461

CHAPTER 22 ELECTRIC FORCE AND ELECTRIC CHARGE

Select odd-numbered solutions, marked with a dagger (†), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.

†22-1. According to Coulomb's law, the magnitude of the electrostatic force one charged object exerts

on another is given by 1 22 ,q qF k

r= where q1 and q2 are the magnitudes of the two charges, r is the

distance between them, and k is the Coulomb constant. In this particular problem, we have

( ) ( )( )

29 51 2

22 2

40 C 40CN m8.99 × 10 5.8 × 10 NC 5000 m

q qF kr

−⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

22-2.( ) ( )19 192

9 91 22 2 10 2

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 2.90 × 10 NC (2.82 × 10 m)

q qF kr

− −−

−

− +⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

22-3.( ) ( )19 192

91 22 2 15 2

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 57.7 N 58 NC (2.0 × 10 m)

q qF kr

− −

−

− +⎛ ⎞= = = =⎜ ⎟

⎝ ⎠

i

28 227

57.7 N 3.5 × 10 m/s1.67 × 10 kg

Fam −= = =

22-4.( ) ( )19 192

91 22 2 14 2

2 1.60 × 10 C 92 1.60 × 10 CN m8.99 × 10 17 NC (5.0 × 10 m)

q qF kr

− −

−

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

27 227

17 N 2.5 × 10 m/s4(1.67 × 10 kg)

Fam −= = =

†22-5. The magnitude of the electric force between these two quarks is given by Coulomb's law. Sincethe two particles have the same sign, they will repel each other with a force

( ) ( )19 192 123 391 2

2 2 15 2

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 51 NC (1.0 × 10 m)

q qF kr

− −

−

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

22-6. Each electron carries a charge of 1.602 × 10−19 C. To determine how many electrons, divide thetotal charge by the charge on an individual electron.

612

19

1.0 × 10 C 6.25 × 10 electrons1.60 × 10 C

N−

−

−= =−

†22-7. Each electron carries a charge of 1.60 × 10−19 C. To determine how many electrons, divide thetotal charge by the charge on an individual electron.

2019

25 C 1.6 × 10 electrons1.60 × 10 C

N −

−= =−

22-8.19

31e 11

e

1.602 177 × 10 C 9.109 386 × 10 kg/ 1.758 820 × 10 C/kg em

e m

−−−= = =

−

22-9.( ) ( )19 192

9 71 22 2 10 2

1.60 × 10 C (17) 1.60 × 10 CN m8.99 × 10 2.39 × 10 NC (1.28 × 10 m)

q qF kr

− −−

−

+ +⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

CHAPTER 22

462

22-10. (a)( ) ( )27 272

111 2g 2 2 12 2

40

1.67 × 10 kg 1.67 × 10 kgN m6.67 × 10 kg (1.0 × 10 m)

1.9 × 10 N

m mF Gr

− −−

−

−

⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

i

(b)( ) ( )19 192

91 2e 2 2 12 2

4

436e

40g

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 C (1.0 × 10 m)

2.3 × 10 N

2.3 × 10 N 1.2 × 101.9 × 10 N

q qF kr

FF

− −

−

−

−

−

+ +⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

= =

i

(c)( ) ( )19 192

91 22 40

g

6

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 C 1.9 × 10 N

1.1 × 10 m

q qr kF

− −

−

+ +⎛ ⎞= = ⎜ ⎟

⎝ ⎠

=

i

†22-11. Although the gravitational and electric forces have similar formulas, the magnitudes of theseforces are very different. The electric force is a much stronger force that that of gravity.

( ) ( )

( ) ( )

11 112111 2

g 2 2 10 2

13

19 19291 2

e 2 2 10 2

8

1.0 × 10 kg 1.0 × 10 kgN m6.67 × 10 kg (1.0 × 10 m)

6.7 × 10 N

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 C (1.0 × 10 m)

2.3 × 10 N

m mF Gr

q qF kr

− −−

−

−

− −

−

−

⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

+ +⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

i

i

22-12.( ) ( )19 192

91 22 2 13 2

1.60 × 10 C 82 1.60 × 10 CN m8.99 × 10 0.044 NC (6.5 × 10 m)

q qF kr

− −

−

− +⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

28 231

0.044 N 4.8 × 10 m/s9.11 × 10 kg

Fam −= = =

†22-13. The electric charge on a proton is +1.60 × 10−19 C. One mole contains Avogadro's number ofparticles, NA = 6.02 × 1023 particles. Therefore, the total charge of mole mole of protons isFaraday constant = ( +1.60 × 10−19 C/proton)( 6.02 × 1023 protons) = 9.63 × 104 C

22-14. Each second, there are 1.3 C of electrons moving in the lightbulb filament. Each electron carries acharge of 1.602 × 10−19 C. Therefore, 1.3 C corresponds to

1819

1.3 C 8.1 × 10 electrons1.60 × 10 C

N −

−= =−

†22-15. If there are 7.5 × 10−6 C of excess electrons on the sphere; and each electron carries a charge of1.602 × 10−19 C, then

613

19

7.5 × 10 C 4.7 × 10 electrons1.60 × 10 C

N−

−

−= =−

22-16. The number of moles of iron is30.30 g 5.38 × 10 moles

55.8 g/moln −= =

CHAPTER 22

463

Each iron atom has 26 electrons, so the total number of electrons isN = 26nNA = 26(5.38 × 10−3 mol)(6.022 × 1023 atoms/mol) = 8.4 × 1022 electrons

†22-17. The number of moles of copper is22.7 g 4.25 × 10 moles

63.5 g/moln −= =

Each copper atom has 29 electrons (see the periodic table), so the total number of electrons isN = 29nNA = 29(4.25 × 10−2 mol)(6.022 × 1023 atoms/mol) = 7.42 × 1023 electronsThe total charge for these electrons is

q1 = (7.42 × 1023 electrons)(−1.60 × 10−19 C) = −1.19 × 105 CAn equal quantity of positive charge exists due to the protons in the remaining nucleii,

q2 = +1.19 × 105 CTherefore, the magnitude of the electric force on the electrons due to the nucleii is

( ) ( )5 529 191 2

2 2 2

1.19 × 10 C 1.19 × 10 CN m8.99 × 10 3.2 × 10 NC (2.0 m)

q qF kr

− +⎛ ⎞= = =⎜ ⎟

⎝ ⎠

i

22-18.

ElementPercentageof body (%) Mass (kg) Number of moles

O 70 51.1 O51100 g 3190 moles

16.0 g/moln = =

C 20 14.6 C14600 g 1220 moles

12.0 g/moln = =

H 10 7.3 H7300 g 7230 moles

1.01 g/moln = =

Each mole contains Avogadro's number of atoms with each atom containing it’s atomic numberof electrons and protons. Therefore, the total number of electrons or protons in the body isapproximately,

[ ]23

28

6.022 × 10 atoms/mol (3190 mol O)(8) + (1220 mol C)(6) + (7230 mol H)(1)

2.4 × 10 electrons or protons

N =

=

†22-19. One mole of sodium chloride, NaCl, has a mass which is the sum of the molar masses of Na andCl, (22.99 g Na/mol) + (35.45 g Cl/mol) = 58.44 g NaCl/mol. The number of moles to bedisolved in 100 g of water is

NaCl36 g 0.62 moles

58.44 g/moln = =

Each sodium atom has 11 electrons and 11 protons and each chlorine atom has 17 electrons and17 protons. The total number of electrons or protons added to the water will be

[ ]23NaCl

25

6.022 × 10 atoms/mol (0.62 mol Na)(11) + (0.62 mol Cl)(17)

1.0 × 10 electrons or protons

N =

=The number of electrons or protons in the 100 g of water is found in similar manner. One mole ofwater, H2O, has a mass which is the sum of the molar masses of two hydrogen atoms (2 × 1.01 gH/mol) and one oxygen atom (16.0 g O/mol) = 18.0 g H2O/mol. The number of moles in 100 g ofwater is

CHAPTER 22

464

H2O100 g 5.6 moles

18.0 g/moln = =

Each hydrogen atom has 1 electron and 1 proton and each oxygen atom has 8 electrons and 8protons. The total number of electrons or protons in the water will be

[ ]23H2O

25

6.022 × 10 atoms/mol 2(5.6 mol H)(1) + (5.6 mol O)(8)

3.3 × 10 electrons or protons

N =

=Therefore, the factor by which the number of electrons or protons increases is

25 25H2O NaCl

25H2O

3.3 × 10 1.0 × 10 1.33.3 × 10

N NN

+ += =

22-20.( ) ( )6 62

91 2e 2 2 2

2

2.0 × 10 C 2.0 × 10 CN m8.99 × 10 C (1.0 m)

3.6 × 10 N (repulsive)

q qF kr

− −

−

− −⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

i

22-21. The ratio of the exact value to the simpler value is2

92

29

2

N m8.987 551 787 × 10 C 0.9986 99.9%

N m9.0 × 10 C

= =

i

i

22-22. The total charge on each disk is the product of the area and the charge per unit area.2 8 2 12(0.01 m) (2.5 × 10 C/m ) 7.85 × 10 Cq π − −= =

The electric force that each plate exerts on the other is( ) ( )12 122

91 2e 2 2 2

13

7.85 × 10 C 7.85 × 10 CN m8.99 × 10 C (2.0 m)

1.4 × 10 N (repulsive)

q qF kr

− −

−

⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

i

†22-23. Through the ionization, an atom at one end has one excess electron, giving it a −e net negativecharge; and an atom at the other end has lost an electron, giving it a +e net positive charge. Theelectric force then compresses the chain molecule by ∆L = (−0.012)(1.9 × 10−6 m). The moleculeis to be modeled like a spring. The electric force compresses the spring until the electric force andthe spring force are equal in magnitude, but oppositely directed. Applying Hooke's law, we find

( ) ( )

S e

19 1929

22 6 69

6

1.60 × 10 C 1.60 × 10 CN m8.99 × 10 C (1.9 × 10 m) + ( 0.012)(1.9 × 10 m)

2.9 × 10 N/m( 0.012)(1.9 × 10 m)

F FkL L

− −

− −−

−

= =∆ ∆

− +⎛ ⎞⎜ ⎟

⎡ ⎤⎝ ⎠ −⎣ ⎦= =−

i

22-24.( ) ( )31 152

111 2g 2 2 5 2

35

9.11 × 10 kg 2.0 × 10 kgN m6.67 × 10 kg (1.0 × 10 m)

1.2 × 10 N

m mF Gr

−−

−

⎛ ⎞= = ⎜ ⎟

⎝ ⎠=

i

For the electric force to be equal in magnitude, but oppositely directed to that of the gravitationalforce,

CHAPTER 22

465

( ) ( )

( )

e

1 2 1 22 2

211 15 31

217

29 19

2

N m6.67 × 10 2.0 × 10 kg 9.11 × 10 kgkg

8.4 × 10 C N m8.99 × 10 1.60 × 10 C

C

gF F

q q m mk Gr r

kQe GMm

GMmQke

− −

−

−

=

=

=

⎛ ⎞⎜ ⎟⎝ ⎠= = = −

⎛ ⎞−⎜ ⎟

⎝ ⎠

i

i

The number of electrons in this amount of charge is17

219

8.4 × 10 C 5.3 × 10 electrons1.60 × 10 C

QNe

−

−

−= = =−

†22-25. Charge 1 is located at xy corredinate (2.0 m, 0 m) and charge 2 is located at (0 m, −3.0 m). Sincethe charges have the same sign, the force on charge 2 due to charge 1 will be directed away fromcharge 1 in the direction given by

2 2

( 2.0 m) ( 3.0 m) 2.0 3.013 13( 2.0 m) ( 3.0 m)

− + − − −= = +− + −

i jr i j

The force exerted on charge 2 due to charge 1 is then

( ) ( )

1 212 12 2

8 629

2 2 2

5 5

2.0 × 10 C 3.0 × 10 CN m 2.0 3.08.99 × 10 C ( 2.0 m) ( 3.0 m) 13 13

( 2.3 × 10 N) + ( 3.5 × 10 N)

q qF kr

− −

− −

= =

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

F r r

i j

i j

i

Similarly, the force exerted on charge 1 due to charge 2 is

( ) ( )

1 221 21 2

8 629

2 2 2

5 5

ˆ ˆ r r

2.0 × 10 C 3.0 × 10 CN m 2.0 3.0ˆ ˆ8.99 × 10 x yC ( 2.0 m) ( 3.0 m) 13 13ˆ ˆ(2.3 × 10 N)x + (3.5 × 10 N)y

q qF kr

− −

− −

= =

− −⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟− + − ⎝ ⎠⎝ ⎠=

F

i

22-26. The magnitude of the electric force is to be equal to the weight of the plastic chip.2

2

8 23 12

29

2

(5.0 × 10 kg)(9.81 m/s )(1.0 × 10 m) 7.4 × 10 CN m8.99 × 10

C

qmg kr

mgq rk

−− −

=

= = =i

†22-27. Let Q represent the amount of charge on the Earth. The charge on the moon will be1.746.38

q Q=

If the magnitude of the electric force is to be equal to that of the gravitational force, then

CHAPTER 22

466

( ) ( )

e

1 2 1 22 2

2

211 24 22

2

29

2

14

1.746.38

N m6.38 6.67 × 10 5.98 × 10 kg 7.35 × 10 kgkg6.38

1.74 N m1.74 8.99 × 10 C

1.09 × 10 C

gF F

q q m mk Gr r

kQq GMm

k Q GMm

GMmQk

−

=

=

=

⎛ ⎞ =⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠= =

⎛ ⎞⎜ ⎟⎝ ⎠

=

i

i

The number of electrons in this amount of charge is14

32Earth 19

1.09 × 10 C 6.81 × 10 electrons1.60 × 10 C

QNe −

−= = =−

( )14 13

1332

moon 19

1.74 1.74 1.09 × 10 C 2.97 × 10 C6.38 6.38

2.97 × 10 C 1.86 × 10 electrons1.60 × 10 C

q Q

N −

= = =

= =

22-28.7

1119

1.0 × 10 C 6.3 × 10 ions1.60 × 10 C

QNe

−

−

+= = =+

There are 6.3 × 1011 ions per square meter. The fraction of surface atoms that must be ionized is11 2

819 2

6.3 × 10 ions/m 3.2 × 10 ions/atom2.0 × 10 atoms/m

−=

†22-29.0

0

/1 22e_mod /

1 2e 02

1r r

r r

q qk eF rr eq qF rkr

−

−= = ≈ −

The fractional deviation at r = 1.0 m is9

90

1.0 m 1.0 × 101.0 × 10 m

rr

−= =

The fractional deviation at r = 1.0 × 104 m is4

59

0

1.0 × 10 m 1.0 × 101.0 × 10 m

rr

−= =

22-30. An electron is located at xy coordinate (4.0 × 10−11 m, 2.0 × 10−11 m) and a proton is located at theorigin (0 m, 0 m). Since the charges have opposite signs, the force exerted on the electron due tothe proton will be directed toward the proton in the direction

11 11

11 2 11 2

( 4.0 × 10 m) ( 2.0 × 10 m)( 4.0 × 10 m) ( 2.0 × 10 m)0.89 0.45

− −

− −

− + −=− + −

= − −

i jr

i j

CHAPTER 22

467

The force exerted on the electron due to the proton is then

( )

2

ep ep 2

21929

2 11 2 11 2

7 8

1.60 × 10 CN m8.99 × 10 ( 0.89 0.45 )C ( 4.0 × 10 m) ( 2.0 × 10 m)

(1.0 × 10 N) + (5.2 × 10 N)

qF kr

−

− −

− −

= = −

⎛ ⎞= − − −⎜ ⎟ − + −⎝ ⎠=

F r r

i j

i j

i

Similarly, the force exerted on the electron due to the proton is

( )

2

pe ep 2

21929

2 11 2 11 2

7 8

1.60 × 10 CN m8.99 × 10 ( 0.89 0.45 )C ( 4.0 × 10 m) ( 2.0 × 10 m)

( 1.0 × 10 N) + ( 5.2 × 10 N)

qF kr

−

− −

− −

= − = −

⎛ ⎞= − −⎜ ⎟ − + −⎝ ⎠= − −

F r r

i j

i j

i

†22-31. Under normal circumstances, an oxygen atom with 8 electrons, 8 protons, and 8 neutrons wouldbe electrically neutral. However, if each of these particles has a slight differential charge, themaximum amount of charge the oxygen atom due to a charge differential between the 8 protonsand electrons as well as the 8 neutrons isq = (8 + 8 + 8)(1.0 × 10−21)(1.60 × 10−19 C) = 3.8 × 10−39 CAssuming there are two such oxygen atoms, the ratio of the electric force to the graviational forceis

1 222

e2

1 22

g

q qkF kqrm mF GmG

r

= =

where m is the mass of the electrons, protons, and neutrons in each oxygen atom. Inserting thevalues of the parameters gives:

( )

( ) ( ) ( )

2e

2

2 29 392

2 211 31 27 272

6

N m8.99 × 10 3.8 × 10 CC

N m6.67 × 10 8 9.11 × 10 kg 8 1.673 × 10 kg 8 1.675 × 10 kgkg

2.8 × 10 , The net force is attractive.

g

F kqF Gm

−

− − − −

−

=

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞ ⎡ ⎤+ +⎜ ⎟ ⎣ ⎦⎝ ⎠=

i

i

22-32. The electric force provides the centripetal force to maintain the orbit of the electron in theclassical orbital atomic model.

( )( ) ( )

2 2

2

2 29 1922

631 11

N m8.99 × 10 1.60 × 10 CC

2.2 × 10 m/s9.11 × 10 kg 5.3 × 10 m

mv qkr r

kqvmr

−

− −

=

⎛ ⎞⎜ ⎟⎝ ⎠= = =

i

CHAPTER 22

468

The orbital period is the circumference of the circle divided by the orbital speed.11

166

2 2 (5.3 × 10 m) 1.5 × 10 s2.2 × 10 m/s

rTvπ π −

−= = =

†22-33. The situation is similar to example 6,except that the negative charge on the yaxis has been replaced by a positive chargeQ. Study the drawing. The repulsiveelectric force on the positive charge q dueto the upper charge Q is F1; and that due tothe lower charge Q is F2. The net force isthe sum of these two forces. The ycomponents of these forces are equal inmagnitude, but oppositely directed, so the ycomponent of the net electric force on q iszero newtons. The x components of the twoforces add together to give a net force F =F1 + F2 which is directed along the positivex axis with a magnitude

( ) ( ) ( ) 22 2 3 / 22 2 2 222 2 42

22 cos 2d d dd

Qq Qq x kqxQF k kx x xx

θ= = =+ + ⎡ ⎤++ ⎣ ⎦

The net force is directed in the +x direction.22-34. The three charges are positioned on the y axis, so the net force will be along the y axis. Let F1 be

the force on the +10-µC charge due to the −40-µC charge and F2 be the force on the +10-µCcharge due to the +40-µC charge. The magnitude of the net force on the +10 C charge is

( ) ( ) ( ) ( )

1 31 21 2 2 2

12 13

29

2 3 2 3 2

5

10 C 40 C 10 C 40 CN m8.99 × 10 C (3.0 × 10 m) (8.0 × 10 m)

3.4 × 10 N

q qq qF F F kr r

⎡ ⎤= + = +⎢ ⎥

⎣ ⎦⎡ ⎤+ − + +⎛ ⎞

= +⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠ ⎣ ⎦

=

i

Because F1 has the larger magnitude and it is directed in the +y direction, the net force is also inthe +y direction. Therefore, the net force on the +10-C charge is F = 5+3.4 × 10 N j

†22-35. The two electric forces exerted on the electron due to thehydrogen nucleus (a proton) FH and the chlorine nucleus(17 protons) FCl are both attractive because the chargeshave opposite signs. Because each of these forces point indifferent directions, vector methods must be used todetermine the net force on the electron. Consider thedrawing. To determine the x and y components of FCl, theangle must be determined from the distances d1 and d2,which are given.

111 11

102

5.0 × 10 mtan tan 21.31.28 × 10 m

dd

θ−

− −−

⎛ ⎞ ⎛ ⎞= = = °⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

0.5d

0.5dx

+Q

+Q

+q

F2

F1

F1+F2θ

θ

Fd1

Cl

FCl FH

θ x

y

H

e

d2

CHAPTER 22

469

The x component of the net force is the sum of the horizontal components of the forces FH andFCl.

( ) ( ) ( )H, Cl,

19 1929

2 11 2 10 2

7

1.60 × 10 C 17 1.60 × 10 CN m8.99 × 10 0 cos 21.3C (5.0 × 10 m) (1.28 × 10 m)

1.9 × 10 N

x x xF F F− −

− −

−

= +

⎡ ⎤− +⎛ ⎞ ⎢ ⎥= + °⎜ ⎟ ⎢ ⎥+⎝ ⎠ ⎣ ⎦=

i

This component actually points in the −x direction as may be seen on the drawing.Similarly, the y component of the net force is

( ) ( )

( ) ( ) ( )

H, Cl,

19 19

11 229

2 19 19

11 2 10 2

7

1.60 × 10 C 1.60 × 10 C

(5.0 × 10 m)N m8.99 × 10 C 1.60 × 10 C 17 1.60 × 10 C

sin 21.3(5.0 × 10 m) (1.28 × 10 m)

1.7 × 10 N

y y yF F F− −

−

− −

− −

−

= +

⎡ ⎤− +⎢ ⎥⎢ ⎥⎛ ⎞

= ⎢ ⎥⎜ ⎟⎝ ⎠ − +⎢ ⎥

+ °⎢ ⎥+⎢ ⎥⎣ ⎦

=

i

This component is directed in the −y direction.Therefore, the net electric force is

7 7 = (1.9 × 10 N) (1.7 × 10 N)− −− −F i j

22-36. For each of the five +Q charges, the contribution to the netelectric force on the positive charge q is along a line between thetwo charges, Q and q, and directed away from the individualcharge as shown. Each of the five contributions has the samemagnitude.

The five force vectors are equally distributed in anglearound the center of the pentagon, so the angle between eachvector is 72°. In the force diagram, the relevant angles withrespect to the x and y axes are shown. We may now write downthe x component of the net force on charge q as

1, 2, 3, 4, 5,

2 2 2 2

1, 2, 3, 4, 5,

2 2 2 2 2

2 2

0 cos18 sin18 sin18 cos18

0

sin18 cos18 cos18 sin18

2 sin18 2

x x x x x x

y y y y y y

F F F F F F

Qq Qq Qq Qqkr r r r

F F F F F F

Qq Qq Qq Qq Qqkr r r r rQq Qq Qqkr r r

= + + + +

⎡ ⎤= − ° − ° + ° + °⎢ ⎥⎣ ⎦== + + + +

⎡= − − ° + ° + ° − °⎢⎣

= − − ° + 2 cos 36 0⎡ ⎤° =⎢ ⎥⎣ ⎦

where r is the distance from a vertex to the center of the pentagon. F = 0 N.

+Q

+Q

+Q+Q

+Q

+q

72°x

y

18°

36°

18°F1

F3

F5

F4

F2

CHAPTER 22

470

†22-37. Each ball has three forces acting on it: thegravitational force of the Earth, the tension in thethread, and the electric force due to the other ball.Since each ball is in equilibrium, the sum of theseforces is equal to zero newtons. The sum of forces inthe vertical direction may be used to find the tension:

cos 20 0

cos 20

yF T mg

mgT

= ° − =

=°

∑

The sum of forces in the horizontal direction is2

2

2 2 2

sin 20 0

sin 20 sin 20 tan 20cos 20

xQF T kr

r T r mg r mgQk k k

= ° − =

° ° °= = =°

∑

The distance between the balls is r = (25 cm) + 2(10 cm)(sin 20°) = 31.84 cm = 0.3184 m.

2

2

2 2 4 2

N m9C

7

tan 20 (0.3184 m) (2.5 × 10 kg)(9.81 m/s ) tan 208.99 × 10

1.0 × 10 C

r mgQk

−

−

° °= =

=

i

22-38. The x component of the net electric force is

( ) ( )

( ) ( ) 2

2 22 22 2

2 3 / 222 222 42

2cos cosx d d

d dd

Qq QqF kx x

x Qq kqQxkx xx

θ θ⎡ ⎤⎢ ⎥= −⎢ ⎥+ +⎣ ⎦

− −= =+ ⎡ ⎤++ ⎣ ⎦

The y component of the net electric force is

( ) ( ) ( ) ( ) 22 2 2 3 / 22 2 22 222 2 2 42

2 3 ( / 2) 3sin sin2

y d d d dd

Qq Qq Qq d dkqQF k kx x x xx

θ θ⎡ ⎤ − −⎢ ⎥= − − = =⎢ ⎥+ + + ⎡ ⎤++⎣ ⎦ ⎣ ⎦

Therefore,

2 21 2 3 / 2 3 / 22 24 4

3

2d d

kqQx dkqQ

x x= + = − −

⎡ ⎤ ⎡ ⎤+ +⎣ ⎦ ⎣ ⎦F F F i j

22-39. The x component of the net electric force is2 2

2 2

2

2

2 2 2

1.35

xQ QF kL L

QkL

⎡ ⎤= − −⎢ ⎥

⎣ ⎦

= −

The y component of the net electric force is2 2 2

2 2 2

2 1.352 2y

Q Q QF k kL L L

⎡ ⎤= − − = −⎢ ⎥

⎣ ⎦

20° 20°

mg mg

F21 F12T T

25 cm10 cm10 cm

0.5d

0.5dx

+Q

−2Q

+q

F2F1

F1+F2

θθ

+Q

F2

F1

L

Lx

y

+Q

+Q

−QF3

2L

CHAPTER 22

471

Therefore,2 2

1 2 3 2 21.35 1.35Q Qk kL L

= + + = − −F F F F i j

22-40. Due to the symmetry of the situation, the verticalcomponents are equal in magnitude, but oppositelydirected. Thus they cancel. The x components of each ofthese forces is directed parallel to the positive x axis.

( )2 22

2

2 4 242x

L

qQ kqQF kL

= =

Therefore, the net force on the charge +q is

1 2 3 4 2

4 2kqQL

= + + + =F F F F F i

†22-41. Consider the drawing indicating the forces on the electron.The electric force F1 on the electron due to the positivecharge is directed toward the positive charge, since it has anopposite sign. The force F2 exerted on the electron due tothe negative charge is directed away from the negativecharge, since both charges have the same sign. The netforce on the electron is the vector sum of these two forces.The magnitude of the x component of the net force is

( )( ) ( )

( ) ( )2

2

1 21, 2, 2 2

1 2

19

3 2N m9

C 19

3 2 3 2

15

cos

1.60 × 10 C 40 C

(4.0 × 10 m)8.99 × 10

1.60 × 10 C 30 C 4.0 km (4.0 × 10 m) (6.0 × 10 m) 52 km

3.1 × 10 N

x x xq e q eF F F k kr r

θ

−

−

−

= + = +

⎡ ⎤− +⎢ ⎥⎢ ⎥

= ⎢ ⎥− −⎢ ⎥⎛ ⎞+⎢ ⎥⎜ ⎟+⎢ ⎥⎝ ⎠⎣ ⎦

= −

i

The magnitude of the y component of the net force is

( ) ( )

21, 2, 2

2

1929

2 3 2 3 2

16

0 sin

1.60 × 10 C 30 CN m 6.0 km8.99 × 10 C (4.0 × 10 m) (6.0 × 10 m) 52 km

6.9 × 10 N

y y yq eF F F kr

θ

−

−

= + = +

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

i

Therefore, the net force on the electron is15 16

1 2 (3.1 × 10 N) (6.9 × 10 N)− −= + = − +F F F i j

+Q

F2

F1

L

Lx

y

+Q

−Q

−Q

F3

22

L+q

F4

−30 C

F2

F1

6.0 km

x

y

+40 C −e

4.0 km

F

θ

52 km

CHAPTER 22

472

22-42. Consider the drawing indicating the forces on the electron.The electric force F1 on the electron due to the positivecharge is directed toward the positive charge, since it has anopposite sign. The force F2 exerted on the electron due tothe negative charge is directed away from the negativecharge, since both charges have the same sign. The netforce on the electron is the vector sum of these two forces.The magnitude of the x component of the net force is

1 2 1 21, 2, 2 2

1 2

( )cos cos cosx x xq e q e ke q qF F F k kr r r

θ θ θ+= + = + =

( ) ( )( )

1929 1 3.0 km

4.0 km2 3 2 3 2

16

1.60 × 10 C 40 C 30 CN m8.99 × 10 cos tanC (4.0 × 10 m) (3.0 × 10 m)

4.6 × 10 N

−−

−

− + −⎛ ⎞⎡ ⎤= ⎜ ⎟ ⎣ ⎦+⎝ ⎠

=

i

The magnitude of the y component of the net force is

( ) ( )( )

1 2 1 21, 2, 2 2

1 2

1929 1 3.0 km

4.0 km2 3 2 3 2

15

( )sin sin sin

1.60 × 10 C 40 C 30 CN m8.99 × 10 sin tanC (4.0 × 10 m) (3.0 × 10 m)

2.4 × 10 N

y y yq e q e ke q qF F F k kr r r

θ θ θ

−−

−

+= + = + =

− + + −⎛ ⎞⎡ ⎤= ⎜ ⎟ ⎣ ⎦+⎝ ⎠

=

i

Therefore, the net force on the electron is16 15

1 2 (4.6 × 10 N) i (2.4 × 10 N) j− −= + = − +F F F

22-43. Consider the drawing. The ball on the left has a charge q1 = +2.0 ×10−7 C and the ball on the right has a charge q2 = +6.0 × 10−8 C. Thereare three forces acting on each ball: the gravitational force of theEarth, the tension in the thread, and the electric force due to the otherball. Since each ball is in equilibrium, the sum of these forces is equalto zero newtons. Because the angles are the same for both balls andthe symmetry of the situation, the tension in both threads is identical.Because of this, the vertical component of the tension is the same forboth balls. Thus, the mass of each ball is the same. Therefore, findingthe mass of one ball gives the mass of the other. The distance betweenthe two balls is r = 2(10 cm)sin 25°.The sum of forces in the horizontal direction for ball 1 is:

1 21 2sin 25 0x

q qF T kr

= ° − =∑from which the tension in the left thread can be determined.

( ) ( )( )

7 829 31 2

1 22 2

2.0 × 10 C 6.0 × 10 CN m8.99 × 10 1.28 × 10 Nsin 25 C 2(0.10 m) sin 25q qT k

r

− −−

+ +⎛ ⎞= = =⎜ ⎟° °⎝ ⎠

i

The sum of the forces in the vertical direction for ball 1 is:

25° 25°

m1g m2g

F21 F12T1 T2

10 cm10 cm

−30 C

F2F1

3.0 kmx

y

+40 C

−e

4.0 km

F

θ

3.0 kmθ

CHAPTER 22

473

( )1 1

341

1 2

41 2

cos 25 0

1.28 × 10 N cos 25cos 25 1.2 × 10 kg9.81 m/s

1.2 × 10 kg

yF T m g

Tmg

m m

−−

−

= ° − =

°°= = =

= =

∑

22-44. Consider the three charges at the vertices of an equilateraltriangle with sides of length a. The net force on the positivecharge +q placed at the origin is to be determined. The xcomponent of the net force on this charge is

2 2 2 2

2 2 2 2cos 60 cos 602x

q q q qF k k k ka a a a

= − ° = = °

and the y component of the net force on this charge is2 2

2 20 sin 60 sin 60yq qF k ka a

= − ° = °

The magnitude of the net force is2 22 2 2

2 22 2 2 cos 60 sin 60

2x yq q qF F F k k ka a a

⎛ ⎞ ⎛ ⎞= + = ° + ° =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠†22-45. The sum of electric forces acting on the charge q0 is zero

newtons. Therefore, the x component of the net force must beequal to zero.

( ) ( )0 0

2 22 2

2 2

cos 45 cos 45 = 0xQq QqF k k

a a= ° − °

In the y direction, The distance between the −q charge and q0 is

( )0

32 cos 30 1

2qqar a a= + ° = +

( ) ( ) ( )

( ) ( ) ( ) ( )( )

0 0 02 2 22 3 2 2

2 22

0 0 0 0 02 2 2 2 22 23 2 2 3

2 22 2

23

2

sin 45 sin 451

42 2 02 2 21 1

2 2 1 9.85

yqq Qq QqF k

a aa

qq Qq Qq qq Qqk kaa aa a

qQ

⎡ ⎤⎢ ⎥= − ° − °⎢ ⎥+⎢ ⎥⎣ ⎦⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥= − − = − =⎢ ⎥ ⎢ ⎥+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

= + =

22-46. All of the charges are alike, so all of the forces exerted on one ofthe corner charges by the others are directed away from thecharge. Three of the charges are located a distance a away. Threecharges are located a distance 2a away; and one charge is

3.a Therefore, the net force on the corner charge is

−q

+Q

a

60° 60°

a

+Q0.5a

0.5aq0

+q

+q

−q

F−F+

a

60° 60°a

xy

z

CHAPTER 22

474

2 2 2 2

2 22 2

22 2 22 2 2

2 2 2

2 2 22 3( 2) ( 3)

2 3 2 633 3 3

x y z

x y z

q q q kqF F F ka aa a

kq kq kqF F F Fa a a

⎡ ⎤= = = + + =⎢ ⎥

⎣ ⎦

⎛ ⎞ ⎛ ⎞= + + = = =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

22-47. 2 2 2 2

1 1( ) ( )

Qq QqF k kQqd x x x d x

⎡ ⎤ ⎡ ⎤−= + = −⎢ ⎥ ⎢ ⎥+ +⎣ ⎦ ⎣ ⎦

As x becomes very large, the d term becomesnegligible; and F goes to zero. Thus, it appears to thecharge q that the other two charges, which are equalin magnitude, but opposite in sign, are really oneneutral object.

22-48. Both the charge +q and the line of charge arepositive, so the net electric force on charge q will bedirected to the left, the negative x direction. Consideran infinitesimally small portion of the rod with acharge dq′ = λdx. The magnitude of the force oncharge q due to dq′ is

2 2

dq dxdF kq kqr r

λ′= =

The distance from the infinitesimal charge to charge q depends on how far it is from the end ofthe charged rod, x. From the drawing, we find r = d + L − x. Using this information, we may findthe magnitude of the total force on charge q by integrating over the length of the rod.

20 ( )L dxF kq

d L xλ=

+ −∫Let u = d + L − x, then du = −dx and the limits of integration accordingly change.

2

1 1 1d

d

d Ld l

duF kq kq kqu u d d L

λ λ λ+

+

⎛ ⎞ ⎛ ⎞= − = + = + −⎜ ⎟ ⎜ ⎟+⎝ ⎠ ⎝ ⎠∫22-49. The tetrahedron is a four-sided pyramid with

each side an equilateral triangle. All of thesides of the tetrahedron have length a; andany two sides that meet at a vertex do so witha 60° angle between them. The net force onthe charge shown on the lower left hascomponents in the x, y, and z directions. Letthe base of the pyramid lie in the x-y plane,then

2

1 2

Qka

= −F j

2 2

2 2 2(sin 60 ) (cos 60 )Q Qk ka a

= ° − °F i j

2 2 2

3 2 2 2(sin 30 )(cos 60 ) (cos 30 )(cos 60 ) (sin 60 )Q Q Qk k ka a a

= ° ° − ° ° − °F i j k

+Q−Q +q−d/2 d/2 x

xqdq′

d L

+Q

+Q

+Q

+QF1

F2F3

a

a

a

xy

z

CHAPTER 22

475

Vector adding these three forces together gives,

( )( )

( )

2

1 2 3 2

2 2

2 2

2

2

sin 60 + sin 30 (cos 60 )

1+ cos 60 + (sin 30 )(cos 60 ) (cos 60 )

(1.116 1.75 0.5 )

Qka

Q Qk ka akQa

= + + = ° ° ° −

° ° ° − °

= − −

F F F F i

j k

i j k

22-50. Both lines of charge are positive, so the electricforce on the left rod will be directed to the left, thenegative x direction. The ends of the rod nearesteach other are separated by a constant distance x.Consider an infinitesimally small portion of the rodwith a charge dq1 = λ dx1, where λ = Q/L. On theother rod, there is an identical infinitesimal amountof charge dq2 = λ dx2. The magnitude of the forceon dq2 due to dq1 is

21 2 1 2

2 21 2( )

dq dq dx dxdF k kr x x x

λ= =+ +

To find the total repulsive force of one rod on the other, a double integral is required.

( ) ( )

( ) ( ) ( ) ( )

2 21 2220 0 0

1 2 1 2 0

2 22 2 20 0

2 2

2

2

1( )

1 1 ln ln

ln ln ln 2 ln

ln ln2

LL L L

L L

dx dxF k k dxx x x x x x

k dx k x x L x xx x L x x

k L x x L x L x

Q L x L xkL x L x

λ λ

λ λ

λ

= = −+ + + +

⎡ ⎤= − = ⎡ + − + + ⎤⎢ ⎥ ⎣ ⎦+ + +⎣ ⎦= ⎡ + − − + + + ⎤⎣ ⎦

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

∫ ∫ ∫

∫

22-51. The x component of the net electric force is

( )

( ) ( ) 2

2 22

2 3 / 222 222 42

2 cos

2 2

x d

d dd

QqF kx

Qq x kqQxkx xx

θ⎡ ⎤⎢ ⎥=⎢ ⎥+⎣ ⎦

= =+ ⎡ ⎤++ ⎣ ⎦

The y component of the net electric force is

( ) ( )2 22 22 2

sin sin 0y d d

Qq QqF kx x

θ θ⎡ ⎤⎢ ⎥= − + =⎢ ⎥+ +⎣ ⎦

Therefore,

21 2 3 / 224

2d

kqQx

x= + =

⎡ ⎤+⎣ ⎦F F F i

x

+ + +x1

Ldq1

L

+ + +x2

dq2

0.5d

0.5d x

+Q

+Q

+q

F2

F1

F1+F2θθ

CHAPTER 22

476

The magnitude of F is

2 3 / 224

2d

xF kqQx

= −⎡ ⎤+⎣ ⎦

To find the maximum value of F, we take the derivative and equate it to zero.

2

2 2

2 5 / 2243 / 2 3 / 22 2

4 4

2

1 62 2 02

0.358

d

d d

dF d x xkqQ kqQ xdx dx x x

dx d

−⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎡ ⎤= − = − − + =⎣ ⎦⎢ ⎥ ⎢ ⎥⎡ ⎤ ⎡ ⎤+ +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎣ ⎦ ⎣ ⎦

= =

22-52. (a) The x component of the net electric force is

( )

( ) ( ) 2

2 22

2 3 / 222 222 42

2 cos

2 2

x d

d dd

QqF kx

Qq x kqQxkx xx

θ⎡ ⎤−⎢ ⎥=⎢ ⎥+⎣ ⎦

− −= =+ ⎡ ⎤++ ⎣ ⎦

The y component of the net electric force is

( ) ( )2 22 22 2

sin sin 0y d d

Qq QqF kx x

θ θ⎡ ⎤⎢ ⎥= − + =⎢ ⎥+ +⎣ ⎦

Therefore,

21 2 3 / 224

2d

kqQx

x

−= + =⎡ ⎤+⎣ ⎦

F F F i

(b) The magnitude of F is

2 3 / 224

2d

xF kqQx

= −⎡ ⎤+⎣ ⎦

If x is small, then as it is squared in the denominator, the x2 term becomes negligible and we have

( )2 3 / 2 324

22dd

x kqQF kqQ x−= − =⎡ ⎤⎣ ⎦

which is proportional to x.(c) A force that is proportional to x is similar to that of a spring force. Using the answer from part(b), we have

F = Ax where ( )3

2

2d

kqQA −=

The period of the simple harmonic motion that results is

( )

1/ 21/ 2

3 32

2 2 22 4/ d

kqQ kqQTmdA m m

π π π πω

−−⎛ ⎞ ⎛ ⎞⎜ ⎟= = = = ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

22-53. Each of the following processes is impossible because charge is not conserved in them.p + p → n + n + π+ (e + e ≠ 0 + 0 + e)p + p → n + p + π0 (e + e ≠ 0 + e + 0)p + p → n + p + π0 + π− (e + e ≠ 0 + e + 0 + (− e))

0.5d

0.5dx

+Q

+Q

−qF2

F1

F1+F2

θθ

CHAPTER 22

477

22-54. The net charge of the left hand side of the reaction is +2e, since the water molecule is electricallyneutral. The right hand side of the reaction is (−2e) + 8e + ne. If the right hand side is to balancethe left hand side, n = 4.

†22-55. The left hand side of the cobalt reaction has charges: 4e + (−Ne) or (4 − N)e. The right hand sideshows that the Co atom with a charge +3e. Setting the two sides equal to each other gives N = 1.

22-56. 8 8 8outer total inner ( 1.0 × 10 C) ( 2.0 × 10 C) 3.0 × 10 CQ Q Q − − −= − = − − + = −

†22-57. For every silver ion produced, there must also be an electron produced. To deposit a mole ofsilver would require a mole of electrons. For the deposition of 1.0 g of silver, the number of silverions produced is

( ) 23 211 mole ions1.0 g 6.02 × 10 5.6 × 10 ions107.9 g mole

n⎛ ⎞

= =⎜ ⎟⎝ ⎠

Therefore, 5.6 × 1021 electrons flow during the deposition.

22-58. ( ) ( ) ( )( )

2

2

11 11N m9 101 2

22 C

6.0 × 10 C 6.0 × 10 C8.99 × 10 8.1 × 10 N

0.20 mq qF kr

− −−

−= = =i

†22-59. If Fg = Fe, then1 2 1 2

2 2

m m q qG kr r

=

Because the masses of the two grains are equal to m and each has a charge q, we can multiplyboth sides by r2 and make the appropriate subtitutions to get

2

2

2

2

2 2

N m9219 9C

N m11kg

8.99 × 101.60 × 10 C 1.9 × 10 kg

6.67 × 10

Gm ke

ke km eG G

− −−

=

= = = =i

i

22-60. The torque due to the electric force between the probe and the end of the arm is

( ) ( ) ( )2

2

292N m9 6

e 2 2C

2.0 × 10 C8.99 × 10 0.15 m 6.0 × 10 N m

(0.030 m)qF l k lr

τ−

−= = = =i i

†22-61. From the geometry of an equilateral triangle of sides oflength a, the distance from each of the vertices to thecenter is

0.5772 cos 30

ad a= =°

Because all of the charges at the vertices are equal, eachexerts an equal force (in magnitude) on the charge at thecenter. The sum of these forces must equal zero, if thesystem of charges is to remain in equilibrium. The sum ofthe x components and y components is

2 1

1 2 3

cos 30 cos 30 0

sin 30 sin 30 0x

y

F F F

F F F F

= ° − ° =

= − ° − ° =∑∑

−q

+Q

+Q

+Q

F1

F2F3a

a a

x

y

CHAPTER 22

478

The magnitude of each of these three forces is

( )22 0.577Qq QqF k kr a

= =

The geometry of this situation is such that any negative charge q will satisfy the equilibriumrequirement.

22-62. Each ball has three forces acting on it: the gravitational force ofthe Earth, the tension in the thread, and the electric force due tothe other ball. Since each ball is in equilibrium, the sum ofthese forces is equal to zero newtons.

The angle that each thread makes with the vertical is1 5 cmsin 30

10 cmθ − ⎛ ⎞= = °⎜ ⎟

⎝ ⎠The sum of forces in the vertical direction may be used to findthe tension:

cos 30 0

cos 30

yF T mg

mgT

= ° − =

=°

∑

The sum of forces in the horizontal direction is

( ) ( )2

2

2

2

2 2

2 4 28

N m9C

sin 30 0

sin 30 sin 30cos 30

(0.10 m) 2.0 × 10 kg 9.81 m/s tan 303.5 × 10 C

8.99 × 10

xQF T kr

r T r mgQk k

−−

= ° − =

° °= =°

°= =

∑

i

22-63. ( ) ( )( )

2

2

2112N m9 8

22 C

2.0 × 10 C8.99 × 10 3.6 × 10 N

0.010 mQF kr

−−= = =i

To determine the acceleration, the mass must be determined. For each drop the mass is( )

( ) ( )3

343

82 2

3 1000 kg 34 43 31 m

3.6 × 10 N 6.9 × 10 m/s(0.0005 m)

m V R

F Fam R

ρ ρ π

ρ π π

−−

= =

= = = =

22-64. The total charge each hour is

( )4 3600 s1.0 × 10 C/s 0.36 C/h1 h

Q −= =

The total number of electrons flowing into the rod in the hour is18

19

0.36 C 2.3 × 10 electrons1.60 × 10 C

Qne −= = =

θ

mg

F21 T

mg

F12T

20 cm

10 cm

10 cm

10 cm θ

CHAPTER 22

479

22-65.

( ) ( )2

2

2

2

226

N m9C

0.30 N 0.18 m1.0 × 10 C

8.99 × 10

QF kr

FrQk

−

=

= = =i

The number of electrons in excess on one ball an in deficit on the other is6

1219

1.0 × 10 C 6.5 × 10 electrons1.60 × 10 C

Qne

−

−= = =

22-66. The distance between the centers of the balls is(0.20 m)sin(22.5 ) 0.077 mr = ° =

The sum of vertical forces acting on the elevated ball is12

2

12 2

sin 22.5 0

sin 22.5

yF F mg

mg QF kr

= ° − =

= =°

∑

( ) ( ) ( )( )2

2

24 228

N m9C

1.5 × 10 kg 9.81 m/s 0.077 m5.0 × 10 C

sin 22.5 8.99 × 10 sin 22.5mgrQ

k

−−= = =

° °i

†22-67. ?n p e→ + + On the left side the neutron is neutral, therefore, the total charge on the right ofthe arrow must add to zero. Because the proton and electron have charges +e and −e, respectively,the charge on the missing particle must be equal to zero.

0 ?p π++∆ → + + The particle on the left has a charge of +2e. On the right, the proton is +e theπ particle is neutral. The missing particle must have a charge of +e to give a total charge on theright equal to that on the left.

?n+∆ → + The particle on the left has a charge of +e. On the right, the neutron has zero charge,so the missing particle must have a charge of +e to give a total charge on the right equal to that onthe left.

?π µ− −→ + The π and µ particles both have a charge −e, so the missing particle must havezero charge.

45°

F21 mg

F12T

0.10 m

0.10 m

r