Chapter 26
Upcoming SlideShare
Loading in...5
×
 

Chapter 26

on

  • 3,222 views

Chapter 26 solutions Randall D. Knight

Chapter 26 solutions Randall D. Knight

Statistics

Views

Total Views
3,222
Views on SlideShare
3,222
Embed Views
0

Actions

Likes
0
Downloads
18
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Chapter 26 Chapter 26 Document Transcript

  • 26.1. Model: Use the charge model.Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred tothe other because they are relatively free to move. Protons, on the other hand, are tightly bound in nuclei. So,electrons have been removed from the glass rod to make it positively charged.(b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons removed is 8.0 ×10−9 C = 5.0 ×1010 1.60 × 10−19 C
  • 26.2. Model: Use the charge model.Solve: (a) In the process of charging by rubbing, electrons are removed from one material and transferred intothe other because they are relatively free to move. Protons, on the other hand, are tightly bound in the nuclei. So,electrons have been added to the plastic rod to make it negatively charged.(b) Because each electron has a charge of 1.60 × 10−19 C, the number of electrons added is 12 ×10−9 C = 7.5 ×1010 1.60 ×10−19 C
  • 26.3. 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 decreases from −15 nC to −10 nC. That is, −5 nC charge has beenremoved from the plastic. Because it is the negatively charged electrons that are transferred, −5 nC has beenadded to the metal sphere.(b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 5 nC was transferred, the number ofelectrons transferred from the plastic rod to the metal sphere is 5 ×10−9 C = 3.1×1010 1.60 ×10−19 C
  • 26.4. 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 aretransferred, −4.0 nC of electrons has been added to the glass rod. Thus, electrons are removed from the metalsphere and added to the glass rod.(b) Because each electron has a charge of 1.60 ×10−19 C and a charge of 4.0 nC was transferred, number ofelectrons transferred from the metal sphere to the glass rod is 4.0 ×10−9 C = 2.5 ×1010 1.60 × 10−19 C
  • 26.5. Model: Use the charge model.Solve: Each oxygen molecule has 16 protons (8 per atom), and there are 6.02 ×1023 oxygen molecules in 1.0 moleof oxygen. Because one proton has a charge of +1.60 × 10−19 C, the amount of positive charge in 1.0 mole of oxygenis 6.022 ×1023 ×16 ×1.6 × 10−19 C = 1.54 ×106 C
  • 26.6. Model: Use the charge model.Solve: Since the density of water is 1 g/cm3, the mass of 1 L of water is ⎛ 1000 mL ⎞ ⎛ 1 cm ⎞ ⎛ 1 g ⎞ 3(1.0 L ) ⎜ ⎟⎜ ⎟ ⎜ 3 ⎟ = 1.0 kg. Each water molecule (H20) has 10 protons (8 in the oxygen atom and ⎝ L ⎠ ⎝ mL ⎠ ⎝ cm ⎠ 1.0 ×103 gone per hydrogen atom), and thus 10 electrons. The number of moles is = 100 mole. There are 6.02 ×1023 10 g/molewater molecules in 1.0 mole of water. Because one electron has a charge magnitude of 1.60 × 10−19 C, the amount ofnegative charge in 100 mole of water is 100 × 6.022 ×1023 ×10 × 1.6 × 10−19 C = 9.6 × 107 C
  • 26.7. Model: Use the charge model and the model of a conductor as a material through which electronsmove.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 netpositive charge, due to the missing electrons, and thus repel each other. At this point, the electroscope as a wholeis still neutral (no net charge) but has been polarized. On contact, some of the electrons move to the positive rodto neutralize some (but not necessarily all) of the rod’s positive charge. After contact, the electroscope does havea net positive charge. When the rod is removed, the net positive charge on the electroscope quickly spreads tocover the entire electroscope. The net positive charge on the leaves causes them to continue to repel.
  • 26.8. Model: Use the charge model.Solve: (a) No, we cannot conclude that the wall is charged. Attractive electric forces occur between (i) twoopposite charges, or (ii) a charge and a neutral object that is polarized by the charge. Rubbing the balloon doescharge the balloon. Since the balloon is rubber, its charge is negative. As the balloon is brought near the wall, thewall becomes polarized. The positive side of the wall is closer to the balloon than the negative side, so there is anet attractive electric force between the wall and the balloon. This causes the balloon to stick to the wall, with anormal force balancing the attractive electric force and an upward frictional force balancing the very smallgravitational force on the balloon.(b)
  • 26.9. Model: Use the charge model and the model of a conductor as a material through which electronsmove.Solve:The first step shows two neutral metal spheres touching each other. In the second step, the negative rodrepels the negative charges which will retreat as far as possible from the top of the left sphere. Note that thetwo spheres are touching and the net charge on these two spheres is still zero. While the rod is there on topof the left sphere, the right sphere is moved away from the left sphere. Because the right sphere has anexcess negative charge by the same amount as the left sphere has an excess positive charge, the separatedright sphere is negatively charged as shown in the third step. As the two spheres are moved apart furtherand the negatively charged rod is moved away from the spheres, the charges on the two spheres redistributeuniformly over the entire sphere surface. Thus, we have oppositely charged the two spheres.
  • 26.10. Model: Use the charge model and the model of a conductor as a material through which electronsmove.Solve:Charging two neutral spheres with opposite but equal charges 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 oneof the spheres (say, the left sphere). Note that the two spheres are still touching and the net charge on them is zero.The right sphere has an excess positive charge of exactly the same amount as the left sphere’s negative charge. (iii)Separate the spheres while the charged rod remains close to the left sphere. The charge separation remains on thespheres. (iv) Take the charged rod away from the two spheres. The separated charges redistribute uniformly over themetal sphere surfaces.
  • 26.11. Model: Use the charge model and the model of a conductor as a material through which electronsmove.Solve:Charging two neutral spheres with like charges of exactly equal magnitude can be achieved through thefollowing six steps. (i) Bring a charged rod (say, negative) near a neutral metal sphere. (ii) Touch the neutralsphere with the negatively charged rod, so that the rod-sphere system has a net negative charge. (iii) Move therod away from the sphere. The sphere is now negatively charged. (iv) Bring this negatively charged sphere closeto the second neutral sphere. (v) Touch these two spheres. The excess negative charge is distributed evenly overthe two spheres. (vi) Separate the spheres. The excess charge will have the same sign as the charge on thecharging rod and will be evenly distributed between the two spheres.
  • 26.12. Model: Use the charge model.Solve: (a)(b) A charged object causes the charges inside a molecule to move apart while still being a part of the molecule.This is because electrons, that is, the negative charges in an insulator such as a piece of paper cannot movefreely. Thus, the negative part of each molecule is slightly closer to a positively charged object and the positivepart of each molecule is slightly farther from the positively charged object. This charge polarization leads to aslightly greater positive-to-negative attractive force than the positive-to-positive repulsive force. Because thereare many molecules in a piece of paper, the total force of attraction is enough to cause a piece of paper to beattracted to the object. The test will also work for a negatively charged object because then the charges inside themolecules of paper will separate in the other direction.
  • 26.13. Model: Model the charged masses as point charges.Visualize:Solve: (a) The charge q1 exerts a force F1 on 2 on q2 to the right, and the charge q2 exerts a force F2 on 1 on q1 tothe left. Using Coulomb’s law, K q1 q2 ( 9.0 ×10 N m /C )(10 ×10 C )(10 ×10 C ) 9 2 2 −6 −6 F1 on 2 = F2 on 1 = = = 0.90 N (1.0 m ) 2 2 r12(b) Newton’s second law on either q1 or q2 is 0.90 N F1 on 2 = m1a1 ⇒ a1 = = 0.90 m/s 2 1.0 kgAssess: Even a micro-Couomb is a lot of charge. That is why F1 on 2 ( or F2 on 1 ) is a measurable force.
  • 26.14. Model: Model the plastic spheres as point charges.Visualize:Solve: (a) The charge q1 = −50.0 nC exerts a force F1 on 2 on q2 = −50.0 nC to the right, and the charge q2 exertsa force F2 on 1 on q1 to the left. Using Coulomb’s law, K q1 q2 ( 9.0 ×10 N m /C )( 50.0 × 10 C )( 50.0 ×10 C ) 9 2 2 −9 −9 F1 on 2 = F2 on 1 = = = 0.056 N ( 2.0 ×10−2 m ) 2 2 r12(b) The ratio is F1 on 2 0.056 N = = 2.9 mg ( 2.0 ×10−3 kg )( 9.8 m/s2 )
  • 26.15. Model: Model the glass bead and the ball bearing as point charges.Visualize:The ball bearing experiences a downward electric force F1 on 2 . By Newton’s third law, F2 on 1 = F1 on 2 .Solve: Using Coulomb’s law, F1 on 2 = K q1 q2 ⇒ 0.018 N = ( 9.0 ×109 N m 2 /C 2 )( 20 ×10−9 C ) q2 ⇒ q2 = 1.0 ×10−8 C (1.0 ×10−2 m ) 2 2 r12Because the force F1 on 2 is attractive and q1 is a positive charge, the charge q2 is a negative charge. Thus,q2 = −1.0 ×10−8 C = −10 nC.
  • 26.16. Model: Charges A, B, and C are point charges.Visualize: Please refer to Figure EX26.16. Charge A experiences an electric force FB on A due to charge B andan electric force FC on A due to charge C. The force FB on A is directed to the right and the force FC on A is directedto the left.Solve: Coulomb’s law yields: FB on A = ( 9.0 ×10 9 N m 2 /C2 )(1.0 ×10−9 C )(1.0 × 10−9 C ) = 9.0 ×10−5 N (1.0 ×10 m ) −2 2 FC on A = ( 9.0 ×10 9 N m /C )(1.0 ×10 C )( 4.0 ×10 2 2 −9 −9 C) = 9.0 ×10−5 N ( 2.0 ×10 m ) −2 2The net force on A is ˆ ˆ ( ) Fon A = FB on A + FC on A = ( 9.0 × 10−5 N ) i + ( 9.0 ×10−5 N ) −i = 0 N
  • 26.17. Model: Charges A, B, and C are point charges.Visualize: Please refer to Figure EX26.17.Solve: The force on A from charge B is directed upward since the two negative charges repel. FB on A = 1 q A qB = ( 9.0 × 109 Nm 2 /C2 ) (1.0 ×10−9 C )( 2.0 ×10−9 C ) = 4.5 ×10−5 N 4πε 0 r 2 ( 2.0 ×10−2 m ) 2So FB on A = ( 4.5 × 10−5 N, up ) .The force on A from charge C is directed downwards since the two opposite charges attract. FC on A = ( 9.0 ×109 Nm 2 /C 2 ) (1.0 ×10 C )( 2.0 ×10 −9 −9 C) = 2.0 ×10−5 N (3.0 ×10 m ) −2 2So FC on A = (2.0 ×10−5 N, down).The net electric force on charge A is FA = FB on A + FC on A = ( 4.5 ×10−5 N − 2.0 × 10−5 N, up ) = ( 2.5 ×10−5 N, up )
  • 26.18. 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 isdue to charge A only.Solve: Coulomb’s law gives the magnitude of the forces between the charge. Thus, FA on B = FB on A = ( 9.0 ×10 9 N m 2 /C 2 )( 20.0 ×10−9 C )(10.0 ×10−9 C ) = 4.5 ×10−3 N ( 2.0 ×10−2 m ) 2Because the charge on object A is positive and on object B is negative, FB on A is upward and FA on B is downward.Thus, FB on A = +4.5 ×10−3 ˆ N j FA on B = −4.5 × 10−3 ˆ N j
  • 26.19. Model: Assume the glass bead, the proton, and the electron are point charges.Visualize:Solve: Coulomb’s law gives Fbead on electron = Fbead on proton = ( 9 ×10 9 N m 2 /C2 )( 20 × 10−9 C )(1.60 ×10−19 C ) = 2.88 ×10−13 N (1.0 ×10 m) −2 2(a) Newton’s second law is F = ma, so Fbead on proton 2.88 × 10−13 N aproton = = = 1.72 ×1014 m/s 2 mproton 1.67 ×10−27 kgIn vector form aproton = (1.72 × 1014 m/s 2, away from bead )(b) Similarly, Fbead on electron 2.88 ×10−13 N aelectron = = = 3.2 × 1017 m/s 2 melectron 9.11×10−31 kgThus aelectron = ( 3.2 ×1017 m/s 2 , toward bead ) .
  • 26.20. Model: Protons and electrons produce electric fields.Solve: (a) The electric field of the proton is ⎡ ⎤ 1 q +1.60 ×10−19 C ⎥ E= r = ( 9.0 ×109 N m 2 /C 2 ) ⎢ ˆ r = (1.44 × 10−3 N/C, away from proton ) ˆ 4πε 0 r 2 ⎢ (1.0 ×10−3 m ) ⎥ 2 ⎣ ⎦(b) The electric field of the electron is E = (1.44 ×10−3 N/C ) ( − r ) = (1.44 ×10−3 N/C, toward electron ) ˆ
  • 26.21. 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, ( ) ( Fproton = e 200i + 400ˆ N/C = 3.20i + 6.40ˆ × 10−17 N ˆ j ˆ j ) −19where we used e = 1.60 ×10 C. ((b) The charge on an electron is q = −e. Thus, Felectron = − Fproton = −3.20i − 6.40 ˆ × 10−17 N ˆ j )(c) From Newton’s second law, Fproton Fx2 + Fy2 7.15 × 10−17 N aproton = = = = 4.28 ×1010 m/s 2 mproton mproton 1.67 ×10−27 kg(d) The electron experiences a force of the same magnitude but it has a different mass. Thus, 7.15 ×10−17 N aelectron = = 7.85 ×1013 m/s 2 9.11×10−31 kgThe 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.
  • 26.22. 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 1 q q E= ⇒ 1.0 N/C = ( 9.0 ×109 N m 2 /C 2 ) ⇒ q = 1.11×10−10 C = 0.111 nC 4πε 0 r (1.0 m ) 2 2
  • 26.23. Model: The electric field is that of a positive charge on the glass bead. The charge is assumed to be apoint charge.Solve: The electric field is 8.0 × 10−9 C E = ( 9.0 ×109 N m 2 /C 2 ) r = 1.80 ×105 r N/C ˆ ˆ ( 2.0 ×10 m) −2 2 ˆwhere r is the unit vector from the charge to the point at which we calculated the field. That is, the direction ofthe electric field is away from the bead.
  • 26.24. Model: The electric field is that of the charge on the object. Assume the charge on the object is apoint charge.Solve: The electric field at a distance r from a point charge q is 1 q E= ˆ r 4πε 0 r 2Because the electric field points toward the object, E = (180,000 N/C )( − r ) . Thus, ˆ q − (180,000 N/C ) = ( 9.0 ×109 N m 2 /C2 ) ⇒ q = −8.0 × 10−9 C = −8.0 nC ( 2.0 ×10−2 m ) 2
  • 26.25. 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 Σ Fnet y = Fon q − FG . To balance the gravitational force withthe electric force, mg (1.0 × 10 kg ) ( 9.8 N/kg ) −3 Fon q = FG ⇒ q E = mg ⇒ E = = = 3.3 × 106 N/C q 3.0 ×10−9 CBecause Fon q must be upward and the charge is negative, the electric field at the location of the plastic ball mustbe pointing downward. Thus E = ( 3.3 ×106 N/C, downward ).Assess: F = qE means the sign of the charge q determines the direction of F or E. For positive q, E and Fare pointing in the same direction. But E and F point in opposite directions when q is negative.
  • 26.26. Model: A field is the agent that exerts an electric force on a charge.Visualize:Solve: ( ) (a) To balance the gravitational force on a proton Σ ( Fnet ) y = Fon p − FG p = 0 N. This means mg (1.67 ×10 kg ) ( 9.8 N/kg ) −27 Fon p = ( FG )p ⇒ q E = mg ⇒ E = = = 1.02 × 10−7 N/C q 1.60 ×10−19 CBecause Fon p must be upward and the proton charge is positive, the electric field at the location of the protonmust also be pointing upward. Thus E = (1.02 × 10−7 N/C, downward ) .(b) In the case of the electron, mg ( 9.11×10 kg ) ( 9.8 N/kg ) −31 E= = = 5.58 ×10−11 N/C q 1.60 ×10−19 CBecause Fon e must be upward and the electron has a negative charge, the electric field at the location of theelectron must be pointing downward. Thus E = ( 5.58 × 10−11 N/C, downward ) .
  • 26.27. 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 ⎛ 1 q ⎞ E =⎜ , away from q ⎟ ⎝ 4πε 0 r 2 ⎠Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = 12 ×10−9 C, ⎛ 108 N m 2 /C ⎞ E =⎜ , away from q ⎟ ⎝ r2 ⎠The electric fields at points A, B, and C are 108 N m 2 /C ˆ ˆ EA = i = 4.3 ×104 i N/C (5.0 ×10 m ) −2 2 108 N m 2 /C ⎡ 1 ˆ ˆ ⎤ EB = ( ) ( 4 ˆ ) 4 ˆ ⎢ 2 −i + j ⎥ = −1.53 ×10 i + 1.53 × 10 j N/C ( −5.0 ×10 m ) + ( 5.0 ×10 m) ⎣ 2 2 −2 −2 ⎦ 108 N m 2 /C ⎡ 1 ˆ ˆ ⎤ EC = ( ) ( 4 ˆ ) 4 ˆ ⎢ 2 −i − j ⎥ = −1.53 ×10 i − 1.53 ×10 j N/C ( −5.0 ×10 m ) + ( −5.0 ×10 m) ⎣ 2 2 −2 −2 ⎦(b) The three vectors are shown in the diagram.Assess: The vectors EA , EB , and EC are pointing away from the positive charge.
  • 26.28. Model: The electric field is that of a negative 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 ⎛ 1 q ⎞ E =⎜ , away from q ⎟ ⎝ 4πε 0 r 2 ⎠Using 1 4πε 0 = 9.0 ×109 N m 2 /C2 and q = −12 ×10−9 C, ⎛ 108 N m 2 /C ⎞ E =⎜ 2 , toward q ⎟ ⎝ r ⎠The electric fields at points A, B, and C are 108 N m 2 /C EA = ( − ˆj ) = −4.3 ×10 4 ˆ N/C j ( 5.0 ×10 m) −2 2 108 N m 2 /C ⎡ 1 ˆ ˆ ⎤ EB = ( ) ( 4 ˆ 4 ˆ ) ⎢ 2 i + j ⎥ = 1.53 × 10 i + 1.53 × 10 j N/C ( −5.0 ×10 m ) + ( −5.0 ×10 m) 2 2 −2 −2 ⎣ ⎦ 108 N m 2 /C ⎡ 1 ˆ ˆ ⎤ EC = ⎢ ( ) ( i − j ⎥ = 1.53 × 104 i − 1.53 × 104 ˆ N/C ˆ j ) ( −5.0 ×10 m ) + ( 5.0 ×10 m) ⎣ 2 2 2 −2 −2 ⎦(b) The three vectors are shown in the diagram.Assess: Note that the vectors EA , EB , and EC are pointing toward the negative charge.
  • 26.29. Model: Use the charge model and the model of a conductor as a material through which electrons move.Solve: Because the metal spheres are identical, the total charge is split equally between the two spheres. Thatis, qA = qB = 5.0 × 1011 electrons. Thus, the charge on metal spheres A and B is( 5.0 ×10 )( −1.60 ×10 11 −19 C ) = −80 nC.
  • 26.30. Model: Use the charge model and the model of a conductor as a material through which electrons move.Solve: Plastic is an insulator and does not transfer charge from one sphere to the other. The charge of metalsphere A is (1.0 ×1012 )( −1.60 ×10−19 C ) = −160 nC and the charge of metal sphere B is 0 C.
  • 26.31. Model: Use the charge model.Solve: The number of moles in the penny is M 3.1 g n= = = 0.04882 mol A 63.5 g/molThe number of copper atoms in the penny is N = nN A = ( 0.04882 mol ) ( 6.02 ×1023 mol−1 ) = 2.939 ×1022Since each copper atom has 29 electrons and 29 protons, the total positive charge in the copper penny is ( 29 × 2.939 ×10 )(1.60 ×10 22 −19 C ) = 1.36 ×105 CSimilarly, the total negative charge is −1.36 × 105 C.Assess: Total positive and negative charges are equal in magnitude.
  • 26.32. 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 byNewton’s third law, and is F=K q1 q2 = ( 9.0 ×10−9 Nm 2 /C2 ) ( 4.0 ×10−9 C )(8.0 ×10−9 C ) = 7.2 ×10−4 N ( 2.0 ×10−2 m ) 2 r2The beads accelerate at different rates because their masses are different. By Newton’s second law, theacceleration of the bead on the left is F ( 7.2 ×10 N ) −4 aleft = = = 0.36 m/s 2 to the right. m ( 2.0 ×10−3 kg )For the bead on the right, the acceleration is aright = ( 7.2 ×10 −4 N) = 0.180 m/s 2 to the left. ( 4.0 ×10 −3 kg )
  • 26.33. Model: The protons are point charges.Solve: (a) The electric force between the protons is q1 q2 ( 9.0 ×10 N m /C )(1.60 × 10 C )(1.60 ×10 C ) 9 2 2 −19 −19 FE = K = = 58 N ( 2.0 ×10−15 m ) 2 r2(b) The gravitational force between the protons is G m1m2 ( 6.67 ×10 N m 2 /kg 2 )(1.67 ×10−27 kg )(1.67 ×10−27 kg ) −11 FG = = = 4.7 ×10−35 N ( 2.0 ×10 m) 2 r2 −15(c) The ratio of the electric force to the gravitational force is FE 58 N = = 1.23 × 1036 FG 4.7 ×10−35 N
  • 26.34. Model: The 125Xe nucleus and the proton are point charges. That is, all the charge on the Xe nucleusis 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: Fnucleus on proton = K qnucleus qproton = ( 9.0 ×10 9 N m 2 /C2 )( 54 ×1.60 ×10−19 C )(1.60 ×10−19 C ) = 5.0 ×102 N ( 5.0 ×10 m) 2 2 r −15(b) Applying Newton’s second law to the proton, 5.0 ×102 N Fon proton = mproton aproton ⇒ aproton = = 3.0 × 1029 m/s 2 1.67 × 10−27 kg
  • 26.35. Model: The two charged spheres are point charges.Solve: The electric force on one charged sphere due to the other charged sphere is equal to the sphere’s masstimes its acceleration. Because the spheres are identical and equally charged, m1 = m2 = m and q1 = q2 = q. Wehave Kq1q2 Kq 2 F2 on 1 = F1 on 2 = = 2 = ma r2 r mar 2 (1.0 × 10 kg )(150 m/s )( 2.0 × 10 m ) −3 2 −2 2 ⇒q =2 = = 6.7 × 10−15 C 2 K 9.0 ×109 N m 2 /C 2 ⇒ q = 8.2 × 10−8 C = 82 nC
  • 26.36. Model: Objects A and B are point charges.Visualize:Solve: (a) It is given that FA on B = 0.45 N. By Newton’s third law, FB on A = FA on B = 0.45 N.(b) Coulomb’s law is KqA qB K ( 2qB )( qB ) FB on A = FA on B = 0.45 N = = r2 r2 ( 0.45 N ) (10 ×10−2 m ) 2 ⇒ qB = ( 0.45 N ) r 2 = = 5.0 × 10−7 C ⇒ qA = 2qB = 1.0 × 10−6 C 2K 2 ( 9.0 ×109 N m 2 /C 2 )(c) Newton’s second law is FB on A = mAaA. Hence, FB on A FA on B 0.45 N aA = = = = 4.5 m/s 2 mA mB 0.100 kg
  • 26.37. Model: The charges are point charges.Visualize: Please refer to Figure P26.37.Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nCcharge, q2 is the 2.0 nC charge, and q3 is the other 2.0 nC charge. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ ⎝ r2 ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ ⎜ (1.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ ( = (1.8 ×10−4 N, away from q2 ) = (1.8 ×10−4 N ) cos60°i + sin 60° ˆ ˆ j) ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ r2 ( , away from q3 ⎟ = (1.8 × 10−4 N, away from q3 ) = (1.8 × 10−4 N ) − cos60°i + sin 60 ˆ ˆ j ) ⎝ ⎠ ⇒ Fon 1 = F2 on 1 + F3 on 1 = 2 (1.8 ×10−4 N ) sin 60° ˆ = 3.1×10−4 ˆ N j jSo the force on the 1.0 nC charge is 3.1×10−4 N directed upward.
  • 26.38. Model: The charges are point charges.Visualize: Please refer to Figure P26.38.Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1 , where q1 is the 1.0 nCcharge, q2 is the 2.0 nC charge, and q3 is the −2.0 nC charges. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ ⎝ r2 ⎠ ⎛ ( 9.0 ×10 N m /C )(1.0 × 10−19 C )( 2.0 × 10−19 C ) 9 2 2 ⎞ =⎜ , away from q2 ⎟ ⎜ (1.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ ( = (1.80 × 10−4 N, away from q2 ) = (1.80 × 10−4 N ) cos60°i + sin 60° ˆ ˆ j ) ⎛ K q1 q2 ⎞ F3 on 1 = ⎜ 2 ( ˆ j) , toward q3 ⎟ = (1.80 ×10−4 N, toward q3 ) = (1.80 × 10−4 N ) cos60°i − sin 60° ˆ . ⎝ r ⎠ ⇒ Fon 1 = F2 on 1 + F3 on 1 = (1.80 ×10−4 N ) ( 2cos60° ) i = 1.80 ×10−4 i N ˆ ˆSo, the force on the 1.0 nC charge is 1.80 × 10–4 N and it is directed to the right.
  • 26.39. Model: The charges are point charges.Visualize:Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ ⎝ r2 ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(10 × 10−9 C )( 5.0 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ ⎜ (1.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ = ( 4.5 ×10−3 N, away from q2 ) = −4.5 × 10−3 ˆ Nj ⎛ K q1 q2 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ ⎝ r2 ⎠ ⎛ ( 9.0 ×10 N m /C )(10 ×10−9 C )(15 ×10−9 C ) 9 2 2 ⎞ =⎜ , toward q3 ⎟ ⎜ ( 3.0 ×10−2 m ) + (1.0 ×10−2 m ) ⎟ 2 2 ⎝ ⎠ ( = (1.35 × 10−3 N, toward q3 ) = (1.35 ×10−3 N ) − cosθ i + sin θ ˆ ˆ j )From the geometry of the figure, ⎛ 1.0 cm ⎞ θ = tan −1 ⎜ ⎟ = 18.4° ⎝ 3.0 cm ⎠This means cos θ = 0.949 and sin θ = 0.316. Therefore, ( F3 on 1 = −1.28 ×10−3 i + 0.43 ×10−3 ˆ N ˆ j ) ( ) ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.28 × 10−3 i − 4.07 × 10−3 ˆ N ˆ jThe magnitude and direction of the resultant force vector are ( −1.28 ×10 N ) + ( −4.07 × 10−3 N ) = 4.3 × 10−3 N −3 2 2Fon 1 = 4.07 ×10−3 Ntanφ = = 3.180 ⇒ φ = tan −1 ( 3.180 ) = 72.5° below the − x axis, 1.28 ×10−3 Nor Fon 1 points 252.5° counterclockwise from the +x-axis.
  • 26.40. Model: The charges are point charges.Visualize:Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ 2 , toward q2 ⎟ ⎝ r ⎠ ⎛ ( 9.0 × 10 N m /C )(10 × 10−9 C )(10 × 10−9 C ) 9 2 2 ⎞ =⎜ , toward q2 ⎟ ⎜ (1.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ = ( 0.90 × 10 N, toward q2 ) = 0.90 × 10 j N −3 −3 ˆ ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ 2 , toward q3 ⎟ ⎝ r ⎠ ⎛ ( 9.0 × 109 N m 2 /C 2 )(10 × 10−9 C ) ( 8.0 × 10−9 C ) ⎞ =⎜ , toward q3 ⎟ ⎜ ( 3.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ = ( 8.0 × 10 N, toward q3 ) = −0.80 × 10 i N −4 −3 ˆ ( ⇒ Fon 1 = F2 on 1 + F3 on 1 = −0.80 × 10−3 i + 0.90 × 10−3 ˆ N ˆ j )The magnitude and direction of the resultant force vector are ( −0.80 ×10 N ) + ( 0.90 × 10−3 N ) = 1.20 × 10−3 N −3 2 2 Fon 1 = 0.90 × 10−3 N tan φ = ⇒ φ = tan −1 (1.13) = 48° above the − x -axis, 0.80 × 10−3 Nor Fon 1 points 132° counterclockwise from the +x-axis.
  • 26.41. Model: The charges are point charges.Visualize:Solve: The electric force on charge q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ , away from q2 ⎟ ⎝ r2 ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )( 5 ×10−9 C )(10 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ ⎜ ( 4.0 ×10−2 m ) + ( 3.0 ×10−2 m ) ⎟ 2 2 ⎝ ⎠ ( = (1.8 ×10−4 N, away from q2 ) = (1.8 × 10−4 N ) − cosθ i − sin θ ˆ ˆ j )From the geometry of the figure, tan θ = 4.0 cm 3.0 cm ˆ ( ⇒ θ = 53.13° ⇒ F2 on 1 = (1.8 ×10−4 N ) −0.6i − 0.8 ˆ j ) ⎛ ( 9.0 ×109 N m 2 /C 2 )( 5 ×10−9 C )( 5 × 10−9 C ) ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = ( 2.5 ×10−4 N, toward q3 ) = 2.5 × 10−4 i N ˆ ⎜ ( 3.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ ( ) ⇒ Fon 1 = F2 on 1 + F3 on 1 = 1.42 × 10−4 i − 1.44 × 10−4 ˆ N ˆ jThe magnitude and direction of the resultant force vector are (1.42 ×10 N ) + ( −1.44 ×10−4 N ) = 2.0 × 10−4 N −4 2 2 Fon 1 = ⎛ 1.44 ×10−4 N ⎞ φ = tan −1 ⎜ −4 ⎟ = 45° clockwise from the +x-axis. ⎝ 1.42 ×10 N ⎠
  • 26.42. Model: The charges are point charges.Visualize:Solve: The electric force on q1 is the vector sum of the forces F2 on 1 and F3 on 1. We have Error! Objects cannot be created from editing field codes. Error! Objects cannot be created from editing field codes.Error! Objects cannot be created from editing field codes.Error! Objects cannot be created from editing field codes.Error! Objects cannot be created from editing field codes. Error! Objects cannot be created from editing field codes.From the geometry of the figure, 4.0 cm tan θ = ⇒ cosθ = 0.6 and sinθ = 0.8 3.0 cm ( ) ( ) ⇒ F3 on 1 = −1.08 × 10−4 i − 1.44 ×10−4 ˆ N ⇒ Fon 1 = F2 on 1 + F3 on 1 = −1.08 × 10−4 i + 1.36 × 10−4 ˆ N ˆ j ˆ jThe magnitude and direction of the resultant force vector are (1.08 ×10 N ) + ( −1.36 × 10−4 N ) = 1.74 × 10−4 N −4 2 2 Fon 1 = 1.36 ×10−4 N tan φ = ⇒ φ = 52° clockwise from the −x-axis, or 128° CCW from +x-axis. 1.08 ×10−4 N
  • 26.43. Model: The charges are point charges.Visualize: Please refer to Figure P26.43.Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3charge is in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant.The electric force on q1 is the vector sum of the forces F2 on 1 , F3 on 1 , F4 on 1 , and F5 on 1. The magnitude of thesefour forces is the same because all four charges are equal and equidistant from q1. So, F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 = ( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 9 2 2 −9 −9 C) = 3.6 ×10−4 N ( 0.50 ×10 m ) + ( 0.50 ×10 m ) −2 2 −2 2Thus, F on 1 = (3.6 × 10−4 N, toward q2) + (3.6 × 10−4 N, toward q3) + (3.6 × 10−4 N, toward q4) + (3.6 × 10−4 N,toward q5). In component form, (⎣ j ) ( j ) ( ˆ j ) ( Fon 1 = Fon 1 ⎡ cos 45°i + sin 45° ˆ + cos 45°i − sin 45° ˆ + − cos 45°i − sin 45° ˆ + − cos 45°i + sin 45° ˆ ⎤ = 0 N ˆ ˆ ˆ j ⎦ )
  • 26.44. Model: The charges are point charges.Visualize: Please refer to Figure P26.44.Solve: Placing the 1.0 nC charge at the origin and calling it q1, the q2 charge is in the first quadrant, the q3 chargeis in the fourth quadrant, the q4 charge is in the third quadrant, and the q5 charge is in the second quadrant. Theelectric force on q1 is the vector sum of the electric forces from the other four charges q2, q3, q4, and q5. Themagnitude of these four forces is the same because all four charges are equal in magnitude and are equidistant fromq1. So, F2 on 1 = F3 on 1 = F4 on 1 = F5 on 1 = ( 9.0 ×10 N m /C )( 2.0 ×10 C )(1.0 ×10 9 2 2 −9 −9 C) = 3.6 ×10−4 N ( 0.50 ×10 m ) + ( 0.50 ×10 m ) −2 2 −2 2Thus, Fon 1 = (3.6 × 10−4 N, away from q2) + (3.6 × 10−4 N, away from q3) + (3.6 × 10−4 N, toward q4) + (3.6 ×10−4 N, toward q5). In component form, (⎣ j ) ( j ˆ) ( j ) ( Fon 1 = Fon 1 ⎡ − cos 45°i − sin 45° ˆ + − cos 45°i + sin 45° ˆ + − cos 45°i − sin 45° ˆ + − cos 45°i + sin 45° ˆ ⎤ ˆ ˆ ˆ j ⎦ ) −4 ( ˆ ) = ( 3.6 ×10 N ) −4cos 45°i = −1.02 × 10 i N −3 ˆ
  • 26.45. Model: The charges are point charges.Visualize: Please refer to Figure P26.45.Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the firstquadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electricforces from the other three charges q2, q3, and q4. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ 2 , away from q2 ⎟ ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , away from q2 ⎟ ⎜ ( 5.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ = ( 0.72 ×10 N, away from q2 ) = ( 0.72 ×10 N ) − cos 45°i −5 −5 ( ˆ − sin 45° ˆ j ) ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ 2 , toward q3 ⎟ ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C2 )(1.0 ×10−9 C )( 6.0 ×10−9 C ) ⎞ =⎜ , toward q3 ⎟ ⎜ ( 5.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ = ( 2.16 ×10 N, away from q3 ) = 2.16 ×10 j N −5 −5 ˆ ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ r2 ( , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°i − sin 45° ˆ ˆ j ) ⎝ ⎠ ⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ⎡( 2.16 ×10 −5 N ) − ( 0.72 ×10 −5 N ) ( 2sin 45° ) ⎤ ˆ = 1.14 × 10−5 ˆ N ⎣ ⎦j j
  • 26.46. Model: The charges are point charges.Visualize: Please refer to Figure P26.46.Solve: Placing the 1.0 nC charge at the origin and calling it q1, the −6.0 nC is q3, the q2 charge is in the firstquadrant, and the q4 charge is in the second quadrant. The net electric force on q1 is the vector sum of the electricforces from the other three charges q2, q3, and q4. We have ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ 2 , toward q2 ⎟ ⎝ r ⎠ ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C )( 2.0 ×10−9 C ) ⎞ =⎜ , toward q2 ⎟ ⎜ ( 5.0 ×10−2 m ) ⎟ 2 ⎝ ⎠ −5 −5 ( = ( 0.72 × 10 N, toward q2 ) = ( 0.72 × 10 N ) cos 45°i ˆ + sin 45° ˆ j ) ⎛ K q1 q3 ⎞ F3 on 1 = ⎜ , toward q3 ⎟ = ( 2.16 ×10−5 N, toward q3 ) = 2.16 × 10−5 ˆ N j ⎝ r2 ⎠ ⎛ K q1 q4 ⎞ F4 on 1 = ⎜ 2 ( , away from q4 ⎟ = ( 0.72 × 10−5 N ) cos 45°i − sin 45° ˆ ˆ j ) ⎝ r ⎠⇒ Fon 1 = F2 on 1 + F3 on 1 + F4 on 1 = ( 0.72 ×10−5 N ) ( 2cos 45° ) i + ( 2.16 ×10−5 N ) j ˆ ( = 1.02 × 10−5 i + 2.2 × 10−5 ˆ N ˆ j )
  • 26.47. Model: The charged particles are point charges.Visualize:Solve: (a) The mathematical problem is to find the position for which the forces F1 on p and F2 on p are equal inmagnitude and opposite in direction. If the proton is at position x, it is a distance x from q1 and d − x from q2,where d = 1.0 cm. The magnitudes of the forces are K q1 qp K q1 qp K q2 qp F1 on p = = F2 on p = (d − x) 2 2 r1p x2Equating the two forces and using d = 1.0 cm, K q1 qp K q2 qp 2.0 nC 4.0 nC = ⇒ = ⇒ 1 + x2 − 2x = 2x2 ⇒ x2 + 2x − 1 = 0 (d − x) (d − x) 2 2 x2 x2The solutions to the equation are x = +0.414 cm and −2.41 cm. Both are points where the magnitudes of the twoforces are equal, but +0.414 cm is a point where the magnitudes are equal but the directions are the same. Thesolution 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 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 Fon p is zero for a proton, the electric field at that point must be zero. Thus, there will be no forceon any charged particle at that point.
  • 26.48. Model: The charged particles are point charges.Visualize:Solve: The two 2.0 nC charges exert an upward force on the 1.0 nC charge. Since the net force on the 1 nCcharge is zero, the unknown charge must exert a downward force of equal magnitude. This implies that q is apositive charge. The force of charges 2 on charge 1 is ⎛ K q1 q2 ⎞ F2 on 1 = ⎜ 2 , away from q2 ⎟ ⎝ r12 ⎠ (9.0 ×10 N m / C )(1.0 × 10−9 C)(2.0 ×10−9 C) 9 2 2 = (cosθ i + sinθ ˆ) ˆ j (0.020 m) 2 + (0.030 m) 2From the figure, θ = tan–1(2/3) = 33.69°. Thus F2 on 1 = (1.152 ×10−5 i + 0.768 ×10−5 ˆ) N ˆ jFrom symmetry, F3 on 1 is the same except the x-component is reversed. When we add F2 on 1 and F3 on 1 , the x-components cancel and the y-components add to give F 2 on 1 + F 3 on 1 = 1.536 × 10−5 ˆ N jFq on 1 must have the same magnitude, pointing in the − ˆ direction, so j K q q1 (1.536 ×10−5 N)(0.020 m) 2 Fq on 1 = 1.536 ×10−5 N = ⇒q= = 0.68 nC r 2 (9.0 ×109 N m 2 / C 2 )(1.0 × 10−9 C)A positive charge q = 0.68 nC will cause the net force on the 1.0 nC charge to be zero.
  • 26.49. Model: The charged particles are point charges.Visualize: Please refer to Figure P26.49.Solve: The charge q2 is in static equilibrium, so the net electric field at the location of q2 is zero. We have 1 q1 1 ( 3.0 ×10 −9 C) Enet = Eq1 + E−3 nC = 4πε 0 ( 0.20 m ) 2 ( ±iˆ ) + 4πε ( 0.10 m ) 2 ( −iˆ ) = 0 N/C 0 ˆWe have used the ± sign to indicate that a positive charge on q1 leads to an electric field along +i and a negative ˆ ˆcharge on q1 leads to an electric field along −i . Because the above equation can only be satisfied if we use +i ,we infer that the charge q1 is a positive charge. Thus, −9 q1 ( ) ˆ 3.0 ×10 C i = 0 N/C ⇒ q1 = 12.0 nC +i − () ˆ ( 0.20 m ) ( 0.10 m ) 2 2
  • 26.50. Model: The charged particles are point charges.Visualize:Solve: The force on q is the vector sum of the force from −Q and +Q. We have ⎛ −Q + q ⎞ KQq F−Q on + q = ⎜ K 2 a +y 2 , toward − Q ⎟ = 2 a + y2 ( − cosθ i − sinθ ˆ ˆ j ) ⎝ ⎠ ⎛ K +Q + q ⎞ KQq F+ Q on + q = ⎜ 2 a + y2 , away from + Q ⎟ = 2 a + y2 ( − cosθ i + sin θ ˆ ˆ j ) ⎝ ⎠ KQq ⇒ Fnet = ( −2cosθ ) iˆ + 0 ˆ N j a2 + y2From the figure we see that cosθ = a a 2 + y 2 . Thus −2 KQqa ( Fnet ) x = (a + y2 ) 2 3/ 2Assess: Note that (Fnet)x = Fnet because the y-components of the two forces cancel each other out.
  • 26.51. Model: The charged particles are point charges.Visualize:Solve: (a) The force on q is the vector sum of the force from −Q and +Q. We have ⎛ K +Q + q ⎞ KQq F+ Q on + q = ⎜ ⎜ (a − x) 2 , away from + Q ⎟ = ⎟ ( a − x )2 ˆ −i ( ) ⎝ ⎠ ⎛ K −Q + q ⎞ KQq F− Q on + q = ⎜ ⎜ ( a + x )2 , toward − Q ⎟ = ⎟ ( a + x )2 ˆ −i ( ) ⎝ ⎠ ⎡ 1 1 ⎤ 2 KQq ( a 2 + x 2 ) ⇒ ( Fnet ) x = − KQq ⎢ + ⎥=− ⎢(a − x) (a + x) ⎥ ( a2 − x2 ) 2 2 2 ⎣ ⎦To arrive at the final expression we used ( a − x ) ( a + x ) = ⎡( a − x )( a + x ) ⎤ = ( a 2 − x 2 ) . 2 2 2 2 ⎣ ⎦(b) There are two cases when x > a. For x > a, ⎛ K +Q + q ⎞ KQq F+ Q on + q = ⎜ ⎜ ( x − a) 2 , away from + Q ⎟ = ⎟ ( x − a )2 ˆ +i ( ) ⎝ ⎠ ⎛ K −Q + q ⎞ KQq F− Q on + q = ⎜ ⎜ ( x + a )2 , toward − Q ⎟ = ⎟ ( x + a )2 ˆ −i ( ) ⎝ ⎠ ⎡ 1 1 ⎤ −4 KQqax ⇒ ( Fnet ) x = KQq ⎢ − 2⎥ = ⎢ ( x − a ) ( x + a ) ⎥ ( x2 − a2 ) 2 2 ⎣ ⎦For x < − a (that is, for negative values of x), KQq KQq F+ Q on + q = ( x − a) 2 ( −iˆ ) F− Q on + q = (a + x) 2 ( +iˆ ) 4 KQqax ⇒ ( Fnet ) x = − (x − a2 ) 2 2
  • That is, the net force is to the right when x > a and to the right when x < − a. We can combine these two casesinto a single equation for x > a : 4 KQqa x ( Fnet ) x = (x − a2 ) 2 2Here, the force is always to the right when x > a.
  • 26.52. Model: The charges are point charges.Solve:We will denote the charges −Q, 4Q and −Q by 1, 2, and 3, respectively. ⎛ K −Q q ⎞ F1 on q = ⎜ L2 KQq ˆ , toward − Q ⎟ = 2 −i L ( ) ⎝ ⎠ ⎛ ⎞ K 4Q q 4 KQq F2 on q = ⎜ ⎜ , away from 4Q ⎟ = ⎟ ( ) +cos 45°i + sin 45° ˆ ⇒ F2 on q = ˆ j 2 KQq = 2 F1 on q ( ) 2 ⎜ 2L ⎟ 2 L2 L2 ⎝ ⎠ ⎛ K −Q q ⎞ KQq F3 on q = ⎜ L2 , toward − Q ⎟ = 2 − ˆ L j( ) ⎝ ⎠The net electric force on the charge +q is the vector sum of the electric forces from the other three charges. Thenet force is KQq ˆ 2 KQq ⎛ iˆ ˆ ⎞ KQq Fnet = L2 ( ) −i + ⎜+ L2 ⎝ 2 + j 2⎠ L j KQq ˆ L ( ) KQq L j ( ⎟ + 2 − ˆ = − 2 i 1− 2 − 2 ˆ 1− 2 ) ( ) 2 2 ⎡ KQq ⎣ L ( ⎤ ⎡ KQq ⎦ ⎣ L ) ⎤ (⇒ Fnet = ⎢ 2 1 − 2 ⎥ + ⎢ 2 1 − 2 ⎥ = 2 − 2 ⎦ ) KQq L2 ( )
  • 26.53. Model: The charges are point charges.Visualize:We must first identify the region of space where the third charge q3 is located. You can see from the figure thatthe forces can’t possibly add to zero if q3 is above or below the axis or outside the charges. However, at somepoint 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 F1 on 3 and F2 on 3 are equal inmagnitude. If q3 is the distance x from q1, it is the distance L − x from q2. The magnitudes of the forces are K q1 q3 Kq q3 K q2 q3 K ( 4q ) q3 F1 on 3 = = F2 on 3 = = ( L − x) 2 2 2 r13 x2 r23Equating the two forces, Kq q3 K ( 4q ) q3 L ⇒ ( L − x ) = 4 x 2 ⇒ x = and − L 2 = ( L − x) 2 2 x 3The solution x = −L is not allowed as you can see from the figure. To find the magnitude of the charge q3, weapply the equilibrium condition to charge q1: K q2 q1 K q3 q1 4 F2 on 1 = F3 on 1 ⇒ = ⇒ 4q = 9 q3 ⇒ q3 = q ( 3 L) 2 L2 1 9We are now able to check the static equilibrium condition for the charge 4q (or q2): q1 q2 K q3 q2 q 4 q q F1 on 2 = F3 on 2 ⇒ K = ⇒ 2= 9 2= 2 L2 ( L − x) 2 L ( 2 L) 3 LThe sign of the third charge q3 must be negative. A positive sign on q3 will not have a net force of zero either onthe charge q or the charge 4q. In summary, a charge of − 9 q placed x = 1 L from the charge q will cause the 3- 4 3charge system to be in static equilibrium.
  • 26.54. 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 ⎛ 4π 3 ⎞ ⎡ 4π 3⎤ M = ρV = ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 ×10−5 kg = 0.03736 g ⎝ 3 ⎠ ⎣ 3 ⎦The number of moles in the sphere is M 0.03742 g n= = = 5.884 ×10−4 mol A 63.5 g/molThe number of copper atoms in the sphere is N = nN A = ( 5.884 × 10−4 mol )( 6.02 ×1023 mol−1 ) = 5.542 ×1020The number of electrons in the copper sphere is thus 29 × 3.542 × 1020 = 1.027 × 1022. The total positive ornegative charge in the sphere is (1.027 × 1022 )(1.60 × 10−19 C ) = 1643 C. Hence, the spheres will have a netcharge of 10−9 × 1643 C = 1.643 × 10−6 C. The force between two such spheres is (9.0 ×10 N m 2 /C2 )(1.643 × 10−6 C ) 9 2 Kq1q2 F= = = 2.4 × 102 N (1.0 ×10 m) (1.0 ×10 m) −2 2 −2 2(b) This is a force that is easily detectable. Since we don’t observe such forces, any difference between theproton charge and the electron charge must be smaller than 1 part in 109.
  • 26.55. Model: The electron and the proton are point charges.Solve: The electric Coulomb force between the electron and the proton provides the centripetal acceleration forthe electron’s circular motion. Thus, K ( e )( e ) mv 2 = = mrω 2 r2 r ( 9.0 ×10 N m /C )(1.60 ×10 C ) 9 2 2 −19 2 Ke 2 1 rev⇒ f = = = 4.12 ×1016 rad/s × = 6.6 × 1015 rev/s mr 3 ( 9.11×10 kg )( 5.3 ×10 ) −31 −11 3 2π rad
  • 26.56. 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 metalplate into you, the plate becomes as much positively charged as you become negatively charged. When enoughcharge difference builds up between you and the metal plate, the gravitational force on you will becounterbalanced by the upward electrical force. You will slowly begin to hang suspended in the air when K q −q Fplate on you = FG ⇒ = mg ( 2.0 m ) 2 mg ( 2.0 m ) ( 60 kg )( 9.8 N/kg )( 2.0 m ) 2 2 ⇒q = = = 5.11× 10−4 C K 9.0 ×10 N m /C 9 2 2Dividing this charge by the charge on an electron yields 3.2 ×1015 electrons.
  • 26.57. Model: The charged plastic beads are point charges and the spring is an ideal spring that obeys Hooke’slaw.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. Thatis, Kq 2 k Δx r 2 2 = k Δx ⇒ q = r KThe spring constant k is obtained by noting that the weight of a 1.0 g mass stretches the spring 1.0 cm. Thus mg = k (1.0 × 10−2 m) ⇒ k = (1.0 ×10 −3 kg ) ( 9.8 N/kg ) = 0.98 N/m 1.0 ×10−2 m ( 0.98 N/m ) ( 4.5 ×10−2 m − 4.0 ×10−2 m )( 4.5 ×10−2 m ) 2 ⇒q= = 33 nC 9.0 ×109 N m 2 /C2
  • 26.58. Solve: (a) Kinetic energy is K = 1 mv 2 , so the velocity squared is v2 = 2K/m. From kinematics, a 2particle moving through distance Δx with acceleration a, starting from rest, finishes with v2 = 2aΔx. To gainK = 2 × 10−18 J of kinetic energy in Δx = 2.0 μm requires an acceleration v2 2K / m K 2.0 × 10−18 J a= = = = = 1.10 × 1018 m/s 2 2Δx 2 Δx mΔx ( 9.11×10−31 kg )( 2.0 ×10−6 m )(b) The force that produces this acceleration is F = ma = ( 9.11×10−31 kg )(1.10 ×1018 m/s 2 ) = 1.00 × 10−12 N(c) The electric field is F 1.00 ×10−12 N E= = = 6.3 × 106 N/C e 1.6 ×10−19 C(d) The force on an electron due to charge q is F = K q e r 2. To have a breakdown, the force on the electronmust be at least 1.00 × 10−12 N. The minimum charge that could cause a breakdown will be the charge that causesexactly a force of 1.00 × 10−12 N: ( 0.010 m ) (1.00 ×10−12 N ) 2 K qe r 2F F = = 1.00 ×10−12 N ⇒ q = = = 6.9 × 10−8 C = 68 nC r2 Ke ( 9.0 ×109 N m 2 / C 2 )(1.6 ×10−19 C )
  • 26.59. Model: The charged spheres are point charges.Visualize:Each sphere is in static equilibrium and the string makes an angle θ with the vertical. The three forces acting oneach sphere are the electric force, the gravitational force on the sphere, and the tension force.Solve: In static equilibrium, Newton’s first law is Fnet = T + FG + Fe = 0. In component form, ( Fnet ) x = Tx + ( FG ) x + ( Fe ) x = 0 N ( Fnet ) = Ty + ( FG ) y + ( Fe ) y = 0 N Kq 2 ⇒ −T sin θ + 0 N + =0 N T cosθ − mg + 0 N = 0 N d2 Kq 2 Kq 2 ⇒ T sin θ = = T cosθ = + mg ( 2 L sinθ ) 2 d2Dividing the two equations, ( 9.0 ×109 N m2/C2 )(100 ×10−9 C ) = 4.59 ×10−4 2 Kq 2 sin θ tan θ = 2 2 = 4 L mg 4 (1.0 m )2 ( 5.0 ×10−3 kg ) ( 9.8 N/kg )For small-angles, tan θ ≈ sin θ . With this approximation we obtain sinθ = 0.07714 rad and θ = 4.4°.
  • 26.60. Model: The charged spheres are point charges.Visualize:Each sphere is in static equilibrium when the string makes an angle of 20° with the vertical. The three forcesacting 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 is Fnet = T + FG + Fe = 0. In component form, ( Fnet ) = Tx + ( FG ) x + ( Fe ) x = 0 N ( Fnet ) y = Ty + ( FG ) + ( Fe ) y = 0 N Kq 2 ⇒ −T sinθ + 0 N + =0N T cosθ − mg + 0 N = 0 N d2 Kq 2 + Kq 2 ⇒ T sin θ = = T cosθ = + mg ( 2 L sinθ ) 2 d2Dividing the two equations and solving for q, 4 ( sin 2 20° tan 20° ) (1.0 m ) ( 3.0 ×10−3 kg ) ( 9.8 N/kg ) 2 4sin 2 θ tanθ L2 mg q= = = 0.75 μ C K 9.0 ×109 N m 2 /C 2
  • 26.61. Model: The electric field is that of a positive point charge located at the origin.Visualize: Please refer to Figure P26.61. Place the 10 nC charge at the origin.Solve: The electric field is ⎞ ⎛ ( 9.0 ×10 N m /C )(10 ×10 C ) ⎞ 9 2 2 −9 ⎛ 1 q E =⎜ , away from q ⎟ = ⎜ , away from q ⎟ ⎝ 4πε 0 r 2 ⎠ ⎝ ⎜ r 2 ⎟ ⎠ ⎛ 90.0 N m 2 /C ⎞ =⎜ 2 , away from q ⎟ ⎝ r ⎠At each of the three points, ⎛ ⎞ 90.0 N m 2 /CE1 = ⎜ , away from q ⎟ = 1.0 ×105 ˆ N/C j ⎜ ( 3.0 ×10−2 m ) 2 ⎟ ⎝ ⎠ ⎛ ⎞ 90.0 N m 2 /CE2 = ⎜ ⎜ ( 5.0 ×10−2 m ) 2 ⎟ ( , away from q ⎟ = ( 3.6 × 104 N/C ) cosθ i + sinθ ˆ ˆ j ) ⎝ ⎠ ( 4 ˆ 3 ) ( j ˆ j ) = ( 3.6 ×104 N/C ) 5 i + 5 ˆ = 2.9 × 104i + 2.2 × 104 ˆ N/C ⎛ ⎞ 90.0 N m 2 /CE3 = ⎜ , away from q ⎟ = 5.6 × 104i N/C ˆ ⎜ ( 4.0 ×10−2 m )2 ⎟ ⎝ ⎠
  • 26.62. Model: The electric field is that of a positive point charge.Visualize: Please refer to Figure P26.62. Place point 1 at the origin.Solve: The electric field is ⎞ ⎛ ( 9.0 ×10 N m /C )( 2 × 10 C ) ⎞ 9 2 2 −9 ⎛ 1 q E =⎜ , toward q ⎟ = ⎜ , toward q ⎟ ⎝ 4πε 0 r 2 ⎠ ⎝ ⎜ r 2 ⎟ ⎠ ⎛ 18.0 N m 2 /C ⎞ =⎜ 2 , toward q ⎟ ⎝ r ⎠The electric fields at the two points are ⎛ ⎞ 18.0 N m 2 /C E1 = ⎜ , toward q ⎟ ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° counter clockwise from the +x-axis or 60° north of east) ⎛ ⎞ 18.0 N m 2 /C E2 = ⎜ , toward q ⎟ ⎜ (1.0 ×10−2 m ) 2 ⎟ ⎝ ⎠ = (1.8 × 105 N/C, 60° clockwise from the −x-axis or 60° north of west)
  • 26.63. Model: The electric field is that of a positive point charge located at the origin.Visualize: Please refer to Figure P26.63. Place the 5.0 nC charge at the origin.Solve: The electric field is ⎞ ⎛ ( 9 ×10 N m / C )( 5.0 × 10 C ) ⎞ 9 2 2 −9 ⎛ 1 q E =⎜ , away from q ⎟ = ⎜ , away from q ⎟ ⎝ 4πε 0 r 2 ⎠ ⎝ ⎜ r 2 ⎟ ⎠ ⎛ 45 N m 2 /C ⎞ =⎜ 2 , away from q ⎟ ⎝ r ⎠At each of the three points, ⎛ ⎞ 45 N m 2 /C E1 = ⎜ ⎜ ( 2.0 ×10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ( , away from q ⎟ = ( 9.0 × 104 N/C ) cosθ i + sin θ ˆ ˆ j ) ⎝ ⎠ ⎛ 1 ˆ 2 ˆ⎞ = ( 9.0 ×10 N/C ) ⎜ 4 ⎝ 5 i+ 5 ⎠ ( ) j ⎟ = 4.0 × 10 i + 8.0 × 104 ˆ N/C 4ˆ j ⎛ ⎞ 45 N m 2 /CE2 = ⎜ , away from q ⎟ = 4.5 ×105i N/C ˆ ⎜ (1.0 ×10−2 m )2 ⎟ ⎝ ⎠ ⎛ 2 ⎞ 45 N m /CE3 = ⎜ ⎜ ( 2.0 × 10−2 m )2 + (1.0 × 10−2 m )2 ⎟ ( ˆ ) , away from q ⎟ = 4.0 × 104 i − 8.0 × 104 ˆ N/C j ⎝ ⎠
  • 26.64. 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 thefield has a particular value by inspecting the signs of Ex and Ey. At point a, the electric field has no y-componentand 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, Ex = E = Kq ⇒ ra = Kq = ( 9.0 ×10 9 N m 2 /C 2 )(10.0 ×10−9 C ) = 0.0200 m = 2.00 cm ra2 Ex 225,000 N/CThus, point a is at the position (xa, ya) = (4 cm, 1 cm).(b) Point b is above and to the left of the charge. The magnitude of the field at this point is (161,000 N/C ) + ( 80,500 N/C ) 2 2 E = Ex2 + E y = 2 = 180,000 N/CUsing the equation for the field of a point charge, E= Kq ⇒ rb = Kq = ( 9.0 ×10 9 N m 2 /C2 )(10 ×10−9 C ) = 2.236 cm rb2 E 180,000 N/CThis gives the total distance but not the horizontal and vertical components. However, we can determine theangle θ because Eb points straight toward the negative charge. Thus, ⎛ Ey ⎞ ⎛ 80,500 ⎞ −1 ⎛ 1 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎜ E ⎟ ⎝ 161,000 ⎠ ⎝ 2⎠ ⎝ x ⎠The horizontal and vertical distances are then d x = rb cosθ = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b isat the position (xb, yb) = (0 cm, 2 cm).(c) Point c, which is below and to the left of the charge, is calculated by following a similar procedure. We firstfind that E = 36,000 N/C . From this we find that the total distance rc = 5.00 cm. The angle φ is ⎛ Ey ⎞ ⎛ 21,600 ⎞ φ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 36.87° ⎜ Ex ⎟ ⎝ 28,000 ⎠ ⎝ ⎠which gives the distances d x = rc cos φ = 4.00 cm and d y = rc sin φ = 3.00 cm. Thus point c is at position (xc, yc) =(−2 cm, − 2 cm).
  • 26.65. 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 locatethe points of interest based simply on whether the signs of Ex and Ey are positive or negative. For point a, theelectric field has no y-component and the x-component points to the left, so point a must be to the left of thecharge along a horizontal line. Using the field of a point charge, Ex = E = Kq ⇒ ra = Kq = ( 9.0 ×10 9 N m 2 /C 2 )(10.0 ×10−9 C ) = 0.0200 m = 2.00 cm ra2 Ex 225,000 N/CThus, (xa, ya) = (−1 cm, 2 cm).(b) Point b is above and to the right of the charge. The magnitude of the field at this point is (161,000 N/C ) + (80,500 N/C ) 2 2 E = Ex2 + E y = 2 = 180,000 N/CUsing the field of a point charge, E= Kq ⇒ rb = Kq = ( 9.0 ×10 9 N m 2 /C2 )(10.0 ×10−9 C ) = 2.236 cm rb2 E 180,000 N/CThis gives the total distance but not the horizontal and vertical components. However, we can determine theangle θ because Eb points straight away from the positive charge. Thus, ⎛ Ey ⎞ ⎛ 80,500 N/C ⎞ −1 ⎛ 1 ⎞ θ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = tan ⎜ ⎟ = 26.57° ⎜ Ex ⎟ ⎝ 161,000 N/C ⎠ ⎝ 2⎠ ⎝ ⎠The horizontal and vertical distances are then d x = rb cosθ = ( 2.236 cm ) cos 26.57° = 2.00 cm and d y = rb sin θ = 1.00 cm. Thus, point b is at position (xb, yb) = (3 cm, 3 cm).(c) To calculate point c, which is below and to the right of the charge, a similar procedure is followed. We firstfind E = 36,000 N/C from which we find the total distance rc = 5.00 cm. The angle φ is ⎛ Ey ⎞ ⎛ 28,800 ⎞ φ = tan −1 ⎜ ⎟ = tan −1 ⎜ ⎟ = 53.13° ⎜ Ex ⎟ ⎝ 21,600 ⎠ ⎝ ⎠which gives distances d x = rc cos φ = 3.00 cm and d y = rc sin φ = 4.00 cm. Thus, point c is at position (xc, yc) =(4 cm, − 2 cm).
  • 26.66. Model: The electric field is that of three point charges.Visualize: (1 cm ) + ( 3 cm ) 2 2Solve: (a) In the figure, the distances are r1 = r3 = = 3.162 cm and the angle isθ = tan −1 (1/ 3) = 18.43°. Using the equation for the field of a point charge, K q1 ( 9.0 ×10 N m /C )(1.0 ×10 C ) 9 2 2 −9 E1 = E3 = = = 9.0 kN/C ( 0.03162 m ) 2 r12We now use the angle θ to find the components of the field vectors: ˆ j ( E1 = E1 cosθ i − E1 sin θ ˆ = 8540i − 2840 ˆ ˆ j ) N/C = (8.5iˆ − 2.8 ˆj ) kN/C E3 = E cosθ i + E sinθ ˆ = ( 8540i + 2840 ˆ ) N/C = ( 8.5i + 2.8 ˆ ) kN/C 3 ˆ 3 j ˆ j ˆ jE2 is easier since it has only an x-component. Its magnitude is K q2 ( 9.0 ×10 N m /C )(1.0 ×10 C ) 9 2 2 −9 E2 = = ˆ ˆ = 10,000 N/C ⇒ E2 = E2 i = 10.0 i kN/C ( 0.0300 m ) 2 r22(b) The electric field is defined in terms of an electric force acting on charge q: E = F q . Since forces obey aprinciple of superposition ( Fnet = F1 + F2 + … ) it follows that the electric field due to several charges also obeys aprinciple of superposition. ˆ(c) The net electric field at a point 3 cm to the right of q2 is Enet = E1 + E2 + E3 = 27 i kN/C. The y-components ofE1 and E2 cancel, giving a net field pointing along the x-axis.
  • 26.67. Model: The charged ball attached to the string is a point charge.Visualize:The ball is in static equilibrium in the external electric field when the string makes an angle θ = 20° with thevertical. The three forces acting on the charged ball are the electric force due to the field, the gravitational forceon the ball, and the tension force.Solve: In static equilibrium, Newton’s second law for the ball is Fnet = T + FG + Fe = 0. In component form, ( Fnet ) x = Tx + 0 N + qE = 0 N ( Fnet ) y = Ty − mg + 0 N = 0 NThe above two equations simplify to T sin θ = qE T cosθ = mgDividing both equations, we get mg tan θ ( 5.0 × 10 kg ) ( 9.8 N/kg ) tan 20° −3 qEtan θ = ⇒q= = = 1.78 × 10−7 C = 178 nC mg E 100,000 N/C
  • 26.68. Model: The charged ball attached to the string is the point charge.Visualize:The charged ball is in static equilibrium in the external electric field when the string makes an angle θ with thevertical. The three forces acting on the charge are the electric force due to the electric field, the gravitational forceon the ball, and the tension force.Solve: In static equilibrium, Newton’s second law for the charged ball is Fnet = T + FG + F3 = 0. In componentform, ( Fnet ) x = Tx + 0 N + qE = 0 N ( Fnet ) y = Ty − mg + 0 N = 0 NThese two equations become T sin θ = qE and T cosθ = mg . Dividing the equations gives qE ( 25 ×10 C ) ( 200,000 N/C ) −9tan θ = = = 0.255 ⇒ θ = 14.3° mg ( 2.0 ×10−3 kg ) ( 9.8 N/kg )
  • 26.69. Solve: (a) How many excess electrons on a dust particle produce an electric field of magnitude 1.0 ×106 N/C a distance of 1.0 μm from the dust particle?(b) The number of electrons is (1.5 ×10 N/C )(1.0 × 10−6 m ) 6 2 N= = 1.04 ×103 (9.0 ×109 N m 2/C2 )(1.60 ×10−19 C )
  • 26.70. 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 ( 0.020 N )( 0.0150 m ) 2 q= = 22 nC 9.0 ×109 N m 2 /C 2The problem does not give the direction of the force. So, the charges could be both positive or both negative.
  • 26.71. Solve: (a) At what distance from a 15 nC charge is the electric field strength 54,000 N/C?(b) The distance is r= ( 9.0 ×10 9 N m 2 /C 2 )(15 ×10−9 C ) = 0.050 m = 5.0 cm 54,000 N/C
  • 26.72. Solve: (a) A 1.0 nC charge is placed at (0 cm, 2 cm) and another 1.0 nC charge is placed at (0 cm, –2cm). A third charge +q is placed along a line halfway between q1 and q2 such that the angle between the forces on ( ˆ )q due to each of the other two charges is 60°. The resultant force on q is 5.0 × 10−5 N, i . What is the magnitudeof the charge q?(b)We have ⎛ ( 9.0 ×109 N m 2 /C 2 )(1.0 ×10−9 C ) q ⎞ F1 on 3 = ⎜ ⎜ ( 0.020 m sin 30°) 2 ⎟ ( , away from q3 ⎟ = ( 5625 N/C ) q cos30°i − sin 30° ˆ ˆ j ) ⎝ ⎠ ( F2 on 3 = ( 5625 N/C ) q cos30°i + sin 30 ˆ ˆ )j ⇒ Fon 3 = 2 × ( 5625 N/C ) q cos30°i ˆ2 ( 5625 N/C ) q cos30° = 5.0 ×10−5 N ⇒ q = ( 5.0 ×10 −5 N) = 5.1 nC 2 ( 5625 N/C ) cos30°
  • 26.73. Model: Use the charge model.Solve: The mass of copper in a 2.0-mm-diameter copper ball is ⎛ 4π 3 ⎞ ⎡ 4π 3⎤ M = ρV = ρ ⎜ r ⎟ = ( 8920 kg/m3 ) ⎢ (1.0 ×10−3 m ) ⎥ = 3.736 × 10−5 kg = 0.03736 g ⎝ 3 ⎠ ⎣ 3 ⎦The number of moles in the ball is M 0.03736 g n= = = 5.884 ×10−4 mol A 63.5 g/molThe number of copper atoms in the ball is N = nN A = ( 5.884 ×10−4 mol )( 6.02 ×1023 mol−1 ) = 3.542 ×1020We note that the number of electrons per atom is the atomic number, and both the atomic number (29) and theaverage atomic mass (63.5 g) are taken from the periodic table in the textbook. The number of electrons in the copper ball is thus 29 × 3.542 × 1020 = 1.03 × 1022. The number of electronsremoved from the copper ball is 50 ×10−9 C = 3.13 ×1011 1.60 ×10−19 CSo, the fraction of electrons removed from the copper ball is 3.13 × 1011 = 3.0 ×10−11 1.03 × 1022Assess: This is indeed a very small fraction of the available number of electrons in the copper ball.
  • 26.74. 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 forceon each ball is the same. The other forces acting on each ball are the gravitational force on the ball and thetension force.Solve: The force on ball 3 is the sum of the force from ball 1 and ball 2. We have ⎛ K q1 q3 ⎞ Kq 2 F1 on 3 = ⎜ 2 ˆ ( , away from q1 ⎟ = 2 sin 30°i + cos30° ˆ j ) ⎝ r13 ⎠ r Kq 2 F2 on 3 = r2 ( − sin 30°i + cos30 ˆ ˆ j ) 2 Kq 2 2 Kq 2 ⇒ Fon 3 = 2 cos30° ˆ ⇒ Fon 3 = 2 cos30° = Fon 2 = Fon 1 = Fe j r r 2 ( 9.0 ×109 N m 2 /C2 ) q 2 cos30° ⇒ Fe = = ( 3.897 ×1011 q 2 ) N/C 2 ( 0.20 m ) 2The distance l, between one of the balls and the center of the equilateral triangle, is r 0.10 m l cos30° = = 0.10 m ⇒ l = = 0.1155 m 2 cos30°Thus, the angle made by the string with the plane containing the three balls is l 0.1155 m cosθ = = ⇒ θ = 81.70° L 0.80 mFrom the free-body diagram, we have T sin θ − mg = 0 N −T cosθ + Fe = 0 N mg mg ⇒ tanθ = = Fe ( 3.897 ×1011 q 2 ) N/C2⇒ q= mg = ( 3.0 ×10 −3 kg ) ( 9.8 N/kg ) = 1.05 ×10−7 C = 105 nC ( 3.897 ×1011 N/C 2 ) tanθ ( 3.897 ×10 N/C2 ) tan81.70° 11
  • 26.75. 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 inFigure CP26.75. The force FE is due to the external electric field. The force Fe is the attractive force between thepositive and the negative spheres. The tension in the string and the gravitational force are the remaining twoforces on the spheres.Solve: The two electrical forces are calculated as follows: FE = q E = (100 ×10−9 C )(105 N/C ) = 1.00 × 10−2 N K q1 q2 ( 9.0 ×10 N m /C )(100 ×10 C ) 9 2 2 −9 2 9.0 ×10−5 N m 2 /C Fe = 2 = 2 = r r r2From the geometry of Figure CP26.75, 9.0 ×10−5 N m 2 /C r = 2 ( L sin10° ) = 2 ( 0.50 m ) sin10° = 0.174 m ⇒ Fe = = 3.0 ×10−3 N ( 0.174 m ) 2From the free-body diagram, T cos10° = mg T sin10° + Fe = FERearranging and dividing the two equations, sin10° F − Fe = tan10° = E cos10° mg FE − Fe 1.00 ×10−2 N − 0.30 ×10−2 N m= = = 0.41× 10−2 kg = 4.1 g g tan10° ( 9.8 N/kg ) tan10°
  • 26.76. Model: The charges are point charges.Visualize:Solve: The forces on the –1.0 nC charge lie along the line connecting the pairs of charges. Since the 10 nCcharge is positive, F10 points as shown. The angle formed by the dashed lines where they meet is 90°, so Fqmust point towards q in order for F = F10 + Fq to hold.Therefore F10 F= cos30°The distance between the 10 nC and –1.0 nC charges is r = 5.0sin 30° cm = 2.5 cm. Thus F10 = ( 9.0 ×109 N m 2 /C2 ) (10 ×10 C )(1.0 ×10 −9 −9 C) = 1.44 ×10−4 N ( 2.5 ×10 m ) −2 2Hence 1.44 ×10−4 N F= = 1.66 ×10−4 N cos30°
  • 26.77. Model: Charge Q and the dipole charges (q and −q) are point charges.Visualize: Please refer to Figure CP26.77.Solve: (a) The force on the dipole is the vector sum of the force on q and –q. We have ⎛ KQ q ⎞ KQq FQ on + q = ⎜ ⎜ ( r + s / 2 )2 , away from Q ⎟ = ⎟ ( r + s / 2 )2 ˆ −i ( ) ⎝ ⎠ KQq ⎛ ⎞ FQ on −q = ( r − s / 2) 2 ( +iˆ ) ⇒ F net = KQq ⎜ 1 − 1 ⎟i ⎜ ( r − s / 2 )2 ( r + s / 2 )2 ⎟ ˆ ⎝ ⎠(b) The net force Fnet is toward the charge Q, because the attractive force due to Q on the negative charge of the dipole ismore than the repulsive force of Q on the positive charge.(c) Using the binomial approximation, we get −2 ⎛ s ⎞ ⎛ 2s ⎞ KQq ⎛ ⎛ 2s ⎞ ⎛ 2s ⎞ ⎞ 2 KQqs ( r ± s / 2) −2 = r −2 ⎜1 ± ⎟ ≅ r −2 ⎜1 ∓ + ⎟ ⇒ Fnet = 2 ⎜ ⎜1 + ⎟ − ⎜ 1 − ⎟ ⎟ = ⎝ 2r ⎠ ⎝ 2r ⎠ r ⎝ ⎝ 2r ⎠ ⎝ 2r ⎠ ⎠ r3(d) Coulomb’s law applies only to the force between two point charges. A dipole is not a point charge, so there’s noreason that the force between a dipole and a point charge should be an inverse-square force.Assess: Note that when s → 0 m, Fnet → 0 N. In this limit the dipole is a point with zero charge. 26-1