Tài liệu Physics exercises_solution: Chapter 24 pdf

Since the inner conductor is at a higher potential it is positively charged, and the magnitude is: C... The magnitude of the charge for capacitors in series is equal, while the charge is

24.1: QCV (25.0V)(7.28μF)1.8210 4 C.

m00328.0

m00122

F1029.3

C1035.4

m00328.0

V102

F1045.2

C10148.0

b) (2.45 10 F)(0328 10 m) 0.0091m2

0

3 10

Cd A

m10328.0

d) (1.84 106 V m) 1.63 10 5C/m2

0 0 0

ε

σ Ed V

2 12

NmC1085.8

)m00180.0)(

mC1060.5(

2means

d

c) r2rmeansA4A,C4C,andQ4Q480μC

24.6: (a) 12.0 V since the plates remain charged.

)m010.0(

2 0

2 0

d

A ε

)CN10(V

A ε C

2 0

Cd

)C

Nm109)(

m10)(

F1000.5(

R

cm24.4m1024

2 0

a

b r r

πε L

C 

F1035.4)50.000.5(ln

2)m180.0

24.10: a) 1.77 5.84.

mF105.31

22

)(ln)(ln

2

12 0 0

b a

r πε

L C

πε r

r r

r

πε L

C

b)  (2.60V)(31.510 12 F m)8.1910 11 C m

L

C V L Q

)mm5.1/mm5.3(ln

2)

(ln

r r

πε L

C

a b

b) The charge on each conductor is equal but opposite Since the inner conductor

is at a higher potential it is positively charged, and the magnitude is:

C

1043.6)mm1.5mm5.3(ln

)V35.0)(

m8.2(2)(ln



r r

LV ε CV

Q

a b

b a a b a

b

b a

r kC

kCr r

r r kCr kCr r

r

r r k

)m150.0)(

F10116(

)m125.0)(

m148.0(1

r r k

C

a b

a b

b) The electric field at a distance of 12.6 cm:

.N/C6082)

m126.0(

)V120)(

F1094.8(

2

11 2

r

kCV r

kQ E

c) The electric field at a distance of 14.7 cm:

N/C

4468)

m147.0(

)V120)(

F1094.8(

2

11 2

r

kCV r

kQ E

d) For a spherical capacitor, the electric field is not constant between the

surfaces

24.14: a)

)F100.6(

1)

F10)0.50.3((

11

11

6 6

3 2 1 eq

.F1042

The magnitude of the charge for capacitors in series is equal, while the charge is

distributed for capacitors in parallel Therefore,

.C1021.8)F1042.3)(

V0.24

eq 2

2 2 2

2 1

C

C Q C

Q C

,C1008.3C

1021

2

5 1

5 1

5 1

1 1

2 V Q C (3.08 10 C)/(3.00 10 F) 10.3V.AndV

V

.V7.13V3

1)

F0.4F00.2(

11

)(

11

4 3 1 1

Then, (2.40 10 6 F)(28.0V) 6.72 10 5C

eq total 4

3 12

C1072.63

3 5

5 total

12 3

Q

1 1

(4.48 10 5C) (4.00 10 6 F) 11.2V

3 3

V

V

8.16)F1000.4()C1072.6

4 4

V0.52(

F1088.1F

1033.5

)F100.5(

1)

F100.3(

11

11

5 6

eq

6 eq

1 5

6 6

2 1 eq

C

C C C

2

V

24.17: a) (52.0V)(3.0 10 6 F) 1.56 10 4 C.

1 1

C

106.2)F100.5)(

V0.52

2 2

1

1 1

1 1

A ε A ε

d A ε

d C

C

C         So the combined capacitance for two

capacitors in series is the same as that for a capacitor of area A and separation (d1d2)

2 1

A A ε d A ε d A ε C C

C       So the combined capacitance for two

capacitors in parallel is that of a single capacitor of their combined area (A1  A2) and

common plate separation d.

24.20: a) and b) The equivalent resistance of the combination is 6.0 F, therefore the total charge on the network is: (6.0 F)(36V) 2.16 10 4C

eq eq







charge on the 9 μ capacitor because it is connected in series with the point b So:.0 F

.V24F100.9

C1016.2

6 4

Then V3 V11 V12 V6 V V9 36V24V12V

C

106.3)V12)(

F0.3

3 3 3

F11

11 11 11

C108.4

V

8F100.6

C108.4

6 5

12

12 12

6 5

6

6 6

24.21: Capacitances in parallel simply add, so:

F

57F

72F)15(F0.9

1F)0.411(

1F

μ μ

x μ

μ x μ

C100.40

6 6

Thus (13.33V)(3.00 10 6 F) 80.0 10 6 C

1

parallel combination of C1and C2, its charge must be equal to their combined charge:

C100.120C100.80C

1F

1000.9

11

11

6 6

F21.3

and

V4.37F1021.3

C100.120

6 6

 CV

is doubled, the capacitance halves, and the energy stored doubles So the amount of work done to move the plates equals the difference in energy stored in the capacitor, which is

2

 ε E ε

u

Cd A d

A ε C

c) (3.00 106 V m)(0.0015m) 4500V

max max

)C1080.1(2

6 11

2 8 2

24.27: (4.50 10 4 F)(295V)2 19.6J

2 1 2 2

C

Q C

Q

0

1 1

1

1 2 2

1 2

1

3

23

2so

3

22

3

2)

2(

so2and

V C

Q C

Q V Q Q Q Q

Q Q C

Q C

Q C C C C

3 1 2 3 2

2

2 2 1

2 1

3

13

1)(2)(2

12

1

CV C

Q C

Q C

Q C

Q C

xQ C

U

dx  xε dx A Q  

The change is then Q ε A dx

2

0

2 0

A ε

Q F Fdx A ε

dxQ U

U

d) The reason for the difference is that E is the field due to both plates The force

is QE if E is the field due to one plate is Q is the charge on the other plate.

24.30: a) If the separation distance is halved while the charge is kept fixed, then the

capacitance increases and the stored energy, which was 8.38 J, decreases since

2

Q

U  Therefore the new energy is 4.19 J

b) If the voltage is kept fixed while the separation is decreased by one half, then the doubling of the capacitance leads to a doubling of the stored energy to 16.76 J, using

J0.25(2

C1020

Since  for a parallel plate capacitor

m10734

6

farad10

417.3

)m1060.2)(

mN/C1085.8)(

00.1(

3

12

2 3 2

2 12 0

d

The energy density is thus

3

7 3

2 3

2 12

2 1 2 2

1

m

J1063.5)m10734.6)(

m1060.2(

V)40.2)(

farad10

42.3

)J1020.3(222

)(ln

2

0 0

r

r r r

πε L

C

b

a b a

24.34: a) For a spherical capacitor:

.V7.38)F1053.8()C1030.3(

F1053.8)m100.0m115.0(

)m115.0)(

m100.0(11

11 9

k r r

r r k

C

a b

b a

2

)V7.38)(

F1053.8(2

2 0 2 0

)m126.0(

)F1094.8()V120(22

22

r

kq ε E ε u

.mJ1064



 u

b) The same calculation for r 14.7cmu8.8310 5 J m3

c) No, the electric energy density is NOT constant within the spheres

)m120.0(

)C1000.8(32

14

12

12

4

2 9

0 2

2

2 0 0

2 0

q πε ε E ε u

b) If the charge was –8.00 nC, the electric field energy would remain the same

since U only depends on the square of E.

24.37: Let the applied voltage be V Let each capacitor have capacitance 2

2 1

2 2

2

1 p

2

;2)(2

;2

2

.voltagewith

capacitorsingle

afor)

b4

2

Q Q CV CV

Q CV V

C Q

V CV

Q U U

CV CV

24.38: a) CKε0A d gives us the area of the plates:

2 4 2

2 12

3 12

0

m10475.8)mN/C1085.8)(

00.1(

)m1050.1)(

farad10

00.5

2

C)N1000.3)(

m10475.8)(

mNC1085.8)(

00.1

(

10

4 2

4 2

2 12

b) Again, QKε0A(V d)2.70ε0A(V d) If we continue to think of V d as

the electric field, only K has changed from part (a); thus Q in this case is

.C1008.6C)10

dielectric creates the bound charges on its surface

mV1050.2

mV1020.3

E

.mC

σ i

c) U CV uAd Kε E2Ad

0 2 1 2

)V5500)(

F1025.1

7 0

Kε

CV A V

AE Kε d

A Kε

C

24.42: Placing a dielectric between the plates just results in the replacement of  for0in the derivation of Equation (24.20) One can follow exactly the procedure as shown for Equation (24.11)

σ i

24.44: a)    ( 1) ( 1) (2.1)(2.510 7 F)(12V)

0 0 0

Q Q Q

J)1085.1(22

2

1

7 5

0

0 2

V C U

)V1.10)(

F1060.3(

)J1085.11032.2(22

1

2 7

5 5

2 0

U K V

KC U

24.46: a) The capacitance changes by a factor of K when the dielectric is inserted Since

V is unchanged (The battery is still connected),

80.1pC

0.25

pC0.45

before

after before

Q

Q C

C

b) The area of the plates is πr2  π(0.0300m)2 2.82710 3m2, and the separation between them is thus

m10002

2

farad10

5.12

)m10827.2)(

mNC1085.8)(

00.1(

3

12

2 3 2

2 12 0

d

Before the dielectric is inserted,

V000

2

)m10827.2)(

mNC1085.8)(

00.1(

)m1000.2)(

C100.25(

2 3 2

2 12

3 12

0 0

Qd

V

V

Q d

A Kε

)m10827.2)(

00.1)(

mNC1085.8(

C100.25

2 3 2

2 12

Q E

Again, since the voltage is unchanged after the dielectric is inserted, the electric field is also unchanged

24.47: a) before: (9.00 10 6 C) (3.00 10 6 F) 3.00 V

0 0

V

after: C KC0 15.0F;QQ0

K V

C Q

q πd KE ε

2 0

q q E ε

q q πd E ε

q q ε

b q K 

q

24.49: a) Equation (25.22):        

0 0

Q A Kε

Q ε

Q ε

Q

E KEA

Qd Ed

d

A ε K d

εA V

Q

m107.4

)m16.0

3

2 0

b) Q  CV (4.810 11F)(12V)0.5810 9 C

c) E= V d=(12 V)/(4.710 3m)=2553 V m

d) (4.8 10 11F)(12V)2 3.46 10 9 J

2 1 2 2

a) C 2.4110 11 F b) Q0.5810 9 C

)F1041.2(2

)C1058.0(2

9 11

2 9 2

24.51: If the plates are pulled out as in Problem 24.50 the battery is connected, ensuring

that the voltage remains constant This time we find:

a) C 2.410 11 F b) Q2.910 10 C c)

m

V103.10094.0

2

)V12()F104.2(2

9 2

11 2

24.52: a) System acts like two capacitors in series so 1 1 1

12

0 2

2

2 2

2 0 eq

2 0 2

d Q Q C

Q U d

L ε C d

L ε C C

d L

 for a single plate (not ε Q A

0 since charge Q is only on one face).

0 4

2 0 2

2 0 3

2 0 1

2

Q L

ε

Q L

ε

Q L

ε

Q L

2 0 2

2 0 3

2 0 1

2 0

22

22

Q L

ε

Q L

ε

Q L

ε

Q L

2 0 2

2 0 3

2 0 1

2

Q L

ε

Q L

ε

Q L

ε

Q L

2 2

0

2 2 0

2 new

2 0

2 2

4 2 0

2 4

2 0

2 4

2 0

2 0 2

2 0

new

23

34

2

12

1

L ε

d Q L ε

d Q L ε

d Q U U U

L ε

d Q d L L ε

Q L ε

Q L

ε

Q ε d L E ε

This is the work required to rearrange the plates

24.53: a) The power output is 600 W, and 95% of the original energy is converted

.J421J

400)s1048.1()W1070.2

0 3

)J421(222

1

2 2



V

U C CV U

m1000.7

)m1020.4

4 0 2 5 0

)m1020.4

2 5 0

Aε C

Therefore the key must be depressed by a distance of:

.mm224.0m1076.4m1000

24.55: a) d 2

)1

ln(

π2)

)(ln(

2)

Lε πr r

d

L ε r

r d

L πε r

r

L πε C

a a

a a

Q

eq

4 C,and10

1.12

V)

(28    C C1C2 13.0μF.Sotheoriginalenergystoredis

:isstoredenergy new

theSoF

0.13same,

the

still

isecapacitancequivalent

theand,C104.1is

chargetotal

the

now

so

,capacitorsthe

flipandDisconnectJ

1010.5V)(28F)100.13(

eq

4 1

2

3 2

6 2

1 2

eq

2

1

μ C

Q Q Q

J1054.7F)100.13(2

)C104.1(2

3 3

4

4 6

2 4

eq 2

Q U

24.57: a) Ceq 4.00μF 6.00μF10.00μF, andQ total Ceq V (10.00μF)(660V)6.610 3C The voltage over each is 660 V since they are in parallel So:

C

1096.3)V660()F00.6(

C

1064.2)V660()F00.4(

3 2

2 2

3 1

1 1

C Q

μ V

C Q

b) Q total 3.9610 3 C2.6410 3 C1.3210 3C,andstillCeq 10.00F,

so the voltage is V = Q/C = (1.3210 3 C) (10.00 μF)132V,and the new charges:

C

1092.7)V132)(

F00.6(

.C1028.5)V132)(

F00.4(

4 2

2 2

4 1

1 1

C Q

μ V

C Q

b) If one capacitor is a moderately good conductor, then it can be treated as a

“short” and thus removed from the circuit, and one capacitor will have greater than 600

V over it

24.59: a) 1 1 1  1 1 5 2 2 and

5 1 1 1 2 1

C C C C C

33

53

221

2 1 eq 4 3

C C C C C

V44(V

V88(V

88)66(2220

So

4 4

3 2

1

4 3

4 2

Q

V V μ

Q V

24.60: a) With the switch open:     61F1 4.00 F

F

31

1 F

61F

31

C  μ  μ   μ  μ  

C108.4V)(210F)00.4

capacitor carries 4.20 10 4 C The voltages are then just calculated via V=Q/C.

6)(3

1F

6)(3

μ μ

potential difference of 105 V (again, by symmetry)

c) The only way for the sum of the positive charge on one plate of C2and the negative charge on one plate of C to change is for charge to flow through the switch 1

That is, the quantity of charge that flows through the switch is equal to the charge in

;C315

1F4.8

1F

4.8

μ μ

,C102.27C)

1056.7(3andF21F)4.2F8.4F

Q

V

Q μ

μ μ

μ

C

d) (21 F) (10.8V)2 1.22 10 3 J

2 1 2 2

U

24.62: a) 2.4 10 F

F6.0

1F4.0

b) Disconnecting them from the voltage source and reconnecting them to

themselves we must have equal potential difference, and the sum of their charges must be the sum of the original charges:

C

101.90V)(316F)1000.6(

C

101.26V)(316F)1000.4(

V

316F

1010.0

C)102(1.582

)(

2and

3 6

2

3 6

1

6 3

2 1

2 1 2 1 2

2 1

C C

Q V

V C C Q Q Q V C Q V

C Q

24.63: a) Reducing the furthest right leg yields    1 

F 9

6 1F 9

6 1F 9

.3

b) For the three capacitors nearest points a and b:

C109.7V)(420F)103.2

V(120F)60

111

1

C C C

C654V)(120F)45.5(

F45.5

equiv

μ μ

CV Q

μ C









24.65: a) Q is constant.

with the dielectric: V Q CQ (KC0)

without the dielectric: V0 Q C0

3.91V)

V)/(11.5(45.0

3V)(45.02

32

3

)2()3(

;32,3

0 0

eq

0 2 1 eq 0 2 0 1

C

Q C

Q

V

K C

C C C C C C K

a d A ε a d A ε C

a d

d C a d

d d

A ε a d

A ε C

comparable to the capacitance of a typical capacitor in a circuit

1

0 2 2 2

2 0 0

2 0 2

1

r ε π

Q r

πε

Q ε E ε u R

0

2 2

88

4d) This energy is equal to πε Q R

0

2

4 2

1 which is just the energy required to assemble all the charge into a spherical distribution (Note, being aware of double counting gives the factor of 1 in front of the familiar potential energy formula for a charge Q a distance R2

from another charge Q.)

e) From Equation (24.9): πε Q R

C Q U

0

2 2

8

 from part (c) C4πε0R, as in Problem (24.67)

2 0

82

12

1:

πR

r kQ R

kQr ε E ε u R r

82

12

1

2 2

2 0

2

kQ r

kQ ε E ε u R

4:

2

0

4 6 2

0

2

R

kQ dr r R

kQ udr r π udV U

R

r

R R

32

24

R

kQ U

R

kQ r

dr kQ udr r π udV U

R

r

R R

πε ε E ε

0 2 2 2

0 0

2 0

8

λ2

λ2

12

r

r r πε L

U r

dr πε

Lλ urdr πL udV U

/(ln4

λ)/(ln4

2

0

2 2

U r r

L r

r L

Q C

Q





1

2 1 0

1

2 1

1 1 2

1 1 eq

112

22

2/2

d A

ε

d A

ε

d d

A ε d

A ε C

2 1

2 1 0

eq    



K K

K K d

A ε C

24.72: This situation is analagous to having two capacitors in parallel, each with an

area 2A So:

)

(2

22

2 1 0 2

1 2 1

d

A ε d

A ε d

A ε C C

)4.5(

C/m1050

0

2 3

σ E

b) V Ed (1.0107 V/m)(5.010 9 m)0.052V The outside is at the higher potential

c) volume 10 16 m3 R2.8810 6 m

 shell volume 4πR2d 4π(2.88106 m)2(5.0109m)5.21019 m3

J.1036.1)m102.5(V/m)100.1()4.5(V)

0 2

1 2 0 2

V)3000()m200.0()50.2

2

2 0

b) Q i Q(11/K)(1.3310 6 C)(11/2.50)7.9810 7 C

)m200.0((2.50)

C1033

2 0

Kε

Q ε

σ E

d) (1.33 10 C)(3000V) 2.00 10 J

2

12

J1000.2

1 2 0 2

24.75: a) We are to show the transformation from one circuit to the other:

From Circuit 1:

y

q q

 and V bc  2 3 ,

x C

y z y x

z y x x

y z

q C

q K C

q C

q C C C

C C C q

C

q q C

q q C

q

3 3 2 3 1 3

From Circuit 2:

3

2 3 1

1 3

2 1 1

C

q C C

q C

q q C

2 3

1 3

2 1 2

q C

q C

q q C

q

V bc

Setting the coefficients of the charges equal to each other in matching potential equations from the two circuits results in three independent equations relating the two sets of capacitances The set of equations are:

x x

111,11

111

2 1

3 KC y C x

C From these, subbing in the expression for K we get:,

.)(

.)(

.)(

3 2 1

z x z z y y x

y x z z y y x

x x z z y y x

C C C C C C C C

C C C C C C C C

C C C C C C C C

24.76: a) The force between the two parallel plates is:

.22

2

)(2

2 0 0

2 2

2 2 0 0

V z

A ε A ε

CV A ε

q ε

qσ qE

b) When V 0, the separation is just z0 So:

.04

222

)(4

2 0 0 2 3 2

2 0 0

k

AV ε z z z z

AV ε z z k

m1082.3m)10

displacements from the equilibrium positions above, the 1.014 mm separation is seen to

be stable, but not the 0.537 mm separation

24.77: a) 0(( ) ) 0 ( ( 1) )

D

L ε xKL L x L D

2

)1()

1(2

0 2

D

L V ε K V K D

dx L ε

2

2

12

1

C

C V C CV

U

2

)1()

1(1

2

2 0 0

0 0

2

D

L V ε K U

U U dx

K DC

L ε V

24.78: a) For a normal spherical capacitor: 0 4 0 .

a b

b

r r r πε

C   Here we have, in effect, two parallel capacitors, C and L C U

b a L

r r

r r K

b a U

r r

r r C

b) Using a hemispherical Gaussian surface for each respective half:

0 0

2

22

4

r πKε

Q E

Kε

Q πr

4

2 0 0

2

r πε

Q E

ε

Q πr

Q K Q

KQ K

1(2

2

22

1

Q K r

πKε K

22

πKε K

)1(44

)

K πr

Q πr

Q σ

a a

U U

f r a    and ( ) 4 2 4πr2(1 K).

Q πr

Q σ

b b

u U

)1(44

)

K πr

KQ πr

Q σ

a a

L L

)

K πr

KQ πr

Q σ

b b

L L

41

11

4

)1()11

a a

f

i

πr

Q K

K K

K πr

Q K

K K σ

σ

a r a

11

4

)1()11

b a

K K

K πr

Q K

K K σ

σ

b r

24.79: a)

m105.4

)m120.0()2.4(2

24.80: a) The capacitors are in parallel so:

eff eff

K L

h L

Kh d

WL ε d

Wh Kε d

h L W ε d

WL ε

b) For gasoline, with K 1.95:

4

1full:

2

1

;24.1

2

1

;9