M25 KNIG9404 ISM C25

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25-1

Conceptual Questions

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

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

25.3. 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.

25.4. (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.

25.5. 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.

25.6. 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.

25.7. (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.

ELECTRIC CHARGES AND FORCES

25

25-2 Chapter 25


(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.

25.8.

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.

25.9. 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.

25.10.

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.

25.11.

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.

25.12.

Electric Charges and Forces 25-3


25.13. (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′ ′= =

25.14. (a) We have E(r) = 1000 N/C. For a point charge, 21( ) .E rr

∝ If the distance r to the charge is doubled,

2

2 2

2

11 1 (2 ) 14(2 ) , so .

1( ) 4(2 ) 4E r rE rE rr r

r

⎛ ⎞⎜ ⎟⎝ ⎠∝ = = =⎛ ⎞⎜ ⎟⎝ ⎠

Therefore 1(2 ) (1000 N/C) 250 N/C.4

E r = =

(b) Similarly, ( )

12 2

2

1( ) 4,2 rr

rE E r⎛ ⎞ = =⎜ ⎟⎝ ⎠

so

4(1000 N/C) 4000 N/C.2rE ⎛ ⎞ = =⎜ ⎟

⎝ ⎠

25.15. 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

Section 25.1 Developing a Charge Model

Section 25.2 Charge

25.1. 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

−

−− × = ×

− ×

25.2. 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.−×

25-4 Chapter 25


25.3. 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

−

−− × = ×

− ×

25.4. 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

−

−− × = ×

− ×

25.5. Model: Use the charge model. Solve: Each helium atom has 2 protons and there are 236.02 10× helium molecules in 1.0 mole of helium. Because each proton has a charge of 191.60 10 C,−+ × the amount of charge in 1.0 mole of oxygen is

23 19 5(1.0 mol)(6.022 10 atoms/mol)(2 protons/atom)(1.6 10 C/proton) 1.9 10 C−× × = ×

25.6. 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) 1.0 kg.

L mL cm

⎛ ⎞⎛ ⎞ ⎛ ⎞ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Each water molecule 2(H 0) 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 7

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

Section 25.3 Insulators and Conductors

25.7. 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.

25.8. Model: Use the charge model. Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) two opposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon does charge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, the wall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is a net attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with a normal force balancing the attractive electric force and an upward frictional force balancing the gravitational force on the balloon. (b)

25.9. 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.


25-6 Chapter 25


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 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 25.4 Coulomb’s Law

25.12. 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-Couomb is a lot of charge. That is why 1 on 2 2 on 1(or )F F is a measurable force.



25.13. 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 9

1 21 on 2 2 on 1 2 2 2

12

(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 −= =

×

25.14. 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 q

F K qr

−−

−× ×

= ⇒ = ⇒ = ××

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 −= − × = −

25.15. 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 q

F Kr

− −

−× × ×= = =

×

(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 −= = ×

×

25-8 Chapter 25


25.16. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure EX25.16. 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. Solve: Coulomb’s law yields:

9 2 2 9 9A B 5

B 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 q

F Kr

− −−

−× × ×= = = ×

×

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 q

F Kr

− −−

−× × ×= = = ×

×

The net force on A is 5 5

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

25.17. Model: Charges A, B, and C are point charges. Visualize: Please refer to Figure EX25.17. Solve: The force on B from charge A is directed downward since two negative charges repel. 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 q

F Kr

− −−

−× × ×= = = ×

×

So 5A on B ˆ( 4.5 10 ) NF j−= − ×

The force on B from charge C is directed downwards 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 q

F Kr

− −−

−× × ×= = = ×

×

So 4C on B ˆ(3.6 10 ) NF j−= − ×

The net electric force on charge A is 4

A B on A C on A4

ˆ ˆ(4.5 10 ) N (3.6 10 ) Nˆ(4.1 10 ) N

F F F j j

j

−5

−

= + = − × − ×

= − ×

25.18. 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 4B 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.

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

Solve: Coulomb’s law gives 9 2 2 9 19

13bead 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− −

−−

× × ×= = = ××

(a) Becaue 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,

1317 2bead on electron

electron 31electron

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.

Section 25.5 The Field Model

25.20. 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− −= × − = ×

25.21. 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 −= − = − − ×

25-10 Chapter 25


(c) From Newton’s second law, 2 2 17

proton 10 2proton 27

proton 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.

25.22. 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 q

E K qr

−= ⇒ = × ⇒ = × =

25.23. 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.

25.24. 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 2 ˆ.qE K rr

= Because the electric field points away

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

2 ˆ ˆ270,000 N/CqK r rr

=

2 28

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

(9 0 10 N m /C )q

−−. ×= = . × =

. ×

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

25.25. 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.

25.26. 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

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

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 AE , B,E and CE are pointing away from the positive charge.

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

25-12 Chapter 25


Solve: The electric field for a positive charge is

2 ˆqE K rr

=

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.

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

3.1g 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.

25.29. 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 q

F Kr

− −− −

−× ×= = × = ×

×

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−

−×= =×

25.30. 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:

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 2

nucleus 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 −×= ⇒ = = ×

×

25.31. 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

− −−

−

= = = =

× ×= = = ××

= × =

25.32. 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

−−

= = = =

×= = = ××

25-14 Chapter 25


(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

= = = =

25.33. Model: The charges are point charges. Visualize: Please refer to Figure P25.33. 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

24 4

on 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.

25.34. Model: The charges are point charges. Visualize: Please refer to Figure P25.34. 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.

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



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) , away from (1.0 10 m)

ˆ(4.5 10 N, away from ) 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) , toward (3.0 10 m) (1.0 10 m)

ˆ ˆ(1.35 10 N, toward ) (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 18.43.0 cm

θ − ⎛ ⎞= = °⎜ ⎟

⎝ ⎠

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

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

on 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−

−−

×= = ⇒ = = °×

below the −x axis,

or on 1F points 253° counterclockwise from the +x-axis.


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

− −

−

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞( × ) × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = ×

25-16 Chapter 25


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

31

37.2 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

− −

− −

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟× + ×⎝ ⎠

= × = × − −


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

41

41.44 10 Ntan 451.42 10 N

φ−

−−

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


Solve: The charges are 1 25.0 nC, 10 nC,q q= + = + and 3 10 nC.q = − The electric force on 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

4 42

, away from

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

ˆ(2.81 10 N, away from ) (2.81 10 N)

K q qF q

r

q

q j

− −

−

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = − ×

1 33 on 1 32

9 2 2 9 9

32 2 2 2

4

, toward

(9.0 10 N m /C )(5.0 10 C)(10 10 C) , toward (4.0 10 m) (3.0 10 m)

ˆ ˆ(1.80 10 N)(cos sin )

K q qF q

r

q

i jθ θ

− −

− −

−

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟× + ×⎝ ⎠

= × +

25-18 Chapter 25



14.0 cm 4tan tan 53.133.0 cm 3

θ θ − ⎛ ⎞= ⇒ = = °⎜ ⎟⎝ ⎠

4 43 on 1 ˆ ˆ ˆ ˆ(1.80 10 N)[cos(53.13 ) sin (53.13 ) ] (1.80 1.44 ) 10 NF i j i j− −= × ° + ° = + ×

Therefore, 4 4

on 1 2 on 1 3 on 1 ˆ ˆ(1.80 10 1.37 10 ) NF F F i j− −= + = × − ×

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

on 1 (1.80 10 N) ( 1.37 10 N) 1.7 10 NF − − −= × + − × = × 4

41.37 10 Ntan 521.80 10 N

φ φ−×= ⇒ = °

× 2 clockwise from the +x-axis.

25.39. Model: The charges are point charges. Visualize: Please refer to Figure P25.39. 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, 4on 1 2(3.6 10 N, away from )F q−= × + 4

3(3.6 10 N, away from )q−× + 44(3.6 10 N, toward )q−× + (3.6 × 410 N,−

toward ).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 negative direction, which agrees with the calculation.

25.40. Model: The charges are point charges. Visualize: Please refer to Figure P25.40. 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

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)[cos(45

K q qF q

r

q

q j

K q qF q

r

− −

−

− −

−

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = ×

⎛ ⎞= = ×⎜ ⎟⎝ ⎠

ˆ ˆ) sin (45 ) ]i j° − °

Summing these forces vectorially gives 5 5 5

on 1 2 on 1 3 on 1 4 on 1 ˆ 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.

25.41. Model: The charges are point charges. Visualize: Please refer to Figure P25.41. 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 (2.1

K q qF q

r

q

q i j

K q qF q

r

− −

−

− −

⎛ ⎞= ⎜ ⎟⎝ ⎠⎛ ⎞× × ×= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × = × ° + °

⎛ ⎞= =⎜ ⎟⎝ ⎠

5 53

1 4 54 on 1 42

ˆ6 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

− −

− −

= + + = × ° + ×

= × + ×

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

25-20 Chapter 25


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− −

⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠

× × ×= ++

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.

25.44. Model: The charged particles are point charges. Visualize: Please refer to Figure P25.44. 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.

25.46. Model: The charged particles are point charges.

25-22 Chapter 25


Visualize:

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

on 2 2ˆ, away from ( )

( ) ( )Q qK Q q KQqF Q ia x a x+ +

⎛ ⎞+ += + = −⎜ ⎟⎜ ⎟− −⎝ ⎠

on 2 2ˆ, toward ( )

( ) ( )Q qK Q q KQqF Q ia x a x− +

⎛ ⎞− += − = −⎜ ⎟⎜ ⎟+ +⎝ ⎠

2 2

net 2 2 2 2 21 1 2 ( )( )

( ) ( ) ( )xKQq a xF KQq

a x a x a x

⎡ ⎤ += − + = −⎢ ⎥− + −⎣ ⎦

To arrive at the final expression we used 2 2 2 2 2 2( ) ( ) [( )( )] ( ) .a x a x a x a x a x− + = − + = − (b) There are two cases when .x a> For x > a,


( ) ( )Q qK Q q KQqF Q ix a x a+ +

⎛ ⎞+ += + = +⎜ ⎟⎜ ⎟− −⎝ ⎠

on 2 2ˆ, toward ( )

( ) ( )Q qK Q q KQqF Q ix a x a− +

⎛ ⎞− += − = −⎜ ⎟⎜ ⎟+ +⎝ ⎠

net 2 2 2 2 21 1 4( )

( ) ( ) ( )xKQqaxF KQq

x a x a x a

⎡ ⎤ −= − =⎢ ⎥− + −⎣ ⎦

For x < − a (that is, for negative values of x),

on on 2 2ˆ ˆ( ) ( )

( ) ( )Q q Q qKQq KQqF i F ix a a x+ + − += − = +

− +

net 2 2 24( )

( )xKQqaxFx a

= −−

That is, the net force is to the right when x > a and to the right when x < − a. We can combine these two cases into a single equation for :x a>

net 2 2 24

( )( )xKQqa x

Fx a

=−

Here, the force is always to the right when .x a>



25.47. 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

⎡ ⎤ ⎡ ⎤= − + − = −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

25.48. 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 orgin 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 ) 4Q q qF K i K iL L

α= =

Because the net force is at 45° to the x-axis (or y-axis), the 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= ⇒ =

25-24 Chapter 25



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 2

2 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 2q ):

( )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.

25.50. 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 mol

63.5 g/molMnA

−= = = ×

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 2

21 22 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 .

25.51. 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

−

− −

= =

⎛ ⎞× ×= = = × = ×⎜ ⎟× × ⎝ ⎠

25.52. 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

− −× ×= =×

25-26 Chapter 25


25.53. 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.

25.54. Model: Model the metal plate and yourself as point charges. Solve: At the beginning, both you and the metal plates are neutral. As the electrons are pumped from the metal plate into you, the plate gains as much positive charge as you gain negative charge. When enough charge difference builds up between you and the metal plate, the gravitational force on you will be counterbalanced by the upward electrical force. You will begin to hang suspended in the air when

plate on you 2

2 24

9 2 2

(2.0 m)

(2.0 m) (60 kg)(9.8 N/kg)(2.0 m) 5.11 10 C9.0 10 N m /C

GK q q

F F mg

mgqK

−

−= ⇒ =

= = = ××

Dividing this charge by the charge on an electron yields 153.2 10× electrons.

25.55. 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

− − −× − × ×= =×

25.56. 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.

25.57. 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 Kax 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 q

Ker

−− −

−×= = × ⇒ = = = × =

× ×

25.58. 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

= = = +

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°.

25.59. Model: The charged spheres behave as point charges.

25-28 Chapter 25


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

−° ° ×= = =×

25.60. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.60. Place the +10 nC charge at the origin. 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, away from 1.0 10 N/C(3.0 10 m)

90.0 N m /C ˆ ˆ, away from (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, away from 5.6 10 N/C0 m)

q i−

⎛ ⎞= ×⎜ ⎟⎜ ⎟

⎝ ⎠

25.61. Model: The electric field is that of a negative point charge. Visualize: Please refer to Figure P25.61. Place point 1 at the origin. Solve: The electric field is

9 2 2 9

2 2

2

2

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

18.0 N m /C , toward

qE K q q

r r

qr

−⎛ ⎞⎛ ⎞ × ×= = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠



The electric fields at the two points are 2

1 2 2

5

2

2 2 2

5

18.0 N m /C , toward (1.0 10 m)

(1.8 10 N/C, 60 counter clockwise from the -axis or 60 north of east)

18.0 N m /C , toward (1.0 10 m)

(1.8 10 N/C, 60 clockwise from the -

E q

x

E q

x

−

−

⎛ ⎞= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × ° + °

⎛ ⎞= ⎜ ⎟⎜ ⎟×⎝ ⎠

= × ° − axis or 60 north of west)°

25.62. Model: The electric field is that of a positive point charge located at the origin. Visualize: Please refer to Figure P25.62. 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− −

⎛ ⎞= × − ×⎜ ⎟⎜ ⎟× + ×⎝ ⎠

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

Solve: (a) The electric field of a negative charge points toward the charge, so we can roughly locate where the field has a particular value by inspecting the signs of xE and .yE At point a, the electric field has no y-component and the x-component points to the left, so its location must be to the right of the charge along a horizontal line. Using the equation for 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, point a is at the position a a( , ) (4.0 cm, 1.0 cm).x y =

25-30 Chapter 25


(b) Point b is above and to the left 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 equation for the field of a point charge, 9 2 2 9

b2b

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

K q K qE r

Er

−× ×= ⇒ = = =

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

1 1 180,500 1tan tan tan 26.57161,000 2

y

x

E

Eθ − − −

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

The horizontal and vertical distances are then b cos 2.00 cmxd r θ= = and b sin 1.00 cm.yd r θ= = Thus, point b is at

the position b b( , ) (0.0 cm, 2.0 cm).x y = (c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We first find that 36,000 N/C.E = From this we find that the total distance c 5.00 cm.r = The angle φ is

1 1 21,600tan tan 36.8728,000

y

x

E

Eφ − −

⎛ ⎞ ⎛ ⎞⎜ ⎟= = = °⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

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

25.64. 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

Er

−× ×= ⇒ = = =

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 sin 1.00 cm.yd r θ= =

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).

25.65. 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 9

11 3 2 2

1

(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 θθ 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.

25-32 Chapter 25


25.66. 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−

−× °= ⇒ = = = × =

25.67. 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



25.68. 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

−

−× ×= = ×

× ×

25.69. 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.

25.70. 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−× ×= = =

25.71. 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 3 ˆ ˆ(5625 N/C) [cos(30 ) sin(30 ) ]F q i j= ° + °

on 3 ˆ2 (5625 N/C) cos(30 )F q i= × ° 5

5 (5.0 10 N)2(5625 N/C) cos(30 ) 5.0 10 N 5.1 nC2(5625 N/C)cos(30 )

q q−

− ×° = × ⇒ = =°

25.72. Model: Use the charge model. Solve: The mass of copper in a 2.0-mm-diameter copper ball 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 ρ − −⎛ ⎞ ⎡ ⎤= = = × = × =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦

25-34 Chapter 25


The number of moles in the ball is

40.03736 g 5.884 10 mol63.5 g/mol

MnA

−= = = ×

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

A (5.884 10 mol)(6.02 10 mol ) 3.542 10N nN − −= = × × = ×

Note that the number of electrons per atom is the atomic number, and both the atomic number (29) and the average atomic mass (63.5 g) are taken from the periodic table in the textbook. The number of electrons in the copper ball is thus 20 2229 3.542 10 1.03 10 .× × = × The number of electrons removed from the copper ball is

911

1950 10 C 3.13 10

1.60 10 C

−

−× = ××

So, the fraction of electrons removed from the copper ball is 11

1122

3.13 10 3.0 101.03 10

−× = ××

Assess: This is indeed a very small fraction of the available number of electrons in the copper ball.

25.73. 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 jr 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 Fr 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

−−×= = = × ≈

× × °

25.74. 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 CP25.74. 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 CP25.74, 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 4.1gtan(10 ) (9.8 N/kg) tan10

F Fmg

F Fmg

− −−

° −= ° =°

− × − ×= = = × =° °

25-36 Chapter 25



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 9

9 2 2 410 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−

−×= = ×°

25.76. Model: Charge Q and the dipole charges (q and −q) are point charges. Visualize: Please refer to Figure CP25.76. Solve: (a) The force on the dipole is the vector sum of the force on q and −q. We have


( /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 ir 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.

M25 KNIG9404 ISM C25

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