Tài liệu Physics exercises_solution: Chapter 23 docx

The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south the force and displacement are perpendicular; V D V A.. 23.29: a Poin

23.1: 0.357J

m150.0

1m

354.0

1)C30.4)(

C40.2(111 2 2

J

357.0

U U W

0015.0

J)491.0J608.0(22

1

J608.0m

800.0

)C1050.7)(

C1080.2()sm0.22)(

kg0015.0(21

2 1 2

6 6

f f

i

i i

i

v r

q kq mv E

E

k U

K

E

b) At the closest point, the velocity is zero:

m.323.0J

608.0

)C1080.7)(

C1080.2(J

608

r

q kq

J400.0

C)1020.7)(

C1030.2(J

400.0

6 6

m250.0

)C1020.1()C1060.4

s

m6.37

,J198.0)

iii

(

s

m7.36

J,189.0)

ii

(

s

m6.26kg1080.2

J)0994.0(22

1J0994.0

J0994.0m5.0

1m25.0

1)C1020.1(C)1060.4(J0

(i)

b)

4 2

6 6

f f

f f f

f i i f

v K

v K

v mv K

k

U U K K

m500.0

2m500.0

2 6 2

2 2

23.7: a)

J

10 60 3

) m 100 0 ( ) nC 00 2 )( nC 00 3 (

) m 100 0 ( ) nC 00 2 )( nC 00 4 ( ) m 200 0 ( nC) 00 3 )( nC 00 4 ( 7 23 2 1 13 2 1 12 2 1                               k r q q r q q r q q k U 0 , 0 b) 12 3 2 3 1 12 2 1            x r q q x q q r q q k U If So solving for x we find: m 360 0 , m 074 0 0 6 1 26 60 2 0 6 8 60 0  2           x x x x x Therefore m 074 0  x since it is the only value between the two charges 23.8: From Example 23.1, the initial energy E can be calculated: i J 10 09 5

m 10 ) C 10 20 3 )( C 10 60 1 (

) s m 10 00 3 )(

kg 10 11 9 ( 2 1

19 10

19 19

2 6 31





































i

i i i

E k

U K

E

When velocity equals zero, all energy is electric potential energy, so:

m 10 06 9

2 J

10 09

r

e k

23.9: Since the work done is zero, the sum of the work to bring in the two equal charges

q must equal the work done in bringing in charge Q.

2

2

Q d

kqQ d

kq W

W qq  qQ     

23.10: The work is the potential energy of the combination.

J1031.7

32

2m

105

)C106.1()CNm100.9

(

212

2m105

m105

)2((m105

)(m

1025

)2(

19

10

2 19 2

2 9 10 2

10 10

ke

U U U

1

m1000.8with,

54

12214

18 1

10 2

0 0

e πε r

r r πε

2

12

m1038.1)

sm10(kg1067.1(

)C106.1(CNm109

6 27

2 19 2

2 9 2

0

)m1038.1(

)C106.1()CNm109

(

2 13

2 19 2

2 9 2 2

J0.00550V)

800V(200C)105.00(J00250.0)(

v

V V q K

K

qV K qV K qV

U

B B

B A A

B

B B A A

It is faster at B; a negative charge gains speed when it moves to higher potential.

23.14: Taking the origin at the center of the square, the symmetry means that the

potential is the same at the two corners not occupied by the 5.00μC charges (The work done in moving to either corner from infinity is the same) But this also means that

no net work is done is moving from one corner to the other

23.15: E

points from high potential to low potential, so V B V A andV C V A

The force on a positive test charge is east, so no work is done on it by the electric force when it moves due south (the force and displacement are perpendicular); V D V A

J1050.1J

1050

qEd

23.17: a) Work done is zero since the motion is along an equipotential, perpendicular to

the electric field

m

V104.00nC)0.28

23.18: Initial energy equals final energy:

s

m1089.6

J)102.88J

1004.5(kg1011.922

1J1004.5

2

1m

0.40

C1000.2(m

0.10

C)10(3.00C)

1060.1(

J1088.2m

0.25

C)1000.2(m

0.25

C)10(3.00C)

1060.1(

21

6

17 17

31

2 17

2 9

9 19

17 9

9 19

2 2

2 1

1 2

2 1

f e f

i

f e f

f i

i f

i

v

v m

v m k

E

k

E

v m r

keq r

keq r

keq r

keq E

E

V90.0

C)1050.2

kq V

V30.0

C)1050.2

kq

V

23.20: a)    (0.250m)(48.0V) 1.3310 9 C

k k

rV q r

kq V

m)(0.750

C)1033.1

C1050.6m

05.0

C1040.2:

AAt

9 9

2

2 1

q k

V A

m0.06

C1050.6m

08.0

C1040.2:

B

At

9 9

2

2 1

q k

V B

c) W qV (2.5010 9 C)(33V)8.2510 8 J

The negative sign indicates that the work is done on the charge So the work done by the

field is 8.2510 8 J

23.22: a)

4

12

0 a

q πε

V 

c) Looking at the diagram in (a):

2 2 0

124

12)(

x a

q πε r

q πε x

0 x

q πε V a

23.23: a)

r

q k r

kq

V x

c) The potential along the x-axis is always zero, so a graph would be flat.

d) If the two charges are interchanged, then the results of (b) and (c) still hold The potential is zero

23.24: a) 2

)()(:

|

a y

kqy y

a

kq y

a

kq V

a y

)(:

2)

(:

2 2

2 2

a y

kqa a

y

kq y

a

kq V

a y

a y

kqa a

y

kq y

a

kq V

y

kqa a

y

kq y

a

kq V

23.25: a)

)(

)(2

:

a x x

a x kq a

x

kq x

kq V a x

)3(2

:0

a x x

a x kq x a

kq x

kq V a x

)(2

:0

a x x

a x kq a x

kq x

kq V

| x a

q x q k

kqx V

–q (Note: The two charges must be added with the correct sign.)

23.26:a) 2

|

|

12

kq r

kq y

kq V

3

34

when,

y a y y

a y

y kq V a

J)1072.4(22

V98.4

V E

b)    (4.98V) (0.415m)2.3010 10 C

k k

Vd q d

kq

V

c) The electric field is directed away from q since it is a positive charge.

23.29: a) Point b has a higher potential since it is “upstream” from where the positive

charge moves

0)(

|

|)

(

|

|)

23.30:(a) V V Q V2Q 0, so V is zero nowhere except for infinitely far from the

charges

The fields can cancel only between the charges

2 2 2

2

)(

)2(

x x

d x

d

Q k x

kQ E

0)2()(:

30

)2()(:

E

E

d y y

d

Q k y

Q k B

at

d x x

d

Q k x

Q k A

(

)2(

2

d x x d

Q k x kQ

1 2







 m v K

V

The electron loses kinetic energy when it moves to lower potential

23.32: a) 65.6V.

m0.48

C)1050.3

m0.240

C)1050.3

m)(0.150m)

(0.300

1m

0.150

1C)10(24.0Δ

11

2 2

9

2 2 2

a x a kQ V a

V) (796 C) 10 60 1 ( 2 2

31 19

q W

23.34: Energy is conserved:

.V0117.0C)

1060.1(2

)sm1500()kg1067.1(2

1

19

2 27

)V0117.0(2exp)m180.0(

λ

2expλ

2exp)

ln(

2λ

12 0

0 0

0 0

0 0

V πε r

r V πε r

r r r πε V

m0450.0

b) F  Eq(8000N C) (2.4010 9 C)1.9210 5 N

c) W  Fd (1.9210 5 N) (0.0450m)8.6410 7 J

d) U Vq(360V) (2.4010 9 C)8.6410 7 J

23.36: a) V Ed (480N C)(3.810 2 m)18.2V.

b) The higher potential is at the positive sheet

0 0

mV1000.3

V E

0 0

23.38: a) 47.0 10 C m 5311N C

0

2 9 0

σ E

b) V  Ed (5311N/C) (0.0220m)117V

c) The electric field stays the same if the separation of the plates doubles, while the potential between the plates doubles

23.39: a) The electric field outside the shell is the same as for a point charge at the center

of the shell, so the potential outside the shell is the same as for a point charge:

q V

The electric field is zero inside the shell, so no work is done on a test charge as

it moves inside the shell and all points inside the shell are at the same potential as the

RV q R

kq

V

c) No, the amount of charge on the sphere is very small

23.40: For points outside this spherical charge distribution the field is the same as if all

the charge were concentrated at the center

Therefore

2 0

4πε r

q

E and

C1069.1/

N.m109

)m200.0()CN3800(

2 2 9

2 2

πε q

Since the field is directed inward, the charge must be negative The potential of a point charge, taking  as zero, is

V760m

200.0

)C1069.1()C/N.m109(4

8 2

2 9 0

q V

at the surface of the sphere Since the charge all resides on the surface of a conductor, the field inside the sphere due to this symmetrical distribution is zero No work is therefore done in moving a test charge from just inside the surface to the center, and the potential at the center must also be 760V

23.41: a) EV

Cy Bx Axy z z

V E

C.

Ax Cy

Bx Axy y y

V E

Bx.

Ay Cy

Bx Axy x x

V E

z y x

0)(

)(

2)

(

2 2 2

B y A

C x C

Ax x A

B y Bx

BC A

C E

A

BC

,

2,at

0,

2

2 2

23.42: a) EV

.)

2 2

kQx z

y x

kQx z

y x

kQ x

z r

y r

x r

23.43: a) There is no dependence of the potential on x or y, and so it has no

components in those directions However, there is z dependence:

.0

,

C E C z

V E

since the potential is constant there

(b) Infinite parallel plates of opposite charge could create this electric field, where the surface charge is σ Cε0

kq r

kq V r r

(ii)   :    1 1

b b

b a

r r

kq r

kq r

kq V r r r

(iii) rr b :V 0, since outside a sphere the potential is the same as for point charge Therefore we have the identical potential to two oppositely charged point charges

at the same location These potentials cancel

b b

a a

r r

q πε V

V r

q r

q πε

4

10

and 4

1

0 0

114

11

14

0

r r

V r

q πε r

r r πε

q r

V E

r r

r

b a

ab b

b a

d) From Equation (24.23): E 0, since V is zero outside the spheres

e) If the outer charge is different, then outside the outer sphere the potential is no

4

14

14

1

0 0

Q q πε r

Q πε r

q πε

All potentials inside the outer

4

1

0 r b

Q πε

shells are not affected Thus (b) and (c) do not change However, now that the potential does vary outside the spheres, there is an electric field there:

.1

kq r

kQ r

kq r r

V E

1m

012.01

a x a a

kQ x x

V

E x

2 2

2 2

ln2

.1

λ2

11

4

)λ2(

)

()

(2

)ln(

)ln(

2

2 2 0

2 2 0

2 2 2

2

2 / 1 2 2 2

2

2 2 2

2 2 2

2

a x x πε a

x xa πε

a E

x a x

kQ a

x a

x a x a x a

x a x a kQ

a x a x a x a x a

kQ E

x x

b) The potential was evaluated at y and z equal to zero, and thus shows no

dependence on them However, the electric field depends upon the derivative of the

potential and the potential could still have a functional dependence on the variables y and

z, and hence E y and E may be non-zero z

a) Equipotentials and electric field lines of two large parallel plates are shown above b) The electric field lines and the equipotential lines are mutually perpendicular

2 6 2

2 9

2 13

2 12 2

3 2

1

1 2

13

3 1 2

12

2

1

)m16.0(

1m)

08.0(

1)

C100.2()CNm109()

a

r r kq

ma

q q q

q

a m r

q kq

12

3 3

2 1 1 23

3 2 13

3 1 12

2

r

q kq r

q kq r

1m

16.0

1m

08.0

1kg

020.0

)C102()CNm109

(

111

and,,

2 6 2

2 9

23 13 12 1

2 1

3 2 1 3

1 3

kq v

q q q q m

m v

v

23.49: a) W E KW F 4.3510 5 J6.5010 5 J2.1510 5 J

C1060.7

J1015.2

q

–2829 V with respect to the final point

m

V1054.3m08.0

23.50: a) 2 22 2.

mr

ke v r

ke r

eV.13.6J

1017.2m

1029.5

)C1060.1(2

12

12

1c)

.2

12

12

1

b)

18 11

2 19 2

2 2

ke U

U K

E

U r

ke mv

c)

cathode

toward

,mVm

V1005.1)

mV1085.7(3

43

4b)

15 19

3 / 1 5

3 / 1 4/3 5 3

/ 1 3 / 4 4 3

/ 1

Cx x

V

E

23.52: From Problem 22.51, the electric field of a sphere with radius R and q distributed

4andfor

0

3 0

R r r πε

q E R r R

πε

q dr r πε

q dr

R πε

qr

0

2 0

3 0

38

44

Set q2e to get V for the sphere The work done by the attractive force of the sphere r

when one electron is removed from r  d tois

πε

e eV

The total work done by the attractive force of the sphere when both electrons are

removed is twice this, 2Wsphere The work done by the repulsive force of the two electrons

is

)2(

2

d πε

R R

πε e

We can check this result in the special case of d = R, when the electrons initially sit on

the surface of the sphere The potential due to the sphere is the same as for a point charge

e

2

 at the center of the sphere

R πε

e R

πε

e R πε

e R

πε

e U

U

U U W

a b

b a b q

0

2 0

2 0

2 0

2

8

74

124

)2(44

22

23.53: a)

d πε q d

kq d

d d

kq U

kq d d

kq d d

kq d d

d kq

d d

d

kq d d

d

kq d d

d kq

U

0 2

2 2

2 2

2 2

2 2

2

46.133

12

11

123

42

1212

2

112

123

12

21

3

12

223

12

323

12

33

b) The fact that the electric potential energy is less than zero means that it is

energetically favourable for the crystal ions to be together

3

12

112

i

i i d

kq d

d d kq U

)2(ln)C1060.1(2then

,m1082

10

2 19

)C1060.1(2

10

2 19 2

J1059.7

2J101.02J1061.8

11 2

22 2

2

2 18

ke K

23.56: F e mg tanθ (1.5010 3 kg)(9.80m s2)tan(30)0.0085N (Balance

forces in x and y directions.) But also:

V

8.47C

1090.8

m)0500.0(N)0085.0(

Vq Eq

F e

2

λ))(ln)(ln(

2

λ

0 0

a b πε b

b a

b πε

2

λ))ln(

)(ln(

2

λ

0 0

r b πε b

b r

b πε

(

0

a b πε b

V a V

V ab   

c) Between the cylinders:

1)ln(

))(ln(

)ln(

)ln(

)(ln)ln(

2

λ

0

r a b

V r

b r a b

V r

V E

r b a b

V r

b πε V

ab ab

V r a b

23.59: a) F  Eq(1.10103V m)(1.6010 19C)1.7610 16 N,downward.

b)  (1.7610 16 N) (9.1110 31 kg)1.931014 m s2,downward

e m F

1s

1023.9sm1050

6

m060

d) Angle θ arctan(v y v x)arctan(at v x)arctan(1.78 6.50)15.3

e) The distance below center of the screen is:

m

0.0411s

m106.50

m0.120)

sm101.78(m10

πε

dr πε dl

E V

2

λ2

λ:

λgetto

0 0

a b r

V r

πε

a b V πε r πε E

a b

V πε

ln 2

ln2

2λln

2λ

0

0 0

)(

V

) ( cm 00 1 2

m 0.000127

2 cm 0127 0



E

23.61: a) From Problem 23.57,

m

V1072.9

m0.070

1)109.000.140(ln

V50,0001

)(ln

V

mV/

109.72

)sm(9.80kg)10(3.0010

4

2 8

24

ln

2 2

2 2

a x a

a x a a

kQ V

a) For a square with two sets of oppositely charged sides, the potentials cancel and

24

ln

4

2 2

2 2

a x a a

kQ

.)12(

)12(ln

44

4ln

4

2 2

2 2

x a

a x a a

kQ V

(22

22

2 2

0

2 2 2 2

/ 1 2 2 2 2

2 2 2 2

2 2

2 2 2

x R x

ε

σ

x R x R

kQ z

R

kQ r

x

dr r R

kQ

dV

V

r x

dr r R

kQ πR

dr πr r

21

2

2 2 0

2 2 2

σ R

x

x R

kQ x

V

E x

23.64: a) From Example 23.12:

1

1ln2

ln2)(

2 2

2 2 2

2

2 2

x a x a a

kQ a

x a

a x a a

kQ x V

222

12

12

)

(

2

1)

1(ln and ,12

111

,

If

2 2

2

2 2

2

x

kQ x

a a

kQ x

a x

a x

a x

a a

kQ

x

V

α α α x

a x

a x

a x

a x a x

11

ln2

ln2)(

2 2

2 2 2

2

2 2

a x a

kQ a

x a

a x a a

kQ x V

)

/2(ln2

λ)2

ln(

4

)2(In1

4ln2)

2(

)22

(ln2

)

(

2

1)

1(lnand,2

11111

,

0 0

2

2 2

2

2 2

2

2 2

2

x a πε

x a

a

πε

Q

x a a

kQ x

a a

kQ a

x

a x a

kQ

x

V

α α α a

x a

x a

23.65: a) Recall:    r     

R

r R

R r ε

ρ dr

r ε

ρ d

V ε

ρr E R

42

2

0 0

R r

πε r

dr ε

ρR d

V r ε

ρR E R

2

λ2

2

:

0 0

2 0

C)102.00(

m0300.0

C)1000.5

)C1000.2(m)0500.00300.0(

)C1000.5(

2 2

9 2

Qx πε

Q πε

u πε

Q x d a x

x ε

π

Q

V

2 2 0

2 / 1 0 2

/ 3 2 2

14

)(

4

2 2

Equation (23.16)

dθ Q πε

V πa

dθ Q πε a

dl πa

Q πε a

dl πε r

0 0

0

.4

14

14

14

1λ

4

14

1

0

3 1

13 3

3 0 0

2 24

33

ˆ)3ˆ5(:

4.V

55

ˆ)3ˆ5(:

and

3 0

0

3 0

0

2 3

0 0

2 56

r d kQ V

r

kQ E R

r R

R

r d r R

kQ R

kQ d

d V

R

kQr E R

3

2 2

1

R

kQr R

kQ R

kQ r

R

kQ R

kQ V

kQ V

b)

23.71: a) Problem 23.70 shows that

R r r ε

Q V

R r R

r R πε

Q

4andfor

)3

(

2 2

R πε

Q V

V R

πε

Q V

R πε

Q

0

0 0

0 0

8and

,4

,8



b) If Q0,V is higher at the center If Q0,V is higher at the surface

23.72: (a) Points a,b,andc are all at the same potential because E 0 inside the spherical shell of charge on the outer surface So V ab V bc V ac 0

m60.0

)C10150()CNm109

 (b) They are all at the same potential

(c) Only V would change; it would be c 2.25106 V

23.73: a) The electrical potential energy for a spherical shell with uniform surface

charge density and a point charge q outside the shell is the same as if the shell is

replaced by a point charge at its center Since F r dU dr, this means the force the shell exerts on the point charge is the same as if the shell were replaced by a point charge

at its center But by Newton’s 3rd law, the force q exerts on the shell is the same as if the shell were a point charge But q can be replaced by a spherical shell with uniform

surface charge and the force is the same, so the force between the shells is the same as if they were both replaced by point charges at their centers And since the force is the same

as for point charges, the electrical potential energy for the pair of spheres is the same as for a pair of point charges

b) The potential for solid insulating spheres with uniform charge density is the same outside of the sphere as for a spherical shell, so the same result holds

c) The result doesn’t hold for conducting spheres or shells because when two charged conductors are brought close together, the forces between them causes the charges to redistribute and the charges are no longer distributed uniformly over the surfaces