= + =

left right 2 left 2 right

left right

9 2 2 6 6

2 2

( ) ( ) cos cos

cos 45 cos 18 4

(8.988 10 N m /C )(4.7 10 C) 4.372 10 N/C

(0.0707 m) (0.1581 m)

x x x Q Q

E k k

d θ d θ

= + = −

° . °

= × ⋅ × ⎢ − ⎥= ×

⎢ ⎥

⎣ ⎦

EG EG

left right 2 left 2 right

left right

9 2 2 6 6

2 2

2 2 6 6

B

( ) ( ) sin sin

sin 45 sin 18.4

(8.988 10 N m /C )(4.7 10 C) 6.509 10 N/C

(0.0707 m) (0.1581 m)

7.841 10 N/C 7.8 10 N/C tan 56.11 5

y y y

y

x y B

x

Q Q

E k k

d d

E

E E E

E

θ θ

θ

−1

= + = +

° °

= × ⋅ × ⎢ + ⎥= ×

⎢ ⎥

⎣ ⎦

= + = × ≈ × = = ° ≈

EG EG

d

left EG

Q

+ +Q

left EG right

EG

d

right EG

left EG

left d

Q

+

Q

+

left

### θ

right

right

(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.

net

2 2

0

19 4

2 6

31

| | | |

| |

2 2 2

(1.602 10 C)(1.45 10 N/C)

2 (1.60 10 m) 9.03 10 m/s

(9.11 10 kg)

q E

F ma q E a

m

q E

a x a x x

m

υ υ υ

− −

= = → =

= + Δ → = Δ = Δ

× ×

= × = ×

×

(b) The value of the acceleration caused by the electric field is compared to g.

19 4

15 2

31

15 2 2 14

| | (1.602 10 C)(1.45 10 N/C)

2.55 10 m/s

(9.11 10 kg)

/ (2.55 10 m/s )/(9.80 m/s ) 2.60 10

q E a

m a g

− −

× ×

= = = ×

×

= × = ×

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:

2

OH (0 4 )(0 2 )10 2 0.08 10 2

(1.80 10 m) (1.80 10 m)

e e ke

F =k . . =

× ×

O–N repulsion:

2

ON (0 4 )(0 2 )10 2 0.08 10 2

(2.80 10 m) (2.80 10 m)

e e ke

F =k . . =

× ×

N–N repulsion:

2

NN (0 2 )(0 2 )10 2 0.04 10 2

(3.00 10 m) (3.00 10 m)

e e ke

F =k . . =

× ×

H–N attraction:

2

HN (0 2 )(0 2 )10 2 0.04 10 2

(2.00 10 m) (2.00 10 m)

e e ke

F =k . . =

× ×

2 2

A T OH ON NN HN 0.082 0.082 0.042 0.042 1 10 2

1.80 2.80 3.00 2.00 1.0 10 m

ke

F F F F F

d

⎛ ⎞

⎛ ⎞

= − − + = − − + ⎟⎜⎜ ⎟

×

⎠⎝

9 2 2 19 2

10 10

10 2

(8.988 10 N m /C )(1.602 10 C)

(0 02004) 4.623 10 N 5 10 N

(1.0 10 m)

− −

× ⋅ ×

= . = × ≈ ×

(16)

(b) The net force between the cytosine and guanine is due to the following forces.

O–H attraction:

2

OH (0.4 )(0.2 )10 2 0.08 10 2 (two of these)

(1.90 10 m) (1.90 10 m)

e e ke

F =k =

× ×

O–N repulsion:

2

ON (0.4 )(0.2 )10 2 0.08 10 2 (two of these)

(2.90 10 m) (2.90 10 m)

e e ke

F =k =

× ×

H–N attraction:

2

HN (0.2 )(0.2 )10 2 0.04 10 2

(2.00 10 m) (2.00 10 m)

e e ke

F =k =

× ×

N–N repulsion:

2

NN (0 2 )(0 2 )10 2 0.04 10 2

(3.00 10 m) (3.00 10 m)

e e ke

F =k . . =

× ×

2 2

C G OH ON NN HN 2 2 2 2 10 2

9 2 2 19 2

10 10

10 2

0.08 0.08 0.04 0.04 1

2 2 2 2

1.90 2.90 3.00 2.00 1.0 10 m

(8.988 10 N m /C )(1.602 10 C)

(0.03085) 7.116 10 N 7 10 N

(1.0 10 m)

ke

F F F F F

d

− −

⎛ ⎞

⎛ ⎞

= − − + = − − + ⎟⎜⎜ ⎟

×

⎠⎝

× ⋅ ×

= = × ≈ ×

×

(c) For the 10 pairs of molecules, we assume that half are A–T pairs and half are C–G pairs. We 5 average the above results and multiply by 10 .5

5 5 10 10

1

net 2 A T C G

5 5

10 ( ) 10 (4.623 10 N 7.116 10 N)

5.850 10 N 6 10 N

F F F − −

− −

= + = × + ×

= × ≈ ×

38. Use Gauss’s law to determine the enclosed charge. Note that the size of the box does not enter into the calculation.

encl encl E 0 2 12 2 2 8

0

(1850 N m /C)(8.85 10 C /N m ) 1.64 10 C

E Qε Q ε − −

Φ = → = Φ = ⋅ × ⋅ = ×

39. (a) Use Gauss’s law to determine the electric flux.

6

5 2

encl

12 2 2

0

1.0 10 C

1.1 10 N m /C

8.85 10 C /N m

E Qε

− −

− ×

Φ = = = − × ⋅

× ⋅

(17)

40. (a) 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.

(b) 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.

2 3 2 2

left

2 2

left right other

(7.50 10 N/C)(8.50 10 m)

54.2 N m /C ; 54.2 N m /C ; 0

EA E

Φ = = = × ×

Φ = ⋅ Φ = − ⋅ Φ =

A

41. Equation 16–10 applies. Note that the separation distance does not enter into the calculation.

0 12 2 2 2 10

0

/ (8.85 10 C /N m )(130 N/C)(0.85 m) 8.3 10 C

Q A

E Q ε EA

ε − −

= → = = × ⋅ = ×

42. The electric field can be calculated by Eq. 16–4a, and that can be solved for the charge.

2 2 2 2

11

2 9 2 2

(3.75 10 N/C)(3.50 10 m) 5.11 10 C

8.988 10 N m /C

Q Er

E k Q

k r

× ×

= → = = = ×

× ⋅

This is about 3 10× 8electrons. 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.

(a) Inside the inner radius of the shell, the field is that of the point charge, E k Q2.

r

=

(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 Q2.

r

=

(d) The shell does not affect the field due to Q alone, except in the shell material, where the field is 0. 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.

2 electric gravitational 2

9 2 2

19

31 2

(8.988 10 N m /C )

(1.602 10 C) 5.08 m

(9.11 10 kg)(9.80 m/s ) e

F F k mg

r k

r e mg

= → = →

× ⋅

= = × =

×

45. 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.

23 19

9 2

6.02 10 H O molecules 10 protons 1.60 10 C

75 kg 4.0 10 C

0.018 kg 1 molecule proton

× ×

= ×

⎜ ⎟ ⎟⎜⎜ ⎟

⎟⎝ ⎠⎝

⎝ ⎠

(18)

46. Calculate the total charge on all electrons in 3.0 g of copper and compare 32 Cμ to that value. The molecular weight of copper is 63.5 g, and its atomic number is 29.

23 19

5

1 mole 6.02 10 atoms 29 1.602 10 C

Total electron charge 3.0 g 1.32 10 C

63.5 g mole atoms 1

e

e

⎛ ⎞ ⎛ ⎞

⎛ ⎞ × ⎛ ⎞ ×

= ⎜ ⎟⎜⎜ ⎟⎜= ×

⎝ ⎠

⎝ ⎠⎝ ⎠ ⎝ ⎠

6

10 5

32 10 C

Fraction lost 2.4 10

1.32 10 C −

×

= = ×

×

47. 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.

2 6 2

5

2 9 2 2

(150 N/C)(6.38 10 m) 6.8 10 C

8.988 10 N m /C

Q Er

E k Q

k r

×

= → = = = ×

× ⋅

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.

E

19

13 2 13 2

31

(1.602 10 C)(150 N/C) 2.638 10 m/s 2.6 10 m/s , up

(9.11 10 kg)

F eE ma eE a

m

− −

= = →

×

= = = × ≈ ×

×

(b) 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.

E

19

10 2 10 2

27

(1.602 10 C)(150 N/C)

1.439 10 m/s 1.4 10 m/s , down

(1.67 10 kg)

F eE ma eE a

m

− −

= = →

×

= = = × ≈ ×

×

(c) For the electron:

13 2

12 2

2.638 10 m/s 2.7 10

9.80 m/s

a g

×

= ≈ ×

For the proton:

10 2

9 2

1.439 10 m/s 1.5 10

9.80 m/s

a g

×

= ≈ ×

49. 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.

3 4

E 3

3 5 3 3 3 2

6 7

19

| |

4 4 (1.8 10 m) (1.00 10 kg/m )(9.80 m/s ) 9.96 10 1.0 10 electrons

3 3(1.602 10 C)(150 N/C)

F q E mg neE r g

r g n

eE

π ρ

π ρ π −

= = → = →

× ×

= = = × ≈ ×

(19)

50. There are four forces to calculate. Call the rightward direction the positive direction. The value of k is

9 2 2

8.988 10 N m /C× ⋅ and the value of e is 1.602 10× −19C.

net CH CN OH ON 9 2 2 2 2 2

10 10

(0 40 )(0 20 ) 1 1 1 1

(1 10 m) (0 30) (0 40) (0 18) (0 28)

2.445 10 N 2.4 10 N

k e e

F F F F F

− −

⎡ ⎤

. .

= + + + = ⎢− + + − ⎥

× ⎢⎣ . . . . ⎥⎦

= × ≈ ×

51. The electric force must be a radial force in order for the electron to move in a circular orbit.

2 2

orbit orbit

2 19 2

9 2 2 11

orbit 2 (8.988 10 N m /C ) (1.602 1031 C) 6 2 5.2 10 m

(9.109 10 kg)(2.2 10 m/s)

Q m

F F k

r r Q

r k

m

υ

υ

− −

= → = →

×

= = × ⋅ = ×

× ×

52. If all of the angles to the vertical (in both cases) are assumed to be small, then the spheres only have horizontal

displacement, and the electric force of repulsion is always horizontal. Likewise, the small-angle condition leads to

tanθ≈sinθ θ≈ for all small angles. See the free-body diagram for each sphere, showing the three forces of gravity, 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 =FT1sinθ1−FE1=0 → FE1=FT1sinθ1

1 1

1 T1 1 1 T1 E1 1 1 1 1 1

1 1

cos sin tan

cos cos

y m g m g

F F θ m g F F θ m g θ m gθ

θ θ

= − → = → = = =

### ∑

A completely parallel analysis would give FE2 =m g2 θ2. 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 → m g1 θ1=m g2 θ2 → θ θ1/ 2=m m2/ 1=1

(b) 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.

( 1 2) 2 1 1

2

d d =A θ θ+ = Aθ → θ =

A

1/3 2

1 1 E1 (2 )2 4

2

kQ Q d kQ

m g F mg d

mg d

θ = = = → =⎛⎜ ⎞⎟

⎝ ⎠

A A

53. Because of the inverse square nature of the electric field, any location where the field is zero must be closer to the weaker charge (Q2). Also, in between the two

charges, the fields due to the two charges are in the same direction and cannot cancel. Thus the only 1

Q

2

1

mgG E1

T1

1

θ

2

mg

E2

T2

### G

2

θ

1 sinθ

l

l 1

θ θ2 l

2

sinθ

(20)

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.

2 1 2 2

2 1

2 ( )2 0 ( )

Q Q

E k k Q d Q

d

= − + = → + = →

+ A A

A A

6

2 2

5 6 1

1 2

1.9 m from ,

5.0 10 C

(2.4 m)

4.3 m from

2.5 10 C 5.0 10 C

Q Q d Q Q Q − − − × = = = − × − × A

54. The electric field at the surface of the pea is given by Eq. 16–4a. Solve that equation for the charge.

2 6 3 2

9 9

2 9 2 2

(3 10 N/C)(3.75 10 m)

5 10 C ( 3 10 electrons)

8.988 10 N m /C

Q Er

E k Q

k r − − × × = → = = = × ≈ × × ⋅

55. 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.

1 2 1 2

6 6

9 2 2 6 4

2

(6.7 10 C)(1.8 10 C)

(8.988 10 N m /C ) (6.7 10 C)(5.3 10 N/C) (0.47 m)

0.14 N, right Q Q

F k Q E

x − − − = − × × = × ⋅ − × × =

### ∑

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.

2 2 3 4 3 2 2

5 3 2 2 5

9

9 2 2

( ) (2 )

16 16(35 kg/m ) (9.80 m/s )(1.0 10 m) 8.0 10 C

3 3(8.99 10 N m /C )

Q kQ

mg k r g

d r

gr Q

k

ρ π

ρπ π −

= → = →

×

= = = ×

× ⋅

57. 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.

1

E T E T

T T 3 2 7 43 cos 38.6 55

sin 0 sin

cos 0 tan

cos

tan (1.0 10 kg)(9.80 m/s ) tan 38.6

8.2 10 C (9500 N/C)

x

y

F F F F F QE

mg

F F mg F QE mg

mg Q

E

θ

θ θ

θ θ θ

θ − − − = = ° = − = → = = = − = → = → = × ° = = = ×

### ∑

55cm = l E FG T FG θ

mgG

(21)

58. (a) The force of sphere B on sphere A is given by Coulomb’s law.

2

AB kQ2 , away from B

F R =

(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 /2.Q Again use Coulomb’s law.

2

AB 2/2 2, away from B 2

QQ kQ

F k

R R

= =

(c) 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 3 /4.Q Again use Coulomb’s law.

2

AB (3 /4)( /2)2 3 2 , away from B 8

Q Q kQ

F k

R R

= =

59. (a) 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.

0

2 2

2 2

10 10

9 2 2 19

0

(5000N/C)(0.50 m)

4.346 10 4 10 particles (8.988 10 N m /C )(20)(1.6 10 C)

NQ Q

E k k

r r

Er N

kQ

= = →

= = = × ≈ ×

× ⋅ ×

(b) According to Example 16–6, each toner particle has a mass of 9.0 10× −16 kg.

m Nm= 0=(4.346 10 )(9.0 10× 10 × −16kg) 3.911 10= × −5≈ ×4 10−5kg

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.

27 2

7

electric gravitational (1.67 10 kg)(9.80 m/s )19 1.02 10 N/C, up

(1.602 10 C)

mg

F F qE mg E

q

− −

×

= → = → = = = ×

×

61. 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.

T E E T

2

6 T

7 0

5.18 N (0.185 kg)(9 80 m/s ) 9.90 10 N/C 3.40 10 C

F F mg F F QE F mg

F mg E

Q

= − − = → = = − →

− − .

= = = ×

×

E

### G

T

(22)

62. (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.

2 2

0

2 2 2 31 6 2

3

0 0 0

19 3

; 2 0

(9.11 10 kg)(5.32 10 m/s)

8.53 10 m 2 2 2 2(1.60 10 C)(9.45 10 N/C)

eE

F ma qE eE a a x

m m x

eE

a eE

m

υ υ

υ υ υ −

− −

= = = − → = − = + Δ = →

× ×

Δ = − = − = = = ×

× ×

⎜ ⎟

⎝ ⎠

(b) 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.

0

31 6

9

0 0 0

19 3

2 2 2(9.11 10 kg)(5.32 10 m/s) 6.41 10 s (1.60 10 C)(9.45 10 N/C)

at

m t

eE

a eE

m υ υ

υ υ υ υ −

− −

= + →

− − × ×

= = = = = ×

× ×

⎜ ⎟

⎝ ⎠

63. 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.

( 2/2) 2 0 2 2 ( )2

2 1

### )

2.41

2 1 ( )

Q Q d

E k k x x d x d d

x x 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.

23 19

2 9

1 mole Al 6.02 10 atoms 13 electrons 1.602 10 C 25 kg Al (25 kg Al)

1 mole 1 molecule electron 2.7 10 kg

1.2 10 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

/2,

A so the torque caused by each force is

### ( )

1

E 2 .

F r QE

τ= = A

Both torques will tend to make the rod rotate counterclockwise in the diagram. The net torque is net 2 ,

2

QE QE

τ = ⎛=

⎝ ⎠

A

A and it is counterclockwise. However, if the charges were switched, the torque would be clockwise.

Q

E

E

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