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22-1

Conceptual Questions

22.1. An insulator can be charged. Plastic is an insulator. A plastic rod can be charged by rubbing it with wool.

22.2. A conductor can be charged. A conductor can be charged by touching it with another charged object.

22.3. B and D are both neutral because they have no effect on each other and neutral is still attracted to either glass or plastic. Since ball A has been touched by plastic it is also now plastic. Since ball C is attracted to plastic (A) and neutral (B) then it must be glass.

22.4. (a) Like charges exert repulsive forces on each other, so the object must also have “plastic” charge. Therefore, it will attract the glass rod, which has the opposite charge (i.e., “glass” charge). (b) You cannot predict this because the object could be glass or neutral. Glass will repel the glass rod but neutral will be attracted to the glass rod.

22.5. Upon touching the charged rod, the metal exchanges charge with the area of the rod touched by the sphere (we are assuming the rod is an insulator). Some of the local excess charge on the rod will spread over the conducting sphere, so that both the rod and the sphere will have an overall excess charge of the same type. Thus, the sphere and the rod will repel each other.

22.6. No, it doesn’t mean the wall is charged. The balloon can stick to the wall by polarizing the wall so that the part closest to the balloon attracts and is attracted by it. The overall charge of the wall is neutral.

22.7. Assume that the basic premise of “like charges repel, unlike charges attract” still holds. Suspend an object with an excess of unknown charge from a string. First, approach a plastic-charged rod, then a glass-charged rod. An object with charge X must be attracted by both of these. However, this is not sufficient because a neutral object would also be attracted by both rods. To determine if the object is neutral or not, approach a neutral object. If the object has charge X, it will be attracted; if the object is neutral, nothing will happen.

22.8.

The final state of each sphere and of the rod is neutral. The conducting rod allows the excess electrons in the negatively charged sphere to move to the positively charged sphere and exactly neutralize the charge there, leaving all three conductors neutral.

ELECTRIC CHARGES AND FORCES

22M

22-2 Chapter 22


22.9. Each sphere ends up with one unit of negative charge. Once they touch, the two spheres become essentially one conductor. The overall net charge is −4 + 2 = −2. Charge is spread uniformly over the surface of a conductor.

22.10. (a)

The negatively charged rod will repel the negative charges on the top of the electroscope, pushing more negative charge down onto the leaves. The leaves will separate more. (b)

The positively charged rod will attract more negative charges to the top of the electroscope. As they depart from the leaves, the leaves will move closer together.

22.11.

The rod will polarize the charges in the combined conductor A + B, attracting negative charges to A and leaving B with excess positive charge. The combined conductor A + B is still neutral, but A alone has net negative charge.

22.12.

Your finger becomes polarized. Positive charge is left on the tip of your finger when negative charges in the finger are repelled by the ball. The excess positive charge in your finger is then nearer to the negatively charged ball than the negative charge in your finger, resulting in a net attractive force that attracts the ball.

Electric Charges and Forces 22-3


22.13.

22.14. (a) The magnitude of the force on A quadruples (increases by a factor of 4), since the force between the charges is proportional to the product of the magnitudes of the charges. Therefore, A 4 .F F′ = (b) By Newton’s third law, the force of A on B is equal in magnitude to the force of B on A; therefore the force on B also quadruples; B A 4 .F F F′ ′= =

22.15. Since the force on a charge in an electric field has magnitude F = qE, the new force is

3 3(3 ) .2 2 2EF q qE F⎛ ⎞′ = = =⎜ ⎟

⎝ ⎠

Exercises and Problems

Exercises Section 22.1 The Charge Model Section 22.2 Charge 22.1. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred to the other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei of atoms and so are essentially not free to move. Thus, electrons have been removed from the glass rod to make it positively charged. (b) Because each electron has a charge of 191.60 10 C,−− × the number of electrons removed is

910

198 0 10 C 5.0 10

1 60 10 C

−

−− . × = ×

− . ×

where the numerator is negative because this is the charge that is removed, so the excess charge left behind is 98.0 10 C.−×

22.2. Model: Use the charge model. Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred into the other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei of atoms and so are essentially not free to move. Thus, electrons have been added to the plastic rod to make it negatively charged. (b) Because each electron has a charge of 191.60 10 C,−− × the number of electrons added is

910

1912 10 C 7.5 10

1.60 10 C

−

−− × = ×

− ×

22.3. Model: Use the charge model and the model of a conductor as material through which electrons move. Solve: (a) The charge of a plastic rod changes from −15 nC to −10 nC. That is, −5 nC charge has been removed from the plastic. Because it is the negatively charged electrons that are transferred, −5 nC has been added to the metal sphere. (b) Because each electron has a charge of 191.60 10 C−− × and a charge of −5.0 nC was transferred, the number of electrons transferred from the plastic rod to the metal sphere is

910

195.0 10 C 3.1 10

1.60 10 C

−

−− × = ×

− ×

22-4 Chapter 22


22.4. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve: (a) The charge of the glass rod decreases from +12 nC to +8.0 nC. Because it is the electrons that are transferred, −4.0 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metal sphere and added to the glass rod. (b) Because each electron has a charge of 191.60 10 C−− × and a charge of −4.0 nC was transferred, number of electrons transferred from the metal sphere to the glass rod is

910

194.0 10 C 2.5 10

1.60 10 C

−

−− × = ×

− ×

22.5. Model: Use the charge model.

Solve: Since the density of water is 31.0 g/cm , the mass of 1.0 L of water is

3

31000 mL 1.0 cm 1.0 g(1.0 L)

L mL cm

⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

1.0 kg. Each water molecule 2(H O) has 10 protons (8 in the oxygen atom and one per hydrogen atom), and thus 10

electrons. The number of moles is 31.0 10 g 100 moles.

10 g/mole× = There are 236.02 10× water molecules in 1.0 mole of

water. Because one electron has a charge of 191.60 10 C,−− × the amount of charge in 100 mole of water is

23 19 72 2(100 mol)(6.022 10 H O/mol)(10 electron/H O)( 1.6 10 C/electron) 9.6 10 C−× − × = − ×

22.6. Model: Assume the aluminum is all of the isotope with 14 neutrons, so Al 27 u.m = Visualize: If N is the number of Al atoms, then (13 )Q e N= since each atom has 13 protons. Solve: N is the total mass divided by the mass of one aluminum atom.

Al Al27

8Al19

(13 )

(27 u)(1.0 C) 1 1.661 10 kg 2.16 10 kg(13 ) 13 1 u1.60 10 C

M MN Q em m

m Q eMe e

−−

−

= ⇒ = ⇒

⎛ ⎞×⎛ ⎞= = = ×⎜ ⎟⎜ ⎟⎜ ⎟×⎝ ⎠⎝ ⎠

Assess: 21 ng is a very small mass of aluminum; this illustrates how small atoms and protons are. The chunk of aluminum would likely be close to neutral because there would be an equal number of electrons.

22.7. Model: The total charge must be conserved. Visualize: Tally up the total charge before the reaction; it must be equal to the total charge after the reaction. Solve:

before 3( 2 ) 2( 3 ) 1( 2 ) 2 Q e e e e= + + − + − = −

If each molecular ion produced has a charge of magnitude e, then there must be 2 of them and they are negative.

22.8. Model: Assume all of the electrons go into the beam dump. Solve:

5 19 13

( 2.0 nC/s)(2.0 h)(3600 s/1 h)

( 1.44 10 C)(1 /1.6 10 C) 9.0 10

QQ tt

e e− −

⎛ ⎞= Δ = −⎜ ⎟Δ⎝ ⎠

= − × × = − ×

So 139.0 10× electrons have traveled down the accelerator. Assess: In a circular accelerator the particles can go around again and have another chance at the target.



Section 22.3 Insulators and Conductors 22.9. Model: Use the charge model and the model of a conductor as a material through which electrons move. Visualize:

The charge carriers in a metal electroscope are the negative electrons. As the positive rod is brought near, electrons are attracted toward it and move to the top of the electroscope. The electroscope leaves now have a net positive charge, due to the missing electrons, and thus repel each other. At this point, the electroscope as a whole is still neutral (no net charge) but has been polarized. On contact, some of the electrons move to the positive rod to neutralize some (but not necessarily all) of the rod’s positive charge. After contact, the electroscope does have a net positive charge. When the rod is removed, the net positive charge on the electroscope quickly covers the entire electroscope (note that no positive charges move, but the electrons distribute themselves over the surface so that there is a net positive charge everywhere on the surface). The net positive charge on the leaves causes them to continue to repel.

22.10. Model: Use the charge model and the model of a conductor as a material through which electrons move. Solve:

The first step shows two neutral metal spheres touching each other. In the second step, the negative rod repels the negative charges which will retreat as far as possible from the top of the left sphere. Note that the two spheres are touching and the net charge on these two spheres is still zero. While the rod is there on top of the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has an excess negative charge then, by charge conservation, the left sphere has the same magnitude of positive charge. Upon separation, the negative charge is trapped on the right sphere, as shown in the third step. As the two spheres are moved apart farther and the negatively charged rod is moved away from the spheres, the charges on the two spheres redistribute uniformly over the entire surface spheres. Thus, we are left with two oppositely charged spheres.


Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through the following six steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutral sphere with the negatively charged

22-6 Chapter 22


rod, so that the rod-sphere system has a net negative charge. (iii) Move the rod away from the sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere close to the second neutral sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly over the two spheres. (vi) Separate the spheres. The excess charge will have the same sign as the charge on the charging rod and will be evenly distributed between the two spheres.


Charging two neutral spheres with opposite charges of equal magnitude can be done through the following four steps. (i) Touch the two neutral metal spheres together. (ii) Bring a charged rod (say, positive) close (but not touching) to one of the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on the pair is zero. The right sphere has an excess positive charge of exactly the same magnitude as the left sphere’s negative charge. (iii) Separate the spheres while the charged rod remains close to the left sphere, so the separated charge remains on the spheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute uniformly over the metal sphere surfaces.

Section 22.4 Coulomb’s Law 22.13. Model: Model the charged masses as point charges. Visualize:

Solve: (a) The charge 1q exerts a force 1 on 2F on 2q to the right, and the charge 2q exerts a force 2 on 1F on 1q to the left. Using Coulomb’s law,

9 2 2 6 61 1

1 on 2 2 on 1 2 212

(9.0 10 N m /C )(10 10 C)(10 10 C) 0.90 N(1.0 m)

K q qF F

r

− −× × ×= = = =

(b) Applying Newton’s second law on either 1q or 2q gives

21 on 2 1 1 1

0.90 N 0.90 m/s1.0 kg

F m a a= ⇒ = =

Assess: Even a micro-Coulomb is a lot of charge. That is why 1 on 2 2 on 1(or )F F is a measurable force.

22.14. Model: Model the plastic spheres as point charges. Visualize:



Solve: (a) The charge 1 50.0 nCq = − exerts a force 1 on 2F on 2 50.0 nCq = − to the right, and the charge 2q exerts

a force 2 on 1F on 1q to the left. Using Coulomb’s law,

9 2 2 9 91 2

1 on 2 2 on 1 2 2 212

(9.0 10 N m /C )(50.0 10 C)(50.0 10 C) 0.056 N(2.0 10 m)

K q qF F

r

− −

−× × ×= = = =

×

(b) The ratio of the electric force to the weight is

1 on 23 2

0.056 N 2.9(2.0 10 kg)(9.8 m/s )

Fmg −= =

×

22.15. Model: Model the glass bead and the ball bearing as point charges. Visualize:

The ball bearing experiences a downward electric force 1 on 2F . By Newton’s third law, 2 on 1 1 on 2.F F= Solve: Using Coulomb’s law,

9 2 2 91 2 2 8

1 on 2 22 2 212

(9.0 10 N m /C )(20 10 C)0.018 N 1.0 10 C

(1.0 10 m)

q q qF K q

r

−−

−× ×

= ⇒ = ⇒ = ××

Because the force 1 on 2F is attractive and 1q is a positive charge, the charge 2q is a negative charge. Thus, 8

2 1.0 10 C 10 nC.q −= − × = −

22.16. Model: The protons are point charges. Solve: (a) The electric force between the protons is

9 2 2 19 191 2

E 2 15 2(9.0 10 N m /C )(1.60 10 C)(1.60 10 C) 58 N

(2.0 10 m)

q qF K

r

− −

−× × ×= = =

×

(b) The gravitational force between the protons is 11 2 2 27 27

351 2G 2 15 2

(6.67 10 N m /kg )(1.67 10 kg)(1.67 10 kg) 4.7 10 N(2.0 10 m)

G m mFr

− −−

−× × ×= = = ×

×

2

(c) The ratio of the electric force to the gravitational force is

36E35

G

58 N 1.2 104.7 10 N

FF −= = ×

×

22.17. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure EX22.17. Charge A experiences an electric force B on AF due to charge B and an

electric force C on AF due to charge C. The force B on AF is directed to the right and the force C on AF is directed to the left.

22-8 Chapter 22


Solve: Coulomb’s law yields: 9 2 2 9 9

A B 5B on A 2 2 2

(9.0 10 N m /C )(1.0 10 C)(1.0 10 C) 9.0 10 N(1.0 10 m)

q qF K

r

− −−

−× × ×= = = ×

×

9 2 2 9 9C A 5

C on A 2 2 2(9.0 10 N m /C )(1.0 10 C)(4.0 10 C) 9.0 10 N

(2.0 10 m)

q qF K

r

− −−

−× × ×= = = ×

×

The net force on A is 5 5

on A B on A C on Aˆ ˆ(9.0 10 N) (9.0 10 N)( ) 0.0 NF F F i i− −= + = × + × − =

22.18. Model: Charges A, B, and C are point charges. Solve: The force on B from charge A is upward downward since upper positive charge attracts the negative charge. Coulomb’s law gives the magnitude of the force as

9 2 2 9 9A B 5

A on B 2 2 2(9.0 10 Nm /C )(1.0 10 C)(2.0 10 C) 4.5 10 N

(2.0 10 m )

q qF K

r

− −−

−× × ×= = = ×

×

So 5A on B

ˆ(4.5 10 ) NF j−= × The force on B from charge C is directed downward since the two opposite charges attract. Coulomb’s law gives the magnitude of the force as

9 2 2 9 9C B 4

C on B 2 2 2(9.0 10 Nm /C )(2.0 10 C)(2.0 10 C) 3.6 10 N

(1.0 10 m)

q qF K

r

− −−

−× × ×= = = ×

×

So 4C on B

ˆ( 3.6 10 ) NF j−= − × The net electric force on charge A is

4A B on A C on A

4

ˆ ˆ(4.5 10 ) N+( 3.6 10 ) Nˆ(3.2 10 ) N

F F F j j

j

−5

−

= + = × − ×

= − ×

22.19. Model: The charges are point charges. Visualize: Please refer to Figure EX22.19. Solve: The electric force on charge 1q is the vector sum of the forces 2 on 1F and 3 on 1,F where 1q is the 1.0 nC charge, 2q is the 2.0 nC charge, and 3q is the other 2.0 nC charge. We have

1 22 on 1 22

9 2 2 9 9

22 2

4 42

, away from

(9.0 10 N m /C )(1.0 10 C)(2.0 10 C) , away from (1.0 10 m)

ˆ ˆ(1.8 10 N, away from ) (1.8 10 N)[cos(60 ) sin(60 ) ]

K q qF q

r

q

q i j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = × ° + °

1 3 4 43 on 1 3 3

2

4 4on 1 2 on 1 3 on 1

ˆ ˆ, away from (1.8 10 N, away from ) (1.8 10 N)[ cos(60 ) sin(60 ) ]

ˆ ˆ2(1.8 10 N)sin(60 ) (3.1 10 ) N

K q qF q q i j

r

F F F j j

− −

− −

⎛ ⎞= = × = × − ° + °⎜ ⎟⎜ ⎟⎝ ⎠

= + = × ° = ×

The force on the 1.0 nC charge is 43.1 10 N−× directed upward.

22.20. Model: The charges are point charges. Visualize: Please refer to Figure EX22.20. Solve: The electric force on charge 1q is the vector sum of the forces 2 on 1F and 3 on 1,F where 1q is the 1.0 nC charge, 2q is the 2.0 nC charge, and 3q is the −2.0 nC charges. We have



1 22 on 1 22

9 2 2 19 19

22 2

4 42

, away from

(9.0 10 N m /C )(1.0 10 C)(2.0 10 C) , away from (1.0 10 m)

ˆ ˆ(1.80 10 N, away from ) (1.80 10 N)[cos(60 ) sin(60 ) ]

K q qF q

r

q

q i j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = × ° + °

1 2 4 43 on 1 3 32

4 4on 1 2 on 1 3 on 1

ˆ ˆ, toward (1.80 10 N, toward ) (1.80 10 N)[cos(60 ) sin(60 ) ]

ˆ ˆ2(1.80 10 N)cos(60 ) 1.8 10 N

K q qF q q i j

r

F F F i i

− −

− −

⎛ ⎞= = × = × ° − °⎜ ⎟⎜ ⎟⎝ ⎠

= + = × ° = ×

So, the force on the 1.0 nC charge is 41.8 10 N−× and it is directed to the right.

22.21. Model: Objects A and B are point charges. Visualize:

Because there are only two charges A and B, the force on charge A is due to charge B only, and the force on B is due to charge A only. Solve: Coulomb’s law gives the magnitude of the forces between the charge:

9 2 2 9 94

A on B B on A 2 2(9.0 10 N m /C )(8.0 10 C)(4.0 10 C) 7.2 10 N

(2.0 10 m)F F

− −−

−× × ×= = = ×

×

Because the charge on object A is positive and on object B is negative, B on AF is upward and A on BF is downward. Thus, 4 4

B on A A on Bˆ ˆ(7.2 10 N) and (7.2 10 N)F j F j− −= + × = − ×

Assess: By Newton’s third law, the two forces have equal magnitudes but opposite directions because they form an action-reaction pair, just as we found.

22.22. Model: Assume the plastic bead, the proton, and the electron are point charges. Visualize:

Solve: Coulomb’s law gives

9 2 2 9 1913

bead on electron bead on proton 2 2(9.0 10 N m /C )(15 10 C)(1.60 10 C) 2.16 10 N

(1.0 10 m)F F

− −−

−× × ×= = = ×

×

22-10 Chapter 22


(a) Because the bead is much more massive than both the electron and the proton, we can ignore any acceleration of the bead. Newton’s second law is F = ma, so

13bead on proton 14 2

proton 27proton

2.16 10 N 1.3 10 m/s1.67 10 kg

Fa

m

−

−×= = = ××

Because opposite charges attract, 14 2

proton (1.3 10 m/s , toward bead)a = ×

(b) Similarly, 13

17 2bead on electronelectron 31

electron

2.16 10 N 2.4 10 m/s9.11 10 kg

Fam

−

−×= = = ××

Thus 17 2electron (2.4 10 m/s , away from bead).a = ×

Assess: Although the force on the proton has the same magnitude as the force on the electron, the electron has a much greater acceleration because it has a much smaller mass.

22.23. Model: The beads are point charges. Visualize:

Solve: The beads are oppositely charged so are attracted to one another. The force on each is the same by Newton’s third law. Coulomb’s law give the force as

9 91 2 9 2 2 4

2 2 2(4.0 10 C)(8.0 10 C)(9.0 10 Nm /C ) 7.2 10 N

(2.0 10 m)

q qF K

r

− −− −

−× ×= = × = ×

×

The beads accelerate at different rates because their masses are different. By Newton’s second law, the acceleration of the plastic bead is

42

plastic 3(7.2 10 N) 0.36 m/s toward glass bead.(2.0 10 kg)

Fam

−

−×= = =×

For the glass bead, the acceleration is 4

2glass 3

(7.2 10 N) 0.18 m/s toward plastic bead.(4.0 10 kg)

a−

−×= =×

22.24. Model: The charges are point charges. Visualize:



We must first identify the region of space where the third charge 3q is located. You can see from the figure that the forces can’t possibly add to zero if 3q is above or below the axis or outside the charges. However, at some point on the x-axis between the two charges the forces from the two charges will be oppositely directed. Solve: The mathematical problem is to find the position for which the forces 1 on 3F and 2 on 3F are equal in magnitude. If 3q is the distance x from 1,q it is the distance L − x from 2.q The magnitudes of the forces are

1 3 3 2 3 31 on 3 2 on 32 2 2 2

13 23

(4 )

( )

K q q Kq q K q q K q qF F

r x r L x= = = =

−

Equating the two forces gives

3 3 2 22 2

(4 )( ) 4 and

3( )

Kq q K q q LL x x x Lx L x

= ⇒ − = ⇒ = −−

The solution x = −L is not allowed as you can see from the figure. To find the magnitude of the charge 3,q we apply the equilibrium condition to charge 1:q

( )3 12 1

2 on 1 3 on 1 3 32 213

44 99

K q qK q qF F q q q q

L L= ⇒ = ⇒ = ⇒ =

We are now able to check the static equilibrium condition for the charge 4q (or 2 ) :q

( )4

3 21 2 91 on 2 3 on 2 2 2 2 2 22

3( )

qK q qq q q qF F KL L x L LL

= ⇒ = ⇒ = =−

The sign of the third charge 3q must be negative. A positive sign on 3q will not have a net force of zero either on

the charge q or the charge 4q. In summary, a charge of 49 q− placed 1

3x L= from the charge q will cause the

3-charge system to be in static equilibrium.

22.25. Model: The charges are point charges. Visualize: We’ll need Hooke’s law for the spring: ( ).F k x= − Δ

Solve: Accounting for both the spring force and the Coulomb force, the net force on the charge on the spring is zero.

1 2 1 2 9 2 2net 2 2 2

1 1 (2.0 C)(4.0 C)( ) (9.0 10 N m /C ) 8.9 kN/m0.012 m (0.026 m)

q q q qF k x K k K

xr rμ μ= − Δ + ⇒ = = × ⋅ =

Δ

Assess: This is a reasonable value for a spring constant of a stiff spring.

22-12 Chapter 22


Section 22.5 The Electric Field 22.26. Model: Protons and electrons produce electric fields. Solve: (a) The electric field of the proton is

199 2 2 3

2 3 21.60 10 Cˆ ˆ(9.0 10 N m /C ) (1.4 10 N/C, away from proton)

(1.0 10 m)qE K r rr

−−

−

⎡ ⎤+ ×= = × = ×⎢ ⎥×⎢ ⎥⎣ ⎦

(b) The electron carries the opposite charge, so the electric field of the electron is 3 3ˆ(1.4 10 N/C)( ) (1.4 10 N/C, toward electron)E r− −= × − = ×

22.27. Model: Model the proton and the electron as point charges. Solve: (a) The force that an electric field E exerts on a charge q is .F qE= A proton has q = e. Thus,

17proton

ˆ ˆ ˆ ˆ(400 100 ) N/C (6.4 1.6 ) 10 NF e i j i j −= + = + ×

where we used 191.602 10 C.e −= ×

(b) The charge on an electron is q = −e. Thus, 17electron proton

ˆ ˆ( 6.4 1.6 ) 10 NF F i j −= − = − − × (c) From Newton’s second law,

2 2 17proton 10 2

proton 27proton proton

6.605 10 N 4.0 10 m/s1.67 10 kg

x yF FFa

m m

−

−

+ ×= = = = ××

(d) The electron experiences a force of the same magnitude but it has a different mass. Thus, 17

13 2electron 31

6.605 10 N 7.3 10 m/s9.11 10 kg

a−

−×= = ×

×

The forces may be the same, but the electron has a much larger acceleration due to its much smaller mass. Assess: The two forces in parts (a) and (b) are equal in magnitude but opposite in direction.

22.28. Model: The electric field is that of a negative charge on the plastic bead. Model the small bead as a point charge. Solve: The electric field is

99 2 2 4

2 2 28.0 10 Cˆ ˆ ˆ(9.0 10 N m /C ) 4.5 10 N/C

(4.0 10 m)qE K r r rr

−

−− ×= = × = − ×

×

where r is the unit vector from the charge to the point at which we calculate the field. Assess: The direction of the electric field is toward the bead, which it should be since the bead is negative.

22.29. Model: The electric field is due to a charge and extends to all points in space. Solve: The magnitude of the electric field at a distance r from a charge q is

9 2 2 102 21.0 N/C (9.0 10 N m /C ) 1.1 10 C 0.11 nC

(1.0 m)

q qE K q

r−= ⇒ = × ⇒ = × =

22.30. Model: A field is the agent that exerts an electric force on a charge. Visualize:



Solve: Newton’s second law on the plastic ball is net on ( ) .y q GF F FΣ = − To balance the gravitational force with the electric force,

36

on q 9(1.0 10 kg)(9.8 N/kg) 3.3 10 N/C

3.0 10 CGmgF F q E mg Eq

−

−×= ⇒ = ⇒ = = = ×

×

Because on qF must be upward and the charge is negative, the electric field at the location of the plastic ball must be

pointing downward. Thus 6(3.3 10 N/C, downward).E = ×

Assess: F qE= means the sign of the charge q determines the direction of F or E. For positive q, E and F are

pointing in the same direction. But E and F point in opposite directions when q is negative.

22.31. Model: Treat the charge on the object is a point charge.

Solve: The electric field at a distance r from a point charge q is 2ˆ.qE K r

r= Because the electric field points away

from the object, ˆ(270,000 N/C) .E r= Thus,

2

2 28

9 2 2

ˆ ˆ270,000 N/C

(270,000 N/C)(2 0 10 m) 1 2 10 C 12 nC(9 0 10 N m /C )

qK r rr

q−

−

=

. ×= = . × =. ×

Assess: Since the field points away from it, we know that the charge must be positive.

22.32. Model: The electric field is that of a positive point charge located at the origin. Visualize:

The positions (5.0 cm, 0.0 cm), (−5.0 cm, 5.0 cm), and (−5.0 cm, −5.0 cm) are denoted by A, B, and C, respectively. Solve: (a) The electric field for a positive charge is

2 , away from qE K qr

⎛ ⎞= ⎜ ⎟⎝ ⎠

Using 9 2 2 99.0 10 N m /C and 12 10 C,K q −= × = ×

2

2108 N m /C , away from E q

r

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

22-14 Chapter 22


The electric fields at points A, B, and C are 2

4A 2 2

108 N m /C ˆ ˆ4.3 10 N/C(5.0 10 m)

E i i−= = ××

24 4

B 2 2 2 2108 N m /C 1 ˆ ˆ ˆ ˆ( ) ( 1.5 10 1.5 10 ) N/C

2( 5.0 10 m) (5.0 10 m)E i j i j− −

⎡ ⎤= − + = − × + ×⎢ ⎥− × + × ⎣ ⎦

24 4

C 2 2 2 2108 N m /C 1 ˆ ˆ ˆ ˆ( ) ( 1.5 10 1.5 10 ) N/C

2( 5.0 10 m) ( 5.0 10 m)E i j i j− −

⎡ ⎤= − − = − × − ×⎢ ⎥− × + − × ⎣ ⎦

(b) The three vectors are shown in the diagram. Assess: The vectors A ,E B,E and CE are pointing away from the positive charge.

22.33. Model: The electric field is that of a negative point charge located at (x, y) = (1.0 cm, 0 cm). Visualize:

Solve: The electric field for a positive charge is

2ˆqE K r

r=

Using 9 2 2 99.0 10 N m /C and 12 10 C,K q −= × = ×

2

2108 N m /C ˆE r

r=

The electric fields at points A, B, and C (see figure above) are 2

4A 2 2

108 N m /C ˆ ˆ6.8 10 N/C(4.0 10 m)

E i i−= = − ××

24

C 2 2108 N m /C ˆ ˆ3.0 10 N/C

(6.0 10 m)E i i−= = ×

×

We need to take components to find B:E

2 2 2

B 2 2 2 2 2 2 2 2 2 2 2 2

3 4

108 N m /C 1.0 10 m 5.0 10 mˆ ˆ(5.0 10 m) (1.0 10 m) (5.0 10 m) (1.0 10 m) (5.0 10 m) (1.0 10 m)

ˆ ˆ(8.1 10 4.1 10 ) N/C

E i j

i j

− −

− − − − − −

⎡ ⎤⎛ ⎞ ⎛ ⎞× ×⎢ ⎥⎜ ⎟ ⎜ ⎟= −⎢ ⎥⎜ ⎟ ⎜ ⎟× + × × + × × + ×⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

= × − ×Assess: Since the field points toward the negative charge, it should have a positive x-component and a negative y-component, as we have found.

22.34. Model: Model the bee as a point charge. Solve: (a) The force on the bee due to gravity is G ,F mg= and the electric force on the bee is e .F Eq= The ratio of the electric force to the bee’s weight is



126e

3 2G

(100 N/C)(23 10 C) 2.3 10(0.10 10 kg)(9.8 m/s )

F EqF mg

−−

−×= = = ×

×

(b) For the bee to be suspended by the electric force, this force must have the same magnitude as the force due to gravity and be directed upward. Equating the magnitudes of the forces and solving for E give

3 27

12(0.10 10 kg)(9.8 m/s ) 4.3 10 N/C

(23 10 C)mgEq mg Eq

−

−×= ⇒ = = = ×

×

The electric field must be directed upward so that the force on a positive charge is upward.

Problems

22.35. Model: Use the charge model. Solve: The number of moles in the penny is

3.1 g 0.04882 mol63.5 g/mol

MnA

= = =

The number of copper atoms in the penny is 23 1 22

A (0.04882 mol)(6.02 10 mol ) 2.939 10N nN −= = × = ×

Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is 22 19 5(29 2.939 10 )(1.60 10 C) 1.4 10 C−× × × = ×

Similarly, the total negative charge is 51.4 10 C.− × Assess: Total positive and negative charges are equal in magnitude.

22.36. Model: Treat the two charged spheres as point charges. Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s mass times its acceleration. Because the spheres are identical and equally charged, 1 2 1 2and .m m m q q q= = = = We have

21 2

2 on 1 1 on 2 2 2

2 3 2 2 22 15 2

9 2 2

8

(1.0 10 kg)(150 m/s )(2.0 10 m) 6.7 10 C9.0 10 N m /C

8.2 10 C 82 nC

Kq q KqF F mar r

marqK

q

− −−

−

= = = =

× ×= = = ××

= × =

22.37. Model: The 125 Xe nucleus and the proton will be treated as point charges. That is, all the charge on the Xe nucleus is assumed to be at its center. Visualize:

22-16 Chapter 22


Solve: (a) The magnitude of the force between the nucleus and the proton is given by Coulomb’s law:

9 2 2 19 19nucleus proton 2nucleus on proton 2 15 2

(9.0 10 N m /C )(54 1.60 10 C)(1.60 10 C) 5.0 10 N(5.0 10 m)

K q qF

r

− −

−× × × ×= = = ×

×

(b) Applying Newton’s second law to the proton, 2

29 2on proton proton proton proton 27

5.0 10 N 3.0 10 m/s1.67 10 kg

F m a a −×= ⇒ = = ×

×

22.38. Model: The charge on the sphere acts as if concentrated at the center. The ball is small enough that we neglect its radius. Visualize: We are given sphere 0.125 mR = and 13

sphere 1.0 10 .q e= × We seek ball .q

Solve: The distance we put in the denominator of Coulomb’s law is sphere 0.50 m 0.625 m.r R= + =

2 4 2sphere ballball2 9 2 2 13 19

sphere

(9.2 10 N)(0.625 m) 1 25 nC(9.0 10 N m /C )(1.0 10 ) 1.6 10 C

q q Fr eF K qr eK q

−

−× ⎛ ⎞= ⇒ = = =⎜ ⎟× ⋅ × ×⎝ ⎠

However, because the force is away from the sphere and the sphere is negatively charged the ball must be negative also, ball 25 nC.q = −

Assess: This is not too big nor too small for a small ball.

22.39. Model: The charger delivers charge to the phone at a rate of –0.75 C/s. Solve:

19 21

( 0.75 C/s)(30 min)(60 s/1 min)

( 1350 C)(1 /1.6 10 C) 8.4 10

QQ tt

e e−

⎛ ⎞= Δ = −⎜ ⎟Δ⎝ ⎠

= − × = − ×

So 218.4 10× electrons are delivered to the phone in 30 min. Assess: This is a very large number of electrons, but there are very many electrons in a coulomb.

22.40. Model: Objects A and B are point charges. Visualize:

Solve: (a) It is given that A on B 0.45 N.F = By Newton’s third law, B on A A on B 0.45 N.F F= = Coulomb’s law gives

1A AA B 2

B on A A on B 2 2

2 2 26

A 9 2 2

( )( )0.45 N

2(0.45 N) 2(0.45 N)(10 10 m) 1.0 10 C(9.0 10 N m /C )

K q qKq qF Fr r

rqK

−−

= = = =

×= = = ××

(b) Newton’s second law is B on A A A.F m a= Hence,

2B on A A on BA

A B

0.45 N 4.5 m/s0.100 kg

F Fam m

= = = =




Solve: The electric force on charge 1q is the vector sum of the forces 2 on 1F and 3 on 1.F We have

1 22 on 1 22

9 2 2 9 9

22 2

3 32

, away from

(9.0 10 N m /C )(10 10 C)(5.0 10 C) , toward (1.0 10 m)

ˆ(4.5 10 N, toward ) 4.5 10 N

K q qF q

r

q

q j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = ×

1 23 on 1 32

9 2 2 9 9

32 2 2 2

3 33

, toward

(9.0 10 N m /C )(10 10 C)(15 10 C) , away from (3.0 10 m) (1.0 10 m)

ˆ ˆ(1.35 10 N, away from ) (1.35 10 N)(cos sin )

K q qF q

r

q

q i jθ θ

− −

− −

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟× + ×⎝ ⎠

= × = × −

From the geometry of the figure,

1 1.0 cmtan 18.433.0 cm

θ − ⎛ ⎞−= = °⎜ ⎟⎝ ⎠

This means cos θ = 0.949 and sin θ = –0.316. Therefore, 3 3

3 on 1ˆ ˆ(1.28 10 0.43 10 ) NF i j− −= × − ×

3 3on 1 2 on 1 3 on 1

ˆ ˆ(1.28 10 4.07 10 ) NF F F i j− −= + = × + ×

The magnitude and direction of the resultant force vector are 3 2 3 2 3

on 1 1.28 10 N 4.07 10 N 4.3 10 NF − − −= ( × ) + ( × ) = × 3

13

4.07 10 Ntan 3.180 tan (3.180) 731.28 10 N

φ φ−

−−

×= = ⇒ = = °×

above the −x axis,

or on 1F points 73° counterclockwise from the +x-axis.


22-18 Chapter 22


Solve: The point charges are 1 210 nC, 8.0 nC,q q= − = + and 3 10 nC.q = The electric force on charge 1q is the

vector sum of the forces 2 on 1F and 3 on 1F . We have

1 22 on 1 22

9 2 2 9 9

22 2

3 32

, toward

9.0 10 N m /C (10 10 C)(8.0 10 C) , toward (1.0 10 m)

ˆ(7.2 10 N, toward ) (7.2 10 N)

K q qF q

r

q

q j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞( × ) × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = ×

1 33 on 1 32

9 2 2 9

32 2

3 33

, toward

(9.0 10 N m /C )(10 10 C)(10 10 C) , toward (3.0 10 m)

ˆ(1.0 10 N, toward ) 1.0 10 N

K q qF q

r

q

q i

−9 −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = −( × )

3 3on 1 2 on 1 3 on 1

ˆ ˆ( 1.0 10 7.2 10 ) NF F F i j− −= + = − × + ×

The magnitude and direction of the resultant force vector are 2 3 2 3

on 1 ( 1.0 10 N) (7.2 10 N) 7.3 10 NF −3 − −= − × + × = × 3

13

7.2 10 Ntan tan (7.2) 82 above the -axis,1.0 10 N

xφ φ−

−−

×= ⇒ = = ° −×

or on 1F points 98° counterclockwise from the +x-axis.




Solve: The electric force on charge 1q is the vector sum of the forces 2 on 1F and 3 on 1F . We have

1 22 on 1 22

9 2 2 9 9

22 2 2 2

4 42

, away from

(9.0 10 N m /C )(5 10 C)(10 10 C) , away from (4.0 10 m) (3.0 10 m)

ˆ ˆ(1.8 10 N, away from ) (1.8 10 N)( cos sin )

K q qF q

r

q

q i jθ θ

− −

− −

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟× + ×⎝ ⎠

= × = × − −

From the geometry of the figure,

42 on 1

4.0 cm ˆ ˆtan 53.13 (1.8 10 N)( 0.60 0.80 )3.0 cm

F i jθ θ −= ⇒ = ° ⇒ = × − −

9 2 2 9 94 4

3 on 1 3 32 2(9.0 10 N m /C )(5 10 C)(5 10 C) ˆ, toward (2.5 10 N, toward ) 2.5 10 N

(3.0 10 m)F q q i− −

−

⎛ ⎞× × ×= = × = ×⎜ ⎟⎜ ⎟×⎝ ⎠

2 2

4 4on 1 2 on 1 3 on 1

ˆ ˆ(1.42 10 1.44 10 ) NF F F i j− −= + = × − ×

The magnitude and direction of the resultant force vector are 4 2 4 2 4

on 1 (1.42 10 N) ( 1.44 10 N) 2.0 10 NF − − −= × + − × = × 4

14

1.44 10 Ntan 451.42 10 N

φ−

−−

⎛ ⎞×= = °⎜ ⎟⎜ ⎟×⎝ ⎠ clockwise from the +x-axis.

22.44. Model: The charges are point charges. Solve: Placing the –1.0 nC charge at the origin and calling it 1,q the 2q charge is in the first quadrant, the 3q charge is in the fourth quadrant, the 4q charge is in the third quadrant, and the 5q charge is in the second quadrant. The electric force on 1q is the vector sum of the electric forces from the other four charges 2 3 4, , ,q q q and 5.q The magnitude of these four forces is the same because all four charges are equal in magnitude and are equidistant from 1.q So,

9 2 2 9 94

2 on 1 3 on 1 4 on 1 5 on 1 2 2 2 2(9.0 10 N m /C )(2.0 10 C)(1.0 10 C) 3.6 10 N

(0.50 10 m) (0.50 10 m)F F F F

− −−

− −× × ×= = = = = ×

× + ×

Thus, 4 4on 1 2 3(3.6 10 N, toward )+(3.6 10 N, toward )F q q− −= × × + 4

4(3.6 10 N, away from )q−× +(3.6 × 410 N,− away from ).q5 In component form,

{}

on 1 on 1

4 3

ˆ ˆ ˆ ˆ[cos(45 ) sin (45 ) ] [cos(45 ) sin (45 ) ]

ˆ ˆ ˆ ˆ[cos(45 ) sin (45 ) ] [cos(45 ) sin (45 ) ]

ˆ ˆ(3.6 10 N) 4cos(45 ) 1.0 10 N

F F i j i j

i j i j

i i− −

= ° + ° + ° − °

+ ° + ° + ° − °

= × [ ° ] = ×

Assess: By symmetry, we see that the vertical forces must cancel, and that the horizontal force must be in the positive direction, which agrees with the calculation.

22.45. Model: The charges are point charges. Visualize: Please refer to Figure P22.45. Solve: Placing the 1.0 nC charge at the origin and calling it 1,q the −6.0 nC is 3,q the 2q charge is in the first quadrant, and the 4q charge is in the second quadrant. The net electric force on 1q is the vector sum of the electric forces from the three charges 2 3, ,q q and 4.q We have

22-20 Chapter 22


1 22 on 1 22

9 2 2 9 9

22 2

5 52

, away from

(9.0 10 N m /C )(1.0 10 C)(2.0 10 C) , away from (5.0 10 m)

ˆ ˆ(0.72 10 N, away from ) (0.720 10 N)[ cos(45 ) sin (45 ) ]

K q qF q

r

q

q i j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = × − ° − °

1 33 on 1 32

9 2 2 9 9

32 2

5 53

1 4 54 on 1 42

, toward

(9.0 10 N m /C )(1.0 10 C)(6.0 10 C) , toward (5.0 10 m)

ˆ(2.16 10 N, away from ) 2.16 10 N

, away from (0.720 10 N)[co

K q qF q

r

q

q j

K q qF q

r

− −

−

− −

−

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = ×

⎛ ⎞= = ×⎜ ⎟⎜ ⎟⎝ ⎠

ˆ ˆs(45 ) sin (45 ) ]i j° − °

Summing these forces vectorially gives 5 5 5

on 1 2 on 1 3 on 1 4 on 1ˆ ˆ[(2.16 10 N) 2(0.720 10 N)sin (45 )] 1.1 10 NF F F F j j− − −= + + = × − × ° = ×

Assess: By symmetry, we see that the horizontal forces must cancel, and that the vertical force must be upward, which agrees with the calculation.

22.46. Model: The charges are point charges. Visualize: Please refer to Figure P22.46. Solve: Place the 1.0 nC charge at the origin and call it 1;q the −6.0 nC is 3,q the 2q charge is in the first quadrant, and the 4q charge is in the second quadrant. The net electric force on 1q is the vector sum of the electric forces from the other three charges 2 3, ,q q and 4.q We have

1 22 on 1 22

9 2 2 9 9

22 2

5 52

1 33 on 1 32

, toward

(9.0 10 N m /C )(1.0 10 C)(2.0 10 C) , toward (5.0 10 m)

ˆ ˆ(0.72 10 N, toward ) (0.72 10 N)[cos(45 ) sin (45 ) ]

, toward

K q qF q

r

q

q i j

K q qF q

r

− −

−

− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = × ° + °

⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠

5 53

1 4 54 on 1 42

ˆ(2.16 10 N, toward ) 2.16 10 N

ˆ ˆ, away from (0.72 10 N)[cos(45 ) sin (45 ) ]

q j

K q qF q i j

r

− −

−

× = ×

⎛ ⎞= = × ° − °⎜ ⎟⎜ ⎟⎝ ⎠

Adding these components together vector-wise gives 5 5

on 1 2 on 1 3 on 1 4 on 15 5

ˆ ˆ2(0.72 10 N)cos(45 ) (2.16 10 N)ˆ ˆ(1.02 10 2.2 10 ) N

F F F F i j

i j

− −

− −

= + + = × ° + ×

= × + ×



22.47. Model: The charged particles are point charges. Visualize:

Solve: (a) The mathematical problem is to find the position for which the forces 1 on pF and 2 on pF are equal in

magnitude and opposite in direction. If the proton is at position x, it is a distance x from 1q and d x− from 2 ,q where d = 1.0 cm. The magnitudes of the forces are

1 p 1 p 2 p1 on p 2 on p2 2 2

1p ( )

K q q K q q K q qF F

r x d x= = =

−

Equating the two forces and using d = 1.0 cm,

1 p 2 p 2 2 22 2 2 2

2.0 nC 4.0 nC 1 2 2 2 1 0( ) ( )

K q q K q qx x x x x

x d x x d x= ⇒ = ⇒ + − = ⇒ + − =

− −

The solutions to the equation are x = +0.414 cm and −2.41 cm. Both are points where the magnitudes of the two forces are equal, but +0.414 cm is a point where the magnitudes are equal and the directions are the same. The solution we want is that the proton should be placed at x = −2.4 cm. (b) Yes, the net force on the electron located at x = −2.4 cm will also be zero. This is because the solution in part (a) does not depend specifically on the type of the charge that experiences zero force from the other two charges. Furthermore, if on pF is zero for a proton, the electric field at that point must be zero. Thus, there will be no force on any charged particle at that point.


Solve: The two 2.0 nC charges exert an upward force on the 1.0 nC charge. Since the net force on the 1 nC charge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is a positive charge. The force of charge 2 on charge 1 is

1 22 on 1 22

129 2 2 9 9

2 2

, away from

(9.0 10 N m /C )(1.0 10 C)(2.0 10 C) ˆ ˆ(cos sin )(0.020 m) (0.030 m)

K q qF q

r

i jθ θ− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

× × ×= ++

22-22 Chapter 22


From the figure, 1tan (2/3) 33.69 .θ −= = ° Thus

5 52 on 1

ˆ ˆ(1.152 10 0.768 10 ) NF i j− −= × + ×

From symmetry, 3 on 1F is the same except the x-component is reversed. When we add 2 on 1F and 3 on 1,F the x-components cancel and the y-components add to give

52 on 1 3 on 1 ˆ1.536 10 NF F j

→ →−+ = ×

on 1qF must have the same magnitude, pointing in the j− direction, so

5 215

on 1 2 9 2 2 9(1.536 10 N)(0.020 m)1.536 10 N 0.68 nC

(9.0 10 N m /C )(1.0 10 C)qK q q

F qr

−−

−×= × = ⇒ = =

× ×

A positive charge 0.68 nCq = will cause the net force on the 1.0 nC charge to be zero.

22.49. Model: The charged particles are point charges. Visualize: Please refer to Figure P22.49. Solve: The charge 2q is in static equilibrium, so the net electric field at the location of 2q is zero. We have

1

91

net 5 nC 2 2(5.0 10 C)ˆ ˆ( ) ( ) 0 N/C

(0.20 m) (0.10 m)qq

E E E K i K i−×= + = ± + =

We have used the ± sign to indicate that a positive charge on 1q leads to an electric field along i+ and a negative

charge on 1q leads to an electric field along i− . Because the above equation can only be satisfied if we use ˆ,i− we infer that the charge 1q is a negative charge. Thus,

91

1 12 25.0 10 Cˆ ˆ( ) ( ) 0 N/C 20 nC 20 nC

(0.20 m) (0.10 m)q i i q q

−×− + = ⇒ = ⇒ = −


Solve: The force on q is the vector sum of the force from −Q and +Q. We have

on 2 2 2 2ˆ ˆ, toward ( cos sin )Q q

Q q KQqF K Q i ja y a y

θ θ+⎛ ⎞− +

= − = − −⎜ ⎟⎜ ⎟+ +⎝ ⎠2

on 2 2 2 2ˆ ˆ, away from ( cos sin )Q q

K Q q KQqF Q i ja y a y

θ θ+ +⎛ ⎞+ +

= + = − +⎜ ⎟⎜ ⎟+ +⎝ ⎠

net 2 2ˆ ˆ( 2cos ) 0 NKQqF i j

a yθ= − +

+



From the figure we see that 2 2cos / .a a yθ = + Thus

net 2 2 3/22( )

( )xKQqaF

a y−=

+

Assess: Note that net net( )xF F= because the y-components of the two forces cancel each other out.

22.51. Model: The charges are point charges. Solve:

We will denote the charges −Q, 4Q and −Q by 1, 2, and 3, respectively.

1 on 2 2ˆ, toward ( )q

K Q q KQqF Q iL L

⎛ ⎞−= − = −⎜ ⎟⎜ ⎟⎝ ⎠

2 on 2 on 1 on 2 2 24 4 2ˆ ˆ, away from 4 [ cos(45 ) sin (45 ) ] 2

( 2 ) 2q q qK Q q KQq KQqF Q i j F F

L L L

⎛ ⎞= = + ° + ° ⇒ = =⎜ ⎟⎜ ⎟⎝ ⎠

3 on 2 2ˆ, toward ( )q

K Q q KQqF Q jL L

⎛ ⎞−= − = −⎜ ⎟⎜ ⎟⎝ ⎠

The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. The net force is

net 2 2 2 2 2

ˆ ˆ2ˆ ˆ ˆ ˆ( ) ( ) (1 2) (1 2)2 2

KQq KQq i j KQq KQq KQqF i j i jL L L L L

⎛ ⎞= − + + + + − = − − − −⎜ ⎟⎜ ⎟

⎝ ⎠

2 2

net 2 2 2(1 2) (1 2) (2 2)KQq KQq KQqFL L L

⎡ ⎤ ⎡ ⎤= − + − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

22.52. Model: The charges are point charges. Solve: Label the charges q, −q, and Q with numbers 1, 2, and 3, respectively. The positive charge q exerts an attractive force on the charge −q at the origin that is given by

21 2

1 on 2 2 2ˆ ˆq q qF K j K j

L L= =

The charge Q exerts an attractive force on −q at the origin that is given by 2

23 on 2 2 2

ˆ ˆ(2 ) 4

Q q qF K i K iL L

α= =

22-24 Chapter 22


Because the net force is at 45° to the x-axis (or y-axis), the i and j components must be of equal magnitude. Setting them equal an solving for α gives

2 2

2 2 44

q qK KL L

α α= ⇒ =

22.53. Model: Use the charge model and assume the copper spheres are point objects with point charges. Solve: (a) The mass of the copper sphere is

3 3 3 3 54 4(8920 kg/m ) (1.0 10 m) 3.736 10 kg 0.03736 g3 3

M V rπ πρ − −⎛ ⎞ ⎡ ⎤= = = × = × =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

The number of moles in the sphere is

40.03742 g 5.884 10 mol63.5 g/mol

MnA

−= = = ×

The number of copper atoms in the sphere is 4 23 1 20

A (5.884 10 mol)(6.02 10 mol ) 5.542 10N nN − −= = × × = ×

The number of electrons in the copper sphere is thus 20 2229 3.542 10 1.027 10 .× × = × The total positive or negative charge in the sphere is 22 19(1.027 10 )(1.60 10 C) 1643 C.−× × = Hence, the spheres will have a net charge of

9 610 1643 C 1.643 10 C.− −× = × The force between two such spheres is

9 2 2 6 221 2

2 2 2 2(9.0 10 N m /C )(1.643 10 C) 2.4 10 N

(1.0 10 m) (1.0 10 m)Kq qF

−

− −× ×= = = ×

× ×

(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between the proton charge and the electron charge must be smaller than 1 part in 910 .

22.54. Model: The electron and the proton are point charges. Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration for the electron’s circular motion. Thus,

22

2

2 9 2 2 19 216 15

3 31 11 3

( )( )

(9.0 10 N m /C )(1.60 10 C) 1 rev(4.12 10 rad/s) 6.6 10 rev/s2 rad(9.11 10 kg)(5.3 10 )

K e e mv mrrr

Kefmr

ω

π

−

− −

= =

⎛ ⎞× ×= = = × = ×⎜ ⎟× × ⎝ ⎠

22.55. Model: Model the charged balls as point charges and ignore friction. Solve: Combining Coulomb’s law and Newton’s second law for the balls at an arbitrary distance d apart gives their acceleration:

2 2

2 2q qF K ma a Kd md

= = ⇒ =

Thus, if we plot acceleration as a function of inverse distance squared, the magnitude of the charge can be calculated from the slope s:

2Kq sms qm K

= ⇒ =



The fit gives a slope of 4 22.97 10 N m /kg,s −= × so the magnitude of the charge is

4 2 3

9 2 2(2.97 10 N m /kg)(2.0 10 kg) 8.1 nC

(9.0 10 N m /C )q

− −× ×= =×

22.56. Model: Even though they are cubes rather than spheres, assume Coulomb’s law applies, and the Coulomb forces are horizontal. Visualize:

Solve: At the point of slipping the static friction will be maximum and the net force on the cube that slips will be zero. The maximum static friction force is smaller on the lighter cube so it will move first.

21 2

net on 1 1 12 2

2 2 21

9 2 2

( ) 0

(0.65)(0.0020 kg)(9.8 m/s )(0.060 m) 71.4 nC9 10 N m /C

q q qF m g K m g Kr r

m grqK

μ μ

μ

= − = ⇒ =

⇒ = = =× ⋅

Now we figure the time it takes at the charging rate given. 71.4 nC 10 s

/ 7.0 nC/sqt

Q tΔ = = =

Δ Δ

So the 2.0 g cube will move after 10 s. Assess: This is a reasonable amount of time.

22-26 Chapter 22


22.57. Model: Assume a circular orbit with a centripetal net force. Visualize:

Solve: The electric force is attractive, as is the gravitational force. 2

2A S A SA S A Sr S r S S2 2

A SA S

A S A S

S S

11 2 2 17 5 9 2 2

5

(6.67 10 N m /kg )(8.7 10 kg)(3.3 10 kg) (9.0 10 N m C )(4400 C)(1.2 C)(3.3 10 kg)(7500 km)

5.2 m/s

q q q qm m v m mF G K m a m G K m v

r r r r r

q qm mG K Gm m K q qr rv

m m r

−

= + = = ⇒ + =

+ +⇒ = =

× ⋅ × × + × ⋅=

×

=

Assess: This is a moderate speed; it would be slower without the electric attraction.

22.58. Model: The charges are point charges. Visualize: Each of the point charges has the same amount of charge q as the other, but q increases with time. Solve: Since the charge increases at a constant rate then the charge after 1.0 s is ( / ) (5.0 nC/s)(1.0 s)q dq dt t= Δ = = 5.0 nC. Take the derivative with respect to time of the Coulomb force law (use the chain rule).

21 2

2 2

9 2 2 42 2

5.0 nC2 2(9.0 10 N m /C ) (5.0 nC/s) 7.2 10 N/s(0.025 m)

q q qF K Kr r

dF q dqKdt dtr

−

= =

= = × ⋅ = ×

Assess: We see that the rate the force is changing itself changes as the charge increases.

22.59. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’s law. Solve: Let q be the charge on each plastic bead. The repulsive force between the beads pushes the beads apart. The spring is stretched until the restoring spring force on either bead is equal to the repulsive Coulomb force:

2 2

2Kq k x rk x q

KrΔ= Δ ⇒ =

The spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus 3

22

(1.0 10 kg)(9.8 N/kg)(1.0 10 m) 0.98 N/m1.0 10 m

mg k k−

−−

×= × ⇒ = =×

2 2 2 2

9 2 2(0.98 N/m)(4.5 10 m 4.0 10 m)(4.5 10 m) 33 nC

9.0 10 N m /Cq

− − −× − × ×= =×



22.60. Model: Take the center of mass of the dipole to be midway between the two charges. Solve: The torque about the center of mass due to each charge is ( )e 2

sFτ = and each is in the same direction, so the

total torque is e .F sτ = The electric force e ,F Eq= so the torque may be written as ,Eqs pEτ = = where we have used p qs≡ in the last step.

22.61. Solve: (a) Kinetic energy is 212 ,K mv= so the velocity squared is 2 / .v K m= 2 From kinematics, a particle

moving through distance Δx with acceleration a, starting from rest, finishes with 2 2 .v a x= Δ To gain 182 10 JK −= × of kinetic energy in Δx = 2.0 μm requires an acceleration of

2 1818 2 18 2

31 62 / 2.0 10 J 1.10 10 m/s 1.1 10 m/s

2 2 (9.11 10 kg)(2.0 10 m)v K m Ka

x x m x

−

− −×= = = = = × ≈ ×

Δ Δ Δ × ×

(b) The force that produces this acceleration is 31 18 2 12 12(9.11 10 kg)(1.10 10 m/s ) 1.00 10 N 1.0 10 NF ma − − −= = × × = × ≈ ×

(c) The electric field required is 12

619

1.00 10 N 6.3 10 N/C1.602 10 C

FEe

−

−×= = = ××

(d) The force on an electron due to a charge q is 2/ .F K q e r= To have breakdown, the force on the electron must

be at least 121.0 10 N.−× The minimum charge that could cause a breakdown will be the charge that causes exactly a

force of 121.0 10 N:−×

2 2 1212 8

2 9 2 2 19(0.010 m) (1.0 10 N)1.0 10 N 6.9 10 C 69 nC

(9.0 10 N m /C )(1.6 10 C)

K q e r FF qKer

−− −

−×= = × ⇒ = = = × =

× ×

22.62. Model: The charged spheres behave as point charges. Visualize:

Each sphere is in static equilibrium and the string makes an angle θ with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. Solve: In static equilibrium, Newton’s first law gives net 0.G eF T F F= + + = In component form,

net G e net G e( ) ( ) ( ) 0 N ( ) ( ) ( ) 0 Nx x x x y y yF T F F F T F F= + + = = + + = 2

2sin 0 N 0 N cos 0 N 0 NKqT T mgd

θ θ− + + = − + =

2 2

2 2sin cos(2 sin )

Kq KqT T mgd L

θ θθ

= = = +

22-28 Chapter 22


Dividing the two equations gives 2 9 2 2 9 2

2 42 2 3

(9.0 10 N m /C )(100 10 C)sin tan 4.59 104 4(1.0 m) (5.0 10 kg)(9.8 N/kg)

KqL mg

θ θ−

−−

× ×= = = ××

For small-angles, tan sin .θ θ≈ With this approximation we obtain sin 0.07714 rad,θ = so θ = 4.4°.

22.63. Model: The electric field is that of a negative point charge located at the origin. Visualize: Place the –10 nC charge at the origin. The field vectors point toward the negative charge. Solve: The electric field is

9 2 2 9 2

2 2 2(9.0 10 N m /C )(10 10 C) 90 N m /C, away from , away from , away from qE K q q q

r r r

−⎛ ⎞ ⎛ ⎞× ×⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

The electric field at each of the three points is

( )

25

1 2 2

24

2 2 2

4 4 4345 5

2

3

90.0 N m /C ˆ, towards 1.0 10 N/C(3.0 10 m)

90.0 N m /C ˆ ˆ, towards ( 3.6 10 N/C)(cos sin )(5.0 10 m)

ˆ ˆ ˆ ˆ( 3.6 10 N/C) ( 2.9 10 2.2 10 ) N/C

90.0 N m /C(4.0 1

E q j

E q i j

i j i j

E

θ θ

−

−

⎛ ⎞= = − ×⎜ ⎟⎜ ⎟×⎝ ⎠⎛ ⎞

= = − × +⎜ ⎟⎜ ⎟×⎝ ⎠

= − × + = − × − ×

=×

42 2

ˆ, towards 5.6 10 N/C0 m)

q i−

⎛ ⎞= − ×⎜ ⎟⎜ ⎟

⎝ ⎠

22.64. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure 22.64. Place the 5.0 nC charge at the origin. Solve: The electric field is

9 2 2 9

2 2

2

2

(9 10 N m /C )(5.0 10 C), away from , away from

45 N m /C , away from

qE K q qr r

qr

−⎛ ⎞× ×⎛ ⎞= = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

The electric field at the three points is 2

41 2 2 2 2

4 4 4

25

2 2 2

3

45 N m /C ˆ ˆ, away from (9.0 10 N/C)(cos sin )(2.0 10 m) (1.0 10 m)

1 2ˆ ˆ ˆ ˆ(9.0 10 N/C) (4.0 10 8.0 10 ) N/C5 5

45 N m /C ˆ, away from 4.5 10 N/C(1.0 10 m)

4

E q i j

i j i j

E q i

E

θ θ− −

−

⎛ ⎞= = × +⎜ ⎟⎜ ⎟× + ×⎝ ⎠

⎛ ⎞= × + = × + ×⎜ ⎟

⎝ ⎠⎛ ⎞

= = ×⎜ ⎟⎜ ⎟×⎝ ⎠

=2

4 42 2 2 2

5 N m /C ˆ ˆ, away from (4.0 10 8.0 10 ) N/C(2.0 10 m) (1.0 10 m)

q i j− −

⎛ ⎞= × − ×⎜ ⎟⎜ ⎟× + ×⎝ ⎠



22.65. Model: The electric field is that of a positive charge at (x, y) = (1.0 cm, 2.0 cm). Visualize:

Solve: (a) The electric field of a positive charge points straight away from the charge, so we can roughly locate the points of interest based simply on whether the signs of xE and yE are positive or negative. For point a, the electric field has no y-component and the x-component points to the left, so point a must be to the left of the charge along a horizontal line. Using the field of a point charge,

9 2 2 9

a2a

(9.0 10 N m /C )(10.0 10 C) 0.020 m 2.0 cm225,000 N/Cx

x

K q K qE E r

Er

−× ×= = ⇒ = = = =

Thus, a a( , ) ( 1.0 cm, 2.0 cm).x y = − (b) Point b is above and to the right of the charge. The magnitude of the field at this point is

2 2 2 2(161,000 N/C) (80,500 N/C) 180,000 N/Cx yE E E= + = + =

Using the field of a point charge, 9 2 2 9

b2b

(9.0 10 N m /C )(10.0 10 C) 2.236 cm180,000 N/C

K q K qE r

r E

−× ×= ⇒ = = =

This gives the total distance but not the horizontal and vertical components. However, we can determine the angle θ because bE points straight away from the positive charge. Thus,

1 1 180,500 N/C 1tan tan tan 26.57161,000 N/C 2

y

x

E

Eθ − − −

⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎜ ⎟= = = = °⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

The horizontal and vertical distances are then b cos (2.236 cm)cos26.57 2.00 cmxd r θ= = ° = and b sinyd r θ= =

1.00 cm. Thus, point b is at position b b( , ) (3.0 cm, 3.0 cm).x y = (c) To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We first find

36,000 N/CE = from which we find the total distance c 5.00 cm.r = The angle φ is

1 1 28,800tan tan 53.1321,600

y

x

E

Eφ − −

⎛ ⎞ ⎛ ⎞⎜ ⎟= = = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

which gives distances c cos 3.00 cmxd r φ= = and c sin 4.00 cm.yd r φ= = Thus, point c is at position c c( , )x y = (4.0 cm, −2.0 cm).

22-30 Chapter 22


22.66. Model: The electric field is that of three point charges. Visualize:

Solve: (a) In the figure, the distances are 2 21 3 (1.0 cm) (3.0 cm) 3.162 cmr r= = + = and the angle is

1tan (1.0/3.0)θ −= = 18.43 .° Using the equation for the field of a point charge,

9 2 2 91

1 3 2 21

(9.0 10 N m /C )(1.0 10 C) 9.0 kN/C(0.03162 m)

K qE E

r

−× ×= = = =

We now use the angle θ to find the components of the field vectors:

1 1 1

3 3 3

ˆ ˆ ˆ ˆ ˆ ˆcos sin (8540 2840 ) N/C (8.5 2.8 ) kN/Cˆ ˆ ˆ ˆ ˆ ˆcos sin (8540 2840 ) N/C (8.5 2.8 ) kN/C

E E i E j i j i j

E E i E j i j i j

θ θ

θ θ

= − = − = −

= + = + = +

2E is easier since it has only an x-component. Its magnitude is

9 2 2 92

2 2 22 22

(9.0 10 N m /C )(1.0 10 C) ˆ ˆ10,000 N/C 10 kN/C(0.0300 m)

K qE E E i i

r

−× ×= = = ⇒ = =

(b) The electric field is defined in terms of an electric force acting on charge : / .q E F q= Since forces obey a

principle of superposition net 1 2( )F F F= + + it follows that the electric field due to several charges also obeys a principle of superposition. (c) The net electric field at a point 3.0 cm to the right of 2q is net 1 2 3

ˆ27 kN/C.E E E E i= + + = The y-components

of 1E and 2E cancel, giving a net field pointing along the x-axis.

22.67. Model: The charged ball attached to the string is a point charge. Visualize:

The ball is in static equilibrium in the electric field when the string makes an angle θ = 20° with the vertical. The three forces acting on the charged ball are the electric force due to the field, the gravitational force on the ball, and the tension force. Solve: In static equilibrium, Newton’s second law for the ball gives net 0.G eF T F F= + + = In component form,

net net( ) 0 N 0 N ( ) 0 N 0 Nx x y yF T qE F T mg= + + = = − + =



The above two equations simplify to sin cosT qE T mgθ θ= =

Dividing the equations, we get 3

7tan (5.0 10 kg)(9.8 N/kg) tan 20tan 1.78 10 C 0.18 C100,000 N/C

qE mgqmg E

θθ μ−

−× °= ⇒ = = = × =

22.68. Model: The charged ball attached to the string is the point charge. Visualize:

The charged ball is in static equilibrium in the electric field when the string makes an angle θ with the vertical. The three forces acting on the charge are the electric force due to the electric field, the gravitational force on the ball, and the tension force. Solve: In static equilibrium, Newton’s second law for the charged ball gives net 3 0.GF T F F= + + = In component form,

net net( ) 0 N 0 N ( ) 0 N 0 Nx x y yF T qE F T mg= + + = = − + =

These two equations become sinT qEθ = and cos .T mgθ = Dividing the equations gives

9

3(25 10 C)(200,000 N/C)tan 0.255 14

(2.0 10 kg)(9.8 N/kg)qEmg

θ θ−×= = = ⇒ = °

×

2

22.69. Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 61.0 10 N/C× a distance of 1.0 μ m from the dust particle? (b) The number of electrons is

6 6 23

9 2 2 19(1.5 10 N/C)(1.0 10 m) 1.0 10

(9.0 10 N m /C )(1.60 10 C)N

−

−× ×= = ×

× ×

22.70. Solve: (a) Two equal charges separated by 1.50 cm exert repulsive forces of 0.020 N on each other. What is the magnitude of the charge? (b) The charge is

2

9 2 2(0.020 N)(0.0150 m) 22 nC

9.0 10 N m /Cq = ± = ±

×

The problem does not give the direction of the force. Thus, the charges could be both positive or both negative.

22.71. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C? (b) The distance is

9 2 2 9(9.0 10 N m /C )(15 10 C) 0.050 m 5.0 cm 54,000 N/C

r−× ×= = =

22-32 Chapter 22


22.72. Solve: (a) A 1.0 nC charge is placed at (0.0 cm, 2.0 cm) and another 1.0 nC charge is placed at (0 cm, −2.0 cm). A third charge +q is placed along a line halfway between 1q and 2q such that the angle between the forces

on q due to each of the other two charges is 60°. The resultant force on q is 5 ˆ(5.0 10 ) N.i−× What is the magnitude of the charge q? (b)

We have 9 2 2 9

1 on 3 32(9.0 10 N m /C )(1.0 10 C) ˆ ˆ, away from (5625 N/C) (cos30 sin30 )

(0.020 m/sin30 )qF q q i j

−⎛ ⎞× ×= = ° − °⎜ ⎟⎜ ⎟°⎝ ⎠

2 on 3ˆ ˆ(5625 N/C) [cos(30 ) sin(30 ) ]F q i j= ° + °

on 3ˆ2 (5625 N/C) cos(30 )F q i= × °

55 (5.0 10 N)2(5625 N/C) cos(30 ) 5.0 10 N 5.1 nC

2(5625 N/C)cos(30 )q q

−− ×° = × ⇒ = =

°

Challenge Problems

22.73. Model: The charged spheres behave as point charges. Visualize:

Each sphere is in static equilibrium when the string makes an angle of 20° with the vertical. The three forces acting on each sphere are the electric force, the gravitational force on the sphere, and the tension force. Solve: In the static equilibrium, Newton’s first law gives net ( ) 0.G eF T F F= + + = In component form, we have

net G e net G e( ) ( ) ( ) 0 N ( ) ( ) ( ) 0 Nx x x x y y yF T F F F T F F= + + = = + + = 2

2sin 0 N 0 N cos 0 N 0 NKqT T mgd

θ θ− + + = − + =

2 2

2 2sin cos(2 sin )

Kq KqT T mgd L

θ θθ

+= = = +



Dividing the two equations and solving for q gives 2 2 2 2 3

9 2 24sin tan 4(sin 20 tan 20 )(1.0 m) (3.0 10 kg)(9.8 N/kg) 0.75 C

9.0 10 N m /CL mgq

Kθ θ μ

−° ° ×= = =×

22.74. Model: The charged balls are point charges. Visualize:

Because of symmetry and the fact that the three balls have the same charge, the magnitude of the electric force on each ball is the same. The other forces acting on each ball are the gravitational force on the ball and the tension force. Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have

21 3

1 on 3 12 213

ˆ ˆ, away from [sin(30 ) cos(30 ) ]K q q KqF q i j

r r

⎛ ⎞= = ° + °⎜ ⎟⎜ ⎟⎝ ⎠

2

2 on 3 2ˆ ˆ[ sin(30 ) cos(30 ) ]KqF i j

r= − ° + °

2 2

on 3 on 3 on 2 on 1 e2 22 2ˆcos(30 ) cos(30 )Kq KqF j F F F F

r r= ° ⇒ = ° = = =

9 2 2 211 2 2

e 22(9.0 10 N m /C ) cos(30 ) (3.897 10 ) N/C

(0.20 m)qF q× °= = ×

The distance l, between one of the balls and the center of the equilateral triangle, is 0.10 mcos(30 ) 0.10 m 0.1155 m

2 cos(30 )rl l° = = ⇒ = =

°

Thus, the angle made by the string with the plane containing the three balls is 0.1155 mcos 81.700.80 m

lL

θ θ= = ⇒ = °

From the free-body diagram, we have sin 0 N cos 0 NeT mg T Fθ θ− = − + =

11 2 2tan(3.897 10 ) N/Ce

mg mgF q

θ = =×

37

11 2 11 2(3.0 10 kg)(9.8 N/kg) 1.05 10 C 0.11 C

(3.897 10 N/C ) tan (3.897 10 N/C ) tan(81.70 )mgq μ

θ

−−×= = = × ≈

× × °

22-34 Chapter 22


22.75. Model: The charged spheres are point charges. Visualize:

The figure shows the free-body diagram of the forces on the sphere with the negative charge that is shown in Figure CP22.75. The force EF is due to the electric field. The force eF is the attractive force between the positive and the negative spheres. The tension in the string and the gravitational force are the remaining two forces on the spheres. Solve: The two electrical forces are calculated as follows:

9 5 2E (100 10 C)(1.00 10 N/C) 1.00 10 NF q E − −= = × × = ×

9 2 2 9 2 5 21 2

e 2 2 2(9.0 10 N m /C )(100 10 C) 9.0 10 N m /CK q q

Fr r r

− −× × ×= = =

From the geometry of Figure CP22.75, 5 2

3e 2

9.0 10 N m /C2 sin(10 ) 2(0.50 m)sin(10 ) 0.174 3.0 10 N(0.174 m)

r L E F−

−×= ° = ° = ⇒ = = ×

From the free-body diagram,

e Ecos10 sin10T mg T F F° = ° + =

Rearranging and dividing the two equations gives

E e

2 22E e

sin(10 ) tan(10 )cos(10 )

1.00 10 N 0.30 10 N 0.41 10 kg 4.1 gtan(10 ) (9.8 N/kg) tan10

F Fmg

F Fmg

− −−

−° = ° =°

− × − ×= = = × =° °


Solve: The forces on the −1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nC charge is positive, 10F points as shown. The angle formed by the dashed lines where they meet is 90°, so qF must point

towards q in order for 10 qF F F= + to hold.



Therefore

10cos(30 )

FF =°

The distance between the 10 nC and −1.0 nC charges is 5.0sin30 cm 2.5 cm.r = ° = Thus

9 99 2 2 4

10 2 2(10 10 C)(1.0 10 C)(9.0 10 N m /C ) 1.44 10 N

(2.5 10 m)F

− −−

−× ×= × = ×

×

Thus, 4

41.44 10 N 1.7 10 Ncos(30 )

F−

−×= = ×°

22.77. Model: Charge Q and the dipole charges (q and −q) are point charges. Visualize: Please refer to Figure CP22.77. Solve: (a) The force on the dipole is the vector sum of the force on q and −q. We have

on 2 2ˆ, away from ( )

( /2) ( /2)Q qK Q q KQqF Q ir s r s+

⎛ ⎞= = −⎜ ⎟⎜ ⎟+ +⎝ ⎠

on net2 2 21 1ˆ ˆ( )

( /2) ( /2) ( /2)Q qKQqF i F KQq i

r s r s r s−⎛ ⎞

= + ⇒ = −⎜ ⎟⎜ ⎟− − +⎝ ⎠

(b) The net force netF is toward the charge Q, because the attractive force due to Q on the negative charge of the dipole is more than the repulsive force of Q on the positive charge. (c) Using the binomial approximation, we get

22 2 2

net 2 32 2 2 2( /2) 1 1 1 1

2 2 2 2s s KQq s s KQqsr s r r Fr r r rr r

−− − − ⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞± = ± ≅ + ⇒ = + − − =⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦∓

(d) Coulomb’s law applies only to the force between two point charges. A dipole is not a point charge, so there’s no reason that the force between a dipole and a point charge should be an inverse-square force. Assess: Note that when net0 m, 0 N.s F→ → In this limit the dipole is a point with zero charge.

m22 knig2461 04 ism c22 - media.physicsisbeautiful.com

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