Tài liệu Physics exercises_solution: Chapter 21 doc

21.18: The charge q must be to the right of the origin; otherwise both 3 q2 and q3would exert forces in the + x direction.. 21.30: a The electric field of the Earth points toward the gro

lead

e 22

21.2: current20,000C sandt 100s10 4 s

Q = It = 2.00 C

.1025.1C1060.1

19 19

n

and

kg101.67

1057.4(44

4

0

2 0 2



n

21.7: a) Using Coulomb’s Law for equal charges, we find:

m)150.0(4

1N220

2 2 0

b) When one charge is four times the other, we have:

C1071.3C10375.1m)

150.0(

44

1N220

2 2 0

So one charge is 3.7110 7 C, and the other is 1.48410 6 C

21.8: a) The total number of electrons on each sphere equals the number of protons.

.1025.7molkg026982

0

kg0250.0

1N

0 2

2 0

r

q πε F

n of the total number

21.9: The force of gravity must equal the electric force.

.m08.5m

8.25)sm8.9(kg)1011.9(

C)1060.1(4

14

31

2 19 0

2 2 2 0

q πε

mg

21.10: a) Rubbing the glass rod removes electrons from it, since it becomes positive.

kg

1027.4)electronkg

1011.9(electrons)10

4.69

(

electrons10

69.4)Celectrons10

25.6(C)107.50(nC

50

7

20 31

10

10 18

The rods mass decreases by 4.2710 20 kg

b) The number of electrons transferred is the same, but they are added to the mass of

the

plastic rod, which increases by 4.2710 20 kg

21.11: F2 isinthex-direction,soF1 mustbein the x-directionandq1 ispositive.

,

2

2 1

2 23

3 2 2

13

3 1 2

r

q q k r

q q k F

F

6 0

221

C)10550.0(4

1N200.04

πε r

q q πε

C)1050.3(

2 0

2 6 2

kq F

21.14: We only need the y-components, and each charge contributes equally.

0.6)

sinsince(N173.0sinm)

500.0(

C)104(C)100.2(4

1

2

6 6

Therefore, the total force is 2F 0.35N, downward

21.15: F2 andF3arebothinthe

x-direction.

N10124.1,

N10749

2 13

3 1 3

5 2

12

2 1 2

q q k

C)100.2(C)10.20()CmN109(

2

6 6

2 2 9

N23.0

The magnitude of the total force is 0.23N 2  0.27N2 0.35N The direction of

the force, as measured from the +y axis is

4027.0

23.0tan 1 

37.3N00.7

N00.7and

N37.3so

,N37.3

2 3

3 2

2 2

12

2 1 2

x x

x x

x

F F F

F F

F

F

F r

q q k

F

For F3x to be negative, q must be on the –x-axis.3

m144.0so

m,144.0so

,

3

3 1 2

3 1

F

q q k x x

q q k F

21.18: The charge q must be to the right of the origin; otherwise both 3 q2 and q3would

exert forces in the + x direction Calculating the magnitude of the two forces:

direction

in theN

375

3

m)200.0(

C)1000.5C)(

1000.3)(

CmN109(4

1

2

6 6

2 2 9 2

12

2 1

0

21

x r

inmN

216

0

C)1000.8(C)1000.3()CmN10

6 6

2 3 9

31

x r

r F

m0208.0N7.00N

3.375

mN216.0

13

2

2 2

1059.2m)

400.0(

C1020.3m)200.0(

C1050.1C)1000

9 9

300.0(

C1000.5m)200.0(

C1000.4C)1000

9 9

21.21: a)

0 2

2

24

1sin)(

4

12,

0

x a

qQa πε

θ x

a

qQ πε

0

0

y a

qQ πε F

d)

21.22: a)

)(

24

1cos

)(

4

1

0 2

2 0

qQx πε

θ x

a

qQ πε

12214

122

4

1

2 2 0 2

2 0 2

2 0

q πε L

q πε

F

positive x-axis

m)250.0(

C)1000.3(4

14

1

2 9 0

2 0





πε r

q πε E

)CN0.12(

C)1000.3(4

14

1CN00

0

2 0

q πε E

21.25: Let +x-direction be to the right Find a x:

 

2 (1.602 10 C) 23.5N CN)

10516.7(

left

thetoissopositive,is

charge,

direction-

lefttheto

is

N10516.7

sm101.132gives

?,s1065.2,sm101.50,

sm1050.1

19 18

18

2 9 0

6 3

3 0

a v

v

a t

v v

x x

x x

x x

x x

5

C106.1

)sm101.00(kg)1011.9(

sm1000.1s)103.00(

m)50.4(22

19

2 12 31

2 12 2

6 - 2

The force is up, so the electric field must be downward since the electron is negative.

(b) The electron’s acceleration is ~ 10 g, so gravity must be negligibly small 11compared to the electrical force

CN650

)sm8.9(kg)00145.0

q

C101.60

)sm9.8(kg)1067.1

19

2 27

C)1060.126(4

14

2 10 19 0

2 0

q πε E

m)1029.5(

C)1060.1(4

14

2 11 19 0

2 0

q πε E

21.29: a) q55.010 6 C,andFis downward with magnitude

C,N1013.1Therefore,

21.30: a) The electric field of the Earth points toward the ground, so a NEGATIVE

charge will hover above the surface

C

92.3C

N150

)sm8.9(kg)0.60

C)92.3(4

14

2 2 0

2 2 0

q πε

too great for practical use

21.31: a) Passing between the charged plates the electron feels a force upward, and just

misses the top plate The distance it travels in the y-direction is 0.005 m Time of flight

s1025

s m

1025.1(m005.0

2 1 2

10 60 1

) s m 10 40 6 kg)(

10 11 9 (

19

2 13 31

b) Since the proton is more massive, it will accelerate less, and NOT hit the plates

To find the vertical displacement when it exits the plates, we use the kinematic equations again:

m

1073.2s)1025.1(2

12

y

c) As mention in b), the proton will not hit one of the plates because although the electric force felt by the proton is the same as the electron felt, a smaller acceleration results for the more massive proton

d) The acceleration produced by the electric force is much greater than g; it is

reasonable to ignore gravity

4 2

9 2

2 9

2 2

9 2

2 9

j i

E

ˆ)CN10(8.64ˆ

)CN106.485(

)9.126sinˆ9.126cosˆ()CN10080.1(

3 3

4 2

j i

E

E

ˆ)CN1095.1(ˆ)C/N10485.6(

ˆ)CN1064.8C/N10813.2(ˆ)CN10485.6(

4 3

3 4

3 2

21.33: Let x  be to the right and y be downward

Use the horizontal motion to find the time when the electron emerges from the field:

sm1079.1

sm1000.8gives2

?,s1025.1,0m,

0.0050

sm1060

1

s1025.1gives

?,sm1060.1,

0,m0200.0

6 2

2

5 0

0

8 0y

0

6

8 2

2

1 0 0

6 0

y y

y

y x

x x

x x

v v

v

v t

v v y

y

v t

v y

y

v

t t

a t v x

x

t v

a x

21.35: a)F m g    31 2    30 Fe eE 

e

g (9.11 10 kg)(9.8m s ) 8.93 10 N

.N1060.1)CN1000.1()C

)1050.1()C1060.1(

)kg1067.1()m0160.0(22

12

1

2 6 19

27 2

t m

eE at

v

p

20

35.1tan 1   

2ˆ,42

6

21.39: a) Let x be east.

E

is west and q is negative, so F

is east and the electron speeds up

sm1033.6gives)(2

?m,375.0,

sm10638.2,

sm1050

4

sm10638.2)kg10109.9()N10403.2(

N10403.2)mV50.1()C10602.1(

|

|

5 0

2

0

2

0 2

11 5

0

2 11 31

19

19 19

x

x

x x

x a v

v

v x

x a



q is west and the proton slows down

sm1059.1gives)(2

?,m375.0,

sm10436.1,

sm1090

1

sm10436.1)kg10673.1()N10403.2(

N10403.2)mV50.1()C10602.1(

|

|

4 0

0

2

0 2

8 4

0

2 8 27

19

19 19

x

x

x x

x a v

v

v x

x a

2 1 2 ) 20 1 ( 2

1

2 2

E E

r

kq r

kq

72.02.17.5

or )

2.1()2.1(2)(

2.1)2

2 1

2 1 1 2 1

2 1 1

21.41: Two positive charges, q , are on the x-axis a distance a from the origin.

a) Halfway between them, E0

a

q x

a

q πε

a x x

a

q x

a

q πε

a x x

a

q x

a

q πε

E x

,)()(41

,)()(41

|

| ,)()(41

,

2 2

0

2 2

0

2 2

0

For graph, see below

21.42: The point where the two fields cancel each other will have to be closer to the

negative charge, because it is smaller Also, it cant’t be between the two, since the two fields would then act in the same direction We could use Coulomb’s law to calculate the actual values, but a simpler way is to note that the 8.00 nC charge is twice as large as the –4.00 nC Charge The zero point will therefore have to be a factor of 2 farther from the 8.00 nC charge for the two fields to have equal magnitude Calling x the distance from

the –4.00 nC charge:

m90.2

220

.1







x x x

21.43: a) Point charge q (2.00 nC) is at the origin and 1 q2 (5.00nC) is at

|

|m)

200.0(

|

|,

m200

|

|m)

400.0(

|

|,

m20

|

|m)

200.0(

|

|,

m200

6.1iii) right,N103.4CN269C

10

6

21.44: A positive and negative charge, of equal magnitude q , are on the x-axis, a

distance a from the origin.

a) Halfway between them, 2 ,

4

12

0 a

q πε

a

q x

a

q πε

a x x

a

q x

a

q πε

a x x

a

q x

a

q πε

E x

,)()(41

,)()(41

|

| ,)()(41

,

2 2

0

2 2

0

2 2

0

with “+” to the right

This is graphed below

21.45: a) At the origin, E 0.

b) At x0.3m, y0:

.CNˆ2667

ˆ)m45.0(

1)

m15.0(

1)

C1000.6(4

1

2 2

9 0

i i

4.0)m5.0(

1ˆ

5.0

3.0)m5.0(

1ˆ

)m4.0(

1)

C1000.6(4

1

2 2

2 9

0

πε



CN5.526C

N)3.510ˆ6

25.0

2.0)C1000.6(24

1:

m2.0,

21.46: Calculate in vector form the electric field for each charge, and add them.

CNˆ150

ˆ)m6.0(

)C1000.6(

4

1

2 9 0

i i

πε



CNˆ8.28ˆ6.21ˆ8.0()m00.1(

1ˆ

6.0()m00.1(

1)C1000.4(

4

1

2 2

9 0

j i

j i

8.28()4.128



4.128

8.28tan 1 

)C100.6(24

1

2 9 0

i i

ˆ)m45.0(

1)

m15.0(

1)

C100.6(4

1

2 2

9 0

i i

4.0)m5.0(

1ˆ

5.0

3.0)m5.0(

1ˆ

)m4.0(

1)

C100.6(4

1

2 2

2 9

0

πε



CN209C

N)5.164ˆ6.129

C)1000.6(24

1,

0:m2.0,

9 0

y

21.48: For a long straight wire, 1.08m.

)CN5.2(2

mC105.12

λ

0

10 0

21.49: a) For a wire of length a centered at the origin and lying along the y-axis, the

electric field is given by Eq (21.10)

i

1

λ2

1

2 2



a x x πε





Graphs of electric field versus position for both are shown below

21.50: For a ring of charge, the electric field is given by Eq (21.8).

)(

4

1

2 3 2 2

4.0andm0.025C,

Q

b) ( 2.50 10 6 C)(7.0ˆN C) 1.75 10-5ˆN

q

on ring

q

21.51: For a uniformly charged disk, the electric field is given by Eq (21.11):

i

1

11

σ



The x -component of the electric field is shown below.

21.52: The earth’s electric field is 150 N C, directly downward So,

,mC1066.23002

0 0

2 9

2 19

2

6

m

C104m

1

cm100e

C106.1cm105

E directed toward the surface

21.54: By superposition we can add the electric fields from two parallel sheets of charge.

E   directed downward

21.56: The field appears like that of a point charge a long way from the disk and an

infinite plane close to the disk’s center The field is symmetrical on the right and left (not shown)

21.57: An infinite line of charge has a radial field in the plane through the wire, and

constant in the plane of the wire, mirror-imaged about the wire:

Cross section through the wire: Plane of the wire:

Length of vector does not depend on angle Length of vector gets shorter at

points further away from wire

21.58: a) Since field lines pass from positive charges and toward negative charges, we

can deduce that the top charge is positive, middle is negative, and bottom is positive b) The electric field is the smallest on the horizontal line through the middle charge,

at two positions on either side where the field lines are least dense Here the

y-components of the field are cancelled between the positive charges and the negative

charge cancels the x-component of the field from the two positive charges.

21.59: a) pqd(4.510 9C)(0.0031m)1.410 11Cm, in the direction from and towards q 2

b) If E

is at 36.9 and the torque , τ pEsin, then:

.CN5.85636.9

sin m)C104.1(

mN102.7

1082

mC106.17m)

1000.3(2

)

0

30 9

dipole 3

p x

molecule (negative x-direction).

)4(

2)

4(

)2(

)2(

)2(

1)

2(

1

d y

yd d

y

d y d

y d

y d

2 2 0 2

2 2

2

p d

y

y πε

qd d

y

yd πε

q

E y

b) This also gives the correct expression for E ysince yappears in the full expression’s denominator squared, so the signs carry through correctly

21.64: a) The magnitude of the field the due to each charge is

,))2(

14

4

1

2 2 0

πε

q r

q πε E

where d is the distance between the two charges The x-components of the forces due to

the two charges are equal and oppositely directed and so cancel each other The two

fields have equal y-components, so:

θ x

d πε

q E

)2(

14

2(

2sin

2

2 x d

d θ



thus

2 3 2 2 0

2 2 2

2 0

2)

2(

14

2

x d

πε

qd x

d

d x

d πε

q E

b) At large x,x2(d 2)2, so the relationship reduces to the approximations

3 0 dipole 4πε x

21.66: a)

The torque is zero when p

is aligned either in the same direction as E

or in the

opposite directions

b) The stable orientation is when p

is aligned in the same direction as Ec)

N6.842sin

N10124.1m)

(0.0200

C)10C)(10.010

00.5(

2 1 2

1 1

3 2

6 6

2 1

y

F F F F

θ F F

k r

q k

1

1 F θ

F x

m,N22.3m)0150.0)(

(

 F x

21.68: a) F i sin ˆj

4

1ˆcos4

1

2 13

3 1 0

2 13

3 1 0

r

q q πε

θ r

q q

i

3m)1016.0)((9.00

nC)nC)(6.0000

.5(4

1ˆ5

4m)1016.0)((9.00

nC)nC)(6.0000

.5(4

1

4 0

4 0

 F

.ˆN)10(6.48ˆ

N)1064.8

j

m)(0.0300

nC)nC)(6.0000

.2(4

1ˆ4

2 0

2 23

3 2 0

πε r

q q πε

0

2 0

2

1)

1(

14

1)(4

1)(4

1

a x a

x a

qQ πε x

a

qQ πε

x a

qQ πε

F q

.4

4

1.)).21( 21(4

1

3 0

2 0

2 0

x a πε

qQ a

x a

qQ πε a

x a

x a

qQ πε

ε mπ

qQ π

f a ε mπ

qQ πf

b) If the charge was placed on the y-axis there would be no restoring force if qand

Q had the same sign It would move straight out from the origin along the y-axis, since the x-components of force would cancel.

21.70: Examining the forces: F xTsinθFe 0andF y Tcosθmg0

θ

2 2

3

2 2

mg πε L q mg

L kq L

21.71: a)

b) Using the same force analysis as in problem 21.70, we find:

θ mg

θ

θ θ

L mg

kq θ

L

d

sin)smkg)(9.8015

.0(m)6.0(4

C)10)(2.79CNm1099.8(sin

)2(

tan2

Therefore θ sin 2θ

331 0tan  Numerical solution of this transcendental equation leads to

r

q kq T

F mg

m.)342.0m)sin20







q q

d) The charges on the spheres are made equal by connecting them with a wire, but

2 0

41

e tan 0.0453N

r

Q πε

θ mg

Q  But the separation r is 2

known: r2 2(0.500m)sin300.500m.Hence: Q q1 2q2  4πε0F e r 2

.C

21.73: a) 0.100molNaClmNa (0.100mol)(22.99g mol)2.30g

mCl(0.100mol)(35.45g mol)3.55g

A 6.02 10mol)

100.0( N   so the charge is:

C

9630C)

1060.1)(

)9630(4

14

2 2 0

2 2 0

q πε F

9 3

2 23

3 2 2

13

3 1 6

3

m)0.2(

C105.4m)

0.3(

C)105.2(N

100

r

q kq r

q kq F

nC

2.3)CN

(1262

N10

b) The force acts on the middle charge to the right

c) The force equals zero if the two forces from the other charges cancel Because of the magnitude and size of the charges, this can only occur to the left of the negative

2

1 23

13

)200.0()300.0

kq x

kq F

from the origin Solving for x we find: x1.76m The other value of x was between

the two charges and is not allowed

21.75: a) 6 ,

4

1)2(

)3(4

1

2 2 0

2

q πε L

q q πε

F   toward the lower the left charge The other two forces are equal and opposite

b) The upper left charge and lower right charge have equal magnitude forces at right angles to each other, resulting in a total force of twice the force of one, directed toward the lower left charge So, all the forces sum to:

N

2

3234

)2(

)3(2)3(4

1

2 0

2 2

2 0

q L

q q L

q q πε F

)()(4

1)

y

q a

y

q πε

p E

4

1)

2 0

64

123

213

214

1)

0 2

2 2

2 2

qa πε y

a y

a y

a y

a y

q πε p

Note that a point charge drops off like 12

y and a dipole like .

13

y

21.77: a) The field is all in the x -direction (the rest cancels) From the q charges:

.)(

4

14

14

1

2 3 2 2 0 2

2 2 2 0 2

2

qx πε

x a

x x

a

q πε

E x a

q πε

1((

24

12

)(

24

12

4

2 0 2

2 3 2 2 0 total

2 0

q x

a

qx πε

E x

q πε

E x

4

112

31

24

1

4 2 0 2

2 2

0

x

qa πε x

a x

q πε

Note that a point charge drops off like 12

x and a dipole like 3

1

21.78: a) 20.0 g carbon 1.67

molg0.12

0

the earth’s poles (negative at north, positive at south), leading to a force:

N

1013.5)m10276.1(

C)10963.0(4

1)2(4

2 7

2 6 0

2 earth

2 0

q πε

C)10963.0(2

44

12

45sin)2(4

12

8 2

7

2 6 0

2 earth

2 0

R

q πε

21.79: a) With the mass of the book about 1.0 kg, most of which is protons and

neutrons, we find: #protons = (1.0kg) (1.67 10 27 kg) 3.0 1026

2

difference present if the electron’s charge was 99.999% of the proton’s is

C

480C)106.1)(

00001.0)(

0

2 2

2

2 2

0 2

2 2

0 2

2 2 0 2

2 0 net

)()(

44

)()(

)()(4

)(

1)

(

14

)(4

1)(4

1charge)central

(on

x b x b

bx πε

q x

b x b

x b x b πε q

x b x

b πε q

x b

q πε x

b

q πε F

Direction

3 0

2 2 2 0

2

b πε

q b b

bx πε

q

2 3

2 3

0 2

3 0

2 2

2

2

12

b ε mπ

q π f f π b ε mπ

q ω

x b ε mπ

q dt

NmC1085.8()kg1066.1

(

12

)C106.1(2

3 10 2

2 12 27

2 19

f

21.81: a)       3 3 

3 4 3 3

3 3

4 ) (8.9 10 kg m )( )(1.00 10 m)

ρ ρV m

kg10728

3   5

mol10867.5)molkg10546.63)(

kg10728.3

5

C10602.1)(

10100.0()

99900.0

e e

q

N103.2)m00.1(

)C6.1

2

2 2

F

21.82: First, the mass of the drop:

kg

1041.13

)m100.15(4)mkg1000

s001.0(

)m1000.3(222

2

4 2

t

d a at d

So:

C.1006.1C

N1000.8

)sm600)(

kg1041.1

4

2 11