20.1 Electric Charge

CHAPTER

2O.—StaUctlectdcity

I

_________________

Section Review20.1 Electric Charge

pages 541—545page 545

1. Charged Objects After a comb is rubbed

on a wool sweater, it is able to pick up

small pieces of paper. Why does the comb

lose that ability after a few minutes?

The comb loses its charge to its sur

roundings and becomes neutral once

again.

2. Types of Charge In the experiments

described earlier in this section, how could

you find out which strip of tape, B or T, is

positively charged?

Bring a positively charged glass rod

near the two strips of tape. The one that

is repelled by the rod is positive.

3. Types of Charge A pith ball is a small

sphere made of a light material, such as plas

tic foam, often coated with a layer of graphite

or aluminum paint. How could you deter

mine whether a pith ball that is suspended

from an insulating thread is neutral, ischarged positively, or is charged negatively?

Bring an object of known charge, such

as a negatively charged hard rubber

rod, near the pith ball. If the pith ball is

repelled, it has the same charge as the

rod. If it is attracted, it may have the

opposite charge or be neutral. To find

out which, bring a positively charged

glass rod near the pith ball, If they

repel, the pith ball is positive; if they

attract, the pith ball must be neutral.

4. Charge Separation A rubber rod can be

charged negatively when it is rubbed with

wool. What happens to the charge of the

wool? Why?

The wool becomes positively chargedbecause it gives up electrons to therubber rod.

5. Conservation of Charge An apple con

tains trillions of charged partides. Whydon’t two apples repel each other when

they are brought together?

Each apple contains equal numbers of

positive and negative charges, so they

appear neutral to each other.

6. Charging a Conductor Suppose you hang

a long metal rod from silk threads so that

the rod is isolated. You then touch a

charged glass rod to one end of the metal

rod. Describe the charges on the metal rod.

The glass rod attracts electrons offthe metal rod, so the metal becomes

positively charged. The charge is dis

tributed uniformly along the rod.

7. Charging by Friction You can charge a

rubber rod negatively by rubbing it with

wool. What happens when you rub a cop

per rod with wool?

Because the copper is a conductor, it

remains neutral as long as it is in con

tact with your hand.

8. Critical Thinking It once was proposed

that electric charge is a type of fluid that

flows from objects with an excess of the

fluid to objects with a deficit. Why is the

current two-charge model better than the

single-fluid model?

The two-charge model can better

explain the phenomena of attraction

and repulsion. It also explains how

objects can become charged when they

are rubbed together. The single-fluid

model indicated that the charge should

be equalized on objects that are in con

tact with each other.

Physics: Principles and Problems Solutions Manual 413

Chapter 20 continued

Practice Problems20.2 Electric Force

pages 546—553page 552

9. A negative charge of —2.OX i0 C and a positive charge of 8.OX i0 C are

separated by 0.30 m. What is the force between the two charges?

FKqq8

— (9.0x109N•m2IC2)(2.0X104C)(8.0x104C)

dAB2 — (0.30 rn)2

= 1.6X10 N

10. A negative charge of —6.0 X l0 C exerts an attractive force of 65 N on a second

charge that is 0.050 m away. What is the magnitude of the second charge?

r- Kqq., 2“AB

— FdAB2 — (65 N)(0.050 rn)2— Kq — (9.0x109N.m2/C2)(6.0X106C)

= 3.0X106C

11. The charge on B in Example Problem 1 is replaced by a charge of +3.00 u,C.

Diagram the new situation and find the net force on A.

Magnitudes of all forces remain the same. The direction changes to 42°

above the —xaxls, or 138°.

‘12. Sphere A is located at the origin and has a charge of +2.OX 10—6 C. Sphere B

is located at +0.60 mon the x-axis and has a charge of —3.6X106C.

Sphere C is located at +0.80 m on the x-axis and has a charge of +4.OX 10—6 C.

Determine the net force on sphere A.

FB onA=

= (9.0x109N.m2IC2) (2.0X10 6 <106 C)= 0.18 N

direction: toward the right

FCOflA=

= (9.0X109N.m2!C2) (2.0X10 <106 C)= 0.1125 = N

direction: toward the left

Fnet = FBOflA — FCOflA = (0.18 N) — (0.1125 N) = 0.068 N toward the right

13. Determine the net force on sphere B in the previous problem.

— qAqBFAOflB—K,, 2

AB

— qAqBFCOnB—K,, 2

AB

Fnet=FC0nBFA0nB 4— Bqc qAqB“,. 2 K,4 2

BC “AB

414 Solutions Manual Physics: Principles and Problems


= 3.1 N toward the right

Section Review20.2 Electric Force

pages 546—553page 55314. Force and Charge How are electric force

and charge related? Describe the force whenthe charges are like charges and the forcewhen the charges are opposite charges.

Electric force is directly related to eachcharge. It is repulsive between likecharges and attractive between opposite charges.

15. Force and Distance How are electric forceand distance related? How would the forcechange if the distance between two chargeswere tripled?

Electric force is inversely related tothe square of the distance betweencharges. lithe distance is tripled, theforce will be one-ninth as great.

16. Electroscopes When an electroscope ischarged, the leaves rise to a certain angleand remain at that angle. Why do they notrise farther?

As the leaves move farther apart, theelectric force between them decreasesuntil it is balanced by the gravitationalforce pulling down on the leaves.

17. Charging an Electroscope Explain how tocharge an electroscope positively using

a. a positive rod.

Touch the positive rod to the electroscope. Negative charges will moveto the rod, leaving the electroscopepositively charged.

b. a negative rod.

Bring the negative rod near, but nottouching the electroscope. Touch(ground) the electroscope with yourfinger, allowing electrons to berepelled off of the electroscope intoyour finger. Remove your finger andthen remove the rod.

18. Attraction of Neutral Objects What twoproperties explain why a neutral object isattracted to both positively and negativelycharged objects?

Charge separation, caused by theattraction of opposite charges and therepulsion of like charges, moves theopposite charges In the neutral bodycloser to the charged object and thelike charges farther away. The inverserelation between force and distancemeans that the nearer, oppositecharges will attract more than the moredistant, like charges will repel. Theoverall effect is attraction.

19. ChargIng by Induction In an electroscopebeing charged by induction, what happenswhen the charging rod is moved away beforethe ground is removed from the knob?

Charge that had been pushed into theground by the rod would return to theelectroscope from the ground leavingthe electroscope neutral.

20. Electric Forces Two charged spheres are helda distance, r, apart. One sphere has a charge of+3C, and the other sphere has a charge of+9p.C. Compare the force of the +3.LCsphere on the +9C sphere with the force ofthe +9C sphere on the +3C sphere

The forces are equal in magnitude andopposite in direction.

21. CrItical Thinking Suppose that you aretesting Coulomb’s law using a small, positively charged plastic sphere and a large,positively charged metal sphere. Accordingto Coulomb’s law, the force depends onl/r2,where r is the distance between thecenters of the spheres. As the spheres get

= (9.0X109N’m2/C2)

(3.6x106C)(4.0x106C) —

(0.20 rn)2

(9.0X109N•m2/C2)

(2.OX106C)(3.6X106C)(0.60 rn)2



close together, the force is smaller thanexpected from Coulomb’s law. Explain.

Some charges on the metal sphere willbe repelled to the opposite side fromthe plastic sphere, making the effectivedistance between the charges greaterthan the distance between the spheres’centers.

Complete the concept map below using thefollowing terms: conduction, distance, elementary charge.

Mastering Conceptspage 55823. If you comb your hair on a dry day, the

comb can become positively charged. Canyour hair remain neutral? Explain. (20.1)

No. By conservation of charge, yourhair must become negatively charged.

24. List some insulators and conductors. (20.1)

Student answers will vary but mayinclude dry air, wood, plastic, glass,cloth, and deionized water as insulators; and metals, tap water, and yourbody as conductors.

25. What property makes metal a good conductor and rubber a good insulator? (20.1)

Metals contain free electrons; rubberhas bound electrons.

26. Laundry Why do socks taken from a clothesdryer sometimes cling to other clothes? (20.2)

They have been charged by contact asthey rub against other clothes, andthus, are attracted to clothing that isneutral or has an opposite charge.

27. Compact Discs If you wipe a compact discwith a dean cloth, why does the CD thenattract dust? (20.2)

Rubbing the CD charges it. Neutralparticles, such as dust, are attracted toa charged object.

28. Coins The combined charge of all electronsin a nickel is hundreds of thousands ofcoulombs. Does this imply anything aboutthe net charge on the coin? Explain. (20.2)

No. Net charge is the difference betweenpositive and negative charges. The coinstill can have a net charge of zero.

29. How does the distance between twocharges impact the force between them?If the distance is decreased while thecharges remain the same, what happensto the force? (20.2)

Electric force is inversely proportionalto the distance squared. As distancedecreases and charges remain thesame, the force increases as the squareof the distance.

30. Explain how to charge a conductor negatively if you have only a positively chargedrod. (20.2)

Bring the conductor close to, but nottouching, the rod. Ground the conductorin the presence of the charged rod; then,remove the ground before removing thecharged rod. The conductor will have anet negative charge.

Applying Conceptspage 55831. How does the charge of an electron differ

from the charge of a proton? How are theysimilar?

Chapter AssessmentConcept Mappingpage 55822.



The charge of the proton is exactly thesame size as the electron, but has theopposite sign.

32. Using a charged rod and an electroscope,how can you find whether or not an objectis a conductor?

Use a known insulator to hold one endof the object against the electroscope.Touch the other end with the chargedrod. If the electroscope indicates acharge, the object is a conductor.

33. A charged rod is brought near a pile of tinyplastic spheres. Some of the spheres areattracted to the rod, but as soon as theytouch the rod, they are flung off in differentdirections. Explain.

The natural spheres are initially attracted to the charged rod, but they acquirethe same charge as the rod when theytouch it. As a result, they are repelledfrom the rod.

34. Lightning Lightning usually occurs when anegative charge in a cloud is transported toEarth. If Earth is neutral, what provides theattractive force that pulls the electronstoward Earth?

The charge in the cloud repels electrons on Earth, causing a charge separation by induction. The side of Earthclosest to the cloud is positive, resulting in an attractive force.

35. Explain what happens to the leaves of apositively charged electroscope when rodswith the following charges are broughtclose to, but not touching, the electroscope.

a. positive

The leaves will move farther apart.

b. negative

The leaves will drop slightly.

36. As shown in Figure 20-13, Coulomb’s lawand Newton’s law of universal gravitationappear to be similar. In what ways are theelectric and gravitational forces similar?How are they different?

u Figure 20-13 (Not to scale)

Similar: inverse-square dependence ondistance, force proportional to productof two masses or two charges; different: only one sign of mass, so gravitational force is always attractive; twosigns of charge, so electric force canbe either attractive or repulsive.

37. The constant, K, in Coulomb’s equation ismuch larger than the constant, G, in theuniversal gravitation equation. Of what significance is this?

The electric force is much larger thanthe gravitational force.

38. The text describes Coulomb’s method forcharging two spheres, A and B, so that thecharge on B was exactly half the charge onA. Suggest a way that Coulomb could haveplaced a charge on sphere B that was exactlyone-third the charge on sphere A.

After changing spheres A and B equally,sphere B is touched to two otherequally sized balls that are touchingeach other. The charge on B will bedivided equally among all three balls,leaving one-third the total charge on it.

39. Coulomb measured the deflection of sphereA when spheres A and B had equal chargesand were a distance, r, apart. He then madethe charge on B one-third the charge on A.How far apart would the two spheres thenhave had to be for A to have had the samedeflection that it had before?

To have the same force with one-thirdthe charge, the distance would haveto be decreased such that d2 = 1/3, or0.58 times as far apart.

Law ofUniversal Gravitation Coulomb’s Law

F=GmABF=Kr r

mA

r r



40. Two charged bodies exert a force of 0.145 N on each other. If they are moved

so that they are one-fourth as far apart, what force is exerted? 4

F x and F _!, so F = 16 times the original force.

41. Electric forces between charges are enormous in comparison to gravitational

forces. Yet, we normally do not sense electric forces between us and oursurroundings, while we do sense gravitational interactions with Earth. Explain.

Gravitational forces only can be attractive. Electric forces can be either

attractive or repulsive, and we can sense only their vector sums, which

are generally small. The gravitational attraction to Earth is larger and

more noticeable because of Earth’s large mass.

Mastering Problems20.2 Electric Force

page 559

Level 142. Two charges, q and qB’ are separated by a distance, r, and exert a force, F, on

each other. Analyze Coulomb’s law and identify what new force would exist

under the following conditions.

a. q is doubled

2qA then new force = 2F

b. qA and q5 are cut in half

qA and qB, then new force = (1)(1)F = 4- F

c. r is tripled

3d,thennewforce= = IF

d. riscutinhalf

d, then new force —c-- = F = 4F2 /1\2 4

ki) Te. q is tripled and r is doubled

3qand2d,thennewforce= 5 = F

43. Lightning A strong lightning bolt transfers about 25 C to Earth. How manyelectrons are transferred?

(—25 C)( —1.6Ox1O19= 1.6X1020 electrons

44. Atoms Two electrons in an atom are separated by 1.5 X 1010 m, the typical sizeof an atom. What is the electric force between them?

F—Kqq

— (9.0X109N.m2IC2)(1.60x1019C)(t60x1019C)— (1.5X1010rn)2 V

= 1.0X108N, away from each other



45. A positive and a negative charge, each of magnitude 2.5 X 10-i C, are separatedby a distance of 15 cm. Find the force on each of the partides.

F— Kqq5 — (9.0x109N•rn2IC2)(2.5X10’5C)(2.5X105C)— d2 — (1.5X101 rn)2

= 2.5x102N, toward the other charge

46. A force of 2.4X 102 N exists between a positive charge of 8.OX i0 C and apositive charge of 3.OX i0 C. What distance separates the charges?

— Kqq5F—

d = / Kqq /(9.0X109 N.m2IC2)(8.0X105C)(3.OX1O5C)1f F 2.4X102N

=O.30m

47. Two identical positive charges exert a repulsive force of 6.4X i0 N whenseparated by a distance of 3.8X 1010 m. Calculate the charge of each.

— Kqq — Kq2F—

d2

— / — /(6.4X109 N)(3.8X1010rn)2— 2X1O9Cq

— / K — 9.0X109N•m2IC2 —

Level 248. A positive charge of 3.0 C is pulled on by two —4.0 tCnegative charges. As shown in Figure 20-14, one

—2 0 3 0 Cnegative charge, —2.0 C, is 0.050 m to the west, ILC + . ,LL

and the other, —4.0 pC, is 0.030 m to the east.What total force is exerted on the positive charge?

0.050 m 0.030 m(9.0x 1 Nm2/C2)(3.0X10—6 C)(2.OX 10—6 C)F1= (0.050 rn)2 I Figure 20-14

=22Nwest

(9.OX1 o N•rn2IC2)(3.OX106 C)(4.0X1 0—6 C)(0.030 rn)2

= 120 N east

Fnet = F2 + F1 = (1.2x102N) — (2.2x101 N)

= 98 N, east

49. Figure 20-15 shows two positively charged q 3qspheres, one with three times the charge of theother. The spheres are 16 cm apart, and the forcebetween them is 0.28 N. What are the charges 16cm ‘Ion the two spheres?

• Figure 20-15

F— —

_____

— d2 d2



— — / (0.28 N)(0.16 rn)2— 2X107C

—‘%J 3K — i,I 3(9.0x109N•m2IC2) —

qB = = 1.5X106C

50. Charge in a Coin How many coulombs of charge are on the electrons in a nickel?

Use the following method to find the answer.

a. Find the number of atoms in a nickel. A nickel has a mass of about 5 g.

A nickel is 75 percent Cu and 25 percent Ni, so each mole of the coin’s

atoms will have a mass of about 62 g.

A coin is = 0.08 mole.

Thus, it has (0.08)(6.02X1023)= 5x1022 atoms

b. Find the number of electrons in the coin. On average, each atom has

28.75 electrons.

(5x1022 atoms)(28.75 electrons/atom) = 1 X1024 electrons

c. Find the coulombs on the electrons.

(1 .6X 10—19 cou lombs/electron)(1 Xl 024 electrons) = 2X 1 o coulombs

51. Three partides are placed in a line. The left particle has a charge of —55 pC, the

middle one has a charge of +45 p.C, and the right one has a charge of —78 C.

The middle partide is 72 cm from each of the others, as shown in Figure 20-16.

—55 jC +45 LC —78 jC

I 72cm ‘I’ 72cm i

• Figure 20-16

a. Find the net force on the middle particle.

Let left be the negative direction

— Kqq1 KqqFnetFi+(Fr) d2

+d2

— —(9.0X109N.m2/C2)(45X106C)(55X106C)— (0.72 rn)2

+

(9.0X109N•m2IC2)(45X106C)(78x106C)(0.72 rn)2

= 18 N, right

b. Find the net force on the right particle.

—

— Kq1q KqqFnet — F1 + (Fm) — +

(2d)2 — d2

— (9.0X109N•m2/C2)(55X106C)(78X106C) +(2(0.72 rn))2

(9.0xl o N.m2iC2)(45x106 C)(78X1 0—6 C)(0.72 rn)2

=—42N,left 4



Mixed Review) page 559

Level 152. A small metal sphere with charge 1.2X 10 C is touched to an identical neutral

sphere and then placed 0.15 m from the second sphere. What is the electric forcebetween the two spheres?The two spheres share the charge equally, so

F= K’IA’IB = (9.0x109N.m2/C2)0<10( <106 C)= 14 N

53. Atoms What is the electric force between an electron and a proton placed5.3X 10h1 m apart, the approximate radius of a hydrogen atom?

F= KLIA(IB= (9.0X109N.m2/C2) (1.60X1 10l9 C)

= 8.2X108N

54. A small sphere of charge 2.4 pC experiences a force of 0.36 N when a second sphereof unknown charge is placed 5.5 cm from it. What is the charge of the second sphere?1,qAqB

‘“

— (0.36 N)(5.5X102rn)2— 5 OX 10—8 CqB

— Kq — (9.0x109N.m2/C2)(2.4X106C) —

55. Two identically charged spheres placed 12 cm apart have an electric force of0.28 N between them. What is the charge of each sphere?

— qAqB —F— Kd2

where qA — qB

— / (p.28 N)(1.2x10-1rn)2q— i.,/ K 1,,! (9.OX io N.rn2/C2)

= 6.7x107C

56. In an experiment using Coulomb’s apparatus, a sphere with a charge of3.6 X 10—8 C is 1.4 cm from a second sphere of unknown charge. The forcebetween the spheres is 2.7X 10—2 N. What is the charge of the second sphere?rj—.

____

UI

d2

Fd2— / (2.7X102N)(1.4x102rn)2qB

— Kq — i (9.0X109N•m2/C2)(3.6X108C)

= 1.6X108C

57. The force between a proton and an electron is 3.5X 1010 N. What is the distancebetween these two particles?

F—

d=1JK

= \/(9.OXlo N.m2/C2)(160X1 1019= 8.1x1010 m



Thinking Criticallypage 56058. Apply Concepts Calculate the ratio of the electric force to the gravitational force

between the electron and the proton in a hydrogen atom.

qqFe —

________

— Kqeqp

Fg— Gmemp — Gmemp

d2

(9.0x109N•m2/C2)(1.60X1019C)2 2 3X1039(6.67X1011N.m2Ikg2)(9.11X1O-31kg)(1.67x1027kg)

dAc2 = dBc2’and 16dAc2 = 64dBc2,or

dAc2 = 4dBc2,so dAc = 2dBC

The third sphere must be placed at +2.00 m on the x-axis so it is twice

as far from the first sphere as from the second sphere.

b. If the third sphere had a charge of +6 C, where should it be placed?

The third charge, q, cancels from the equation, so it doesn’t matter

what its magnitude or sign is.

c. If the third sphere had a charge of —12 p.C, where should it be placed?

As in part b, the magnitude and sign of the third charge, do not matter.

60. Three charged spheres are located at the positions shown in Figure 20-17. Find

the total force on sphere B.

59. Analyze and Condude Sphere A, with a charge of +64 C, is positioned at the

origin. A second sphere, B, with a charge of —16 C, is placed at 11.00 m on

the x-axis.

a. Where must a third sphere, C, of charge + 12 JLC be placed so there is no net

force on it?

The attractive and repulsive forces must cancel, so

— ,.qAqC — ,aqc —

‘AC”,. 2”.i 2FBC,SOAC “BC

I01

—8.2 C

4.0cm 1 B

• Figure 20-17

F1 = FAOnB

— Kqq— (9.0X109N•m2IC2)(4.5X106C)(—8.2X106C)

d2 — (0.040 rn)2



= —208 N = 208 N, to leftThe distance between the other two charges is

V(0.040 rn)2 + (0.030 rn)2 = 0.050 m

—

_jf0.O3O rna\0.040m

= 37° below the negative x-axis, or 2170 from the positive x-axis.F2 = Fc on B

Kqq (9.0x109N•m2/C)(8.2x106C)(6.0X 106 C)d2 (0.050 rn)2

= —177 N = 177 N at 217° from the positive x-axis (37° + 180°)The components of F2 are:F = F2 cos 0 = (177 N)(cos 217°) = —142 N = 142 N to the left

F2 = F2 sin 0 = (177 N)(sin 217°) = —106 N = 106 N down

The components of the net (resultant) force are:Fnet,x —208 N — 142N —350 N = 350 N, to left

Fnet, y = 106 N, down

Fnet = V’(350 N)2 + (106 N)2 = 366 N = 3.7X102N

0 —t 1j106N2 an

\350N

) = 17° below the negative x-axis

Fnet = 3.7X102N at 197° from the positive x-axis

61. The two pith balls in Figure 20-18 each have a mass of 1.0 g and an equalcharge. One pith ball is suspended by an insulating thread. The other is broughtto 3.0 cm from the suspended ball. The suspended ball is now hanging with thethread forming an angle of 30.0° with the vertical. The ball is in equilibriumwith FE, Fg and T• Calculate each of the following.

3.0cm i

• Figure 20-18

a. F on the suspended ball

F9 = mg = (1.0X103kg)(9.80 m/s2) = 9.8x103N

b. FE

Ftan 30.0° = -

9



FE = mg tan 30.00

= (1 .0X103kg)(9.80 mIs2)(tan 30.0°)

= 5.7X103N

c. the charge on the balls

KqqF

d2

Kq2F—-

—/(5.7X103 N)(3.0x102m2)

— 2 4x108Cq— i,J K — i,/ (9.0x109Nm2/C2)

62. Two charges, qA and q’ are at rest near a positive test charge, q, of 7.2 C. The

first charge, q’ is a positive charge of 3.6 C located 2.5 cm away from qT at 350;

qB is a negative charge of —6.6 JLC located 6.8 cm away at 125°.

a. Determine the magnitude of each of the forces acting on

F— Kqq

— (9.0X109N•m2/C2)(7.2X106C)(3.6X106C)A

— — (0.025 rn)2

= 3.7X102N, away (toward q.1.)

F— Kq.q

— (9.0x109N•m2/C2)(7.2X106C)(6.6x 106 C)B

— — (0.068 rn)2

= 92 N, toward (away from q1.)

b. Sketch a force diagram.

¶ “.

I

0

a

p

qB

92 N

qT

F= 3.8x102

C. Graphically determine the resultant force acting on

FB = 92

= 3.7x102N



Writing in Physicspage 56063. History of Science Research several

devices that were used in the seventeenthand eighteenth centuries to study staticelectricity. Examples that you mightconsider include the Leyden jar and theWimshurst mathine. Discuss how theywere constructed and how they worked.Student answers will vary, but shouldinclude information such as the following. The Leyden jar, invented in themid-i 740s, was the earliest capacitor. Itwas used throughout the eighteenth andnineteenth centuries to store charges forelectricity-related experiments anddemonstrations. The Wimshurst machinewas a device used in the nineteenth andearly twentieth centuries to produce anddischarge static charges. Wimshurstmachines, which were replaced by theVan de Graaff generator In the twentiethcentury, used Leyden jars to store thecharges prior to discharge.

64. In Chapter 13, you learned that forces existbetween water molecules that cause water tobe denser as a liquid between 0°C and 4°Cthan as a solid at 0°C. These forces are electrostatic in nature. Research electrostatic intermolecular forces, such as van der Waals forcesand dipole-dipole forces, and describe theireffects on mattet

Answers will vary, but students shoulddescribe the interactions betweenpositive and negative charges at themolecular level. Students shouldnote that the strength of these forcesaccounts for differences in meltingand boiling points and for the unusualbehavior of water between 0°C and 4°C.

Cumulative Reviewpage 56065. Explain how a pendulum can be used to

determine the acceleration of gravity.

) (Chapter 14)

Measure the length and period of thependulum, and use the equation for theperiod of a pendulum to solve for g.

66.. A submarine that is moving 12.0 rn/s sendsa sonar ping of frequency 1.50X Hztoward a seamount that is directly in front ofthe submarine. It receives the echo 1.800 slater. (Chapter 15)a. How far is the submarine from the

seamount?

d = Vt = (1533 m/s)(0.900 s) = 1380 mb What is the frequency of the sonar wave

that strikes the seamount?

fd =f5( = (1.50x1 Hz)

(1533 m!s—0.0 rn/s\ 1533 m/s — 12.0 m/s

1510 Hz

C. What is the frequency of the echoreceived by the submarine?

fd =f5( = (1510 Hz)

(1533 rn/s — (—12.0 mIs)\ 1533 m/s — 0.0 m/s

= 1520 Hz

67. Security Mirror A security mirror is usedto produce an image that is three-fourthsthe size of an object and is located 12.0 cmbehind the mirror. What is the focal lengthof the mirror? (Chapter 17)

-d,m=——

‘Jo

-d

— —(—12.0 cm)34,

= 16.0 cm

1_i +1fd0 d

)

d0d



= (16.0 cm)(—12.016.0 cm + (—12.0 cm)

= —48.0 cm

68.. A 2.00-cm-tall object is located 20.0 cmaway from a diverging lens with a focallength of 24.0 cm. What are the image position, height, and orientation? Is this areal or a virtual image? (Chapter 18)

1_i +1fd0

ddJ

— d0 —f

— (20.0 cm)(—24.0 cm)20.0 cm — (—24.0 cm)

= —10.9 cm

h1 -

m=—=---h0 u0

—dh._ 10“I— d0

— —(—10.9 cm)(2.00 cm)20.0 cm

= 1.09cm

This Is a virtual image that is upright inorientation, relative to the object.

69. Spectrometer A spectrometer contains agrating of 11,500 slits/cm. Find the angleat which light of wavelength 527 nm hasa first-order bright band. (Chapter 19)

The number of centimeters per slit isthe slit separation distance, d.

1 slit= 11,500 slitslcm

Challenge Problempage 552As shown in the figure below, two spheres ofequal mass, m, and equal positive charge, q,are a distance, r, apart.

mass = mcharge = q

• Figure 20-11

1. Derive an expression for the charge, q, thatmust be on each sphere so that the spheresare in equilibrium; that is, so that theattractive and repulsive forces betweenthem are balanced.

The attractive force is gravitation, andthe repulsive force is electrostatic, sotheir expressions may be set equal.

F9 GAB= d2 =Kd2=Fe

The masses and charges are equal,and the distance cancels, so

Gm2 = Kq2, and

q= m./

— / (6.67x1011N.m2/kg2)— (9.0x109N•m2/C2)

= (8.61 X10—ll CIkg)m

2. If the distance between the spheres is doubled, how will that affect the expression forthe value of q that you determined in theprevious problem? Explain.

The distance does not affect the valueof q because both forces are inverselyrelated to the square of the distance,and the distance cancels out of theexpression.

3. If the mass of each sphere is 1.50 kg, determine the charge on each sphere needed tomaintain the equilibrium.

q = (8.61X10 CIkg)(1.50 kg)

= 1.29X100C

Physics: Principles and Problems

mass = mcharge = q

d = 8.70X105cm

A = dsin 0

o = sin1()

—

sin—1(527x109m

— 8.70X103m )= 0.003470

426 Solutions Manual

II

1

20.1 Electric Charge

Documents

Transcript of 20.1 Electric Charge