Solution Manual - College Physics 7th Edition - Serway CH16 Electrical Energy and Capacitance PDF

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Solution Manual - College Physics 7th Edition - Serway CH15 Electric Forces and Electric Fields...

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Chapter 16

Electrical Energy and Capacitance Quick Quizzes 1.

(b). The field exerts a force on the electron, causing it to accelerate in the direction opposite to that of the field. In this process, electrical potential energy is converted into kinetic energy of the electron. Note that the electron moves to a region of higher potential, but because the electron has negative charge this corresponds to a decrease in the potential energy of the electron.

2.

(b), (d). Charged particles always tend to move toward positions of lower potential energy. The electrical potential energy of a charged particle is PE= qV and, for positivelycharged particles, this increases as V increases. For a negatively-charged particle, the potential energy decreases as V increases. Thus, a positively-charged particle located at x = A would move toward the left. A negatively-charged particle would oscillate around x = B which is a position of minimum potential energy for negative charges.

3.

(d). If the potential is zero at a point located a finite distance from charges, negative charges must be present in the region to make negative contributions to the potential and cancel positive contributions made by positive charges in the region.

4.

(c). Both the electric potential and the magnitude of the electric field decrease as the distance from the charged particle increases. However, the electric flux through the balloon does not change because it is proportional to the total charge enclosed by the balloon, which does not change as the balloon increases in size.

5.

(a). From the conservation of energy, the final kinetic energy of either particle will be given by

(

)

(

)

KE f = KEi + PEi − PE f = 0 + qVi − qVf = −q Vf − Vi = − q( ∆ V ) For the electron, q = −e and ∆V = +1 V giving KE f = − ( − e )( +1 V ) = + 1 eV . For the proton, q = +e and ∆V = −1 V , so KE f = − ( e )( −1 V ) = + 1 eV , the same as that of the electron. 6.

(c). The battery moves negative charge from one plate and puts it on the other. The first plate is left with excess positive charge whose magnitude equals that of the negative charge moved to the other plate.

35

36

CHAPTER 16

7.

(a) C decreases. (d) ∆V increases.

(b) Q stays the same. (c) E stays the same. (e) The energy stored increases.

Because the capacitor is removed from the battery, charges on the plates have nowhere to go. Thus, the charge on the capacitor plates remains the same as the plates are pulled σ Q A = , the electric field is constant as the plates are separated. apart. Because E = ∈0 ∈0 Because ∆V = Ed and E does not change, ∆V increases as d increases. Because the same charge is stored at a higher potential difference, the capacitance has decreased. Because Energy stored = Q 2 2C and Q stays the same while C decreases, the energy stored increases. The extra energy must have been transferred from somewhere, so work was done. This is consistent with the fact that the plates attract one another, and work must be done to pull them apart. 8.

(a) C increases. (d) ∆V remains the same.

(b) Q increases. (c) E stays the same. (e) The energy stored increases.

The presence of a dielectric between the plates increases the capacitance by a factor equal to the dielectric constant. Since the battery holds the potential difference constant while the capacitance increases, the charge stored (Q = C ∆V ) will increase. Because the potential difference and the distance between the plates are both constant, the electric field (E = ∆V d) will stay the same. The battery maintains a constant potential difference. With

(

∆V constant while capacitance increases, the stored energy Energy stored = 12 C ( ∆V )

2

) will

increase. 9.

(a). Increased random motions associated with an increase in temperature make it more difficult to maintain a high degree of polarization of the dielectric material. This has the effect of decreasing the dielectric constant of the material, and in turn, decreasing the capacitance of the capacitor.

Electrical Energy and Capacitance

37

Answers to Even Numbered Conceptual Questions 2.

Changing the area will change the capacitance and maximum charge but not the maximum voltage. The question does not allow you to increase the plate separation. You can increase the maximum operating voltage by inserting a material with higher dielectric strength between the plates.

4.

Electric potential V is a measure of the potential energy per unit charge. Electrical potential energy, PE = QV, gives the energy of the total charge Q.

6.

A sharp point on a charged conductor would produce a large electric field in the region near the point. An electric discharge could most easily take place at the point.

8.

There are eight different combinations that use all three capacitors in the circuit. These combinations and their equivalent capacitances are:

 1 1 1  + + All three capacitors in series - Ceq =    C1 C2 C3 

−1

All three capacitors in parallel - Ceq = C1 + C2 + C3 One capacitor in series with a parallel combination of the other two: −1

−1

 1  1  1  1  1 1 + + +  Ceq =   , C eq =   , C eq =   C1 + C2 C3   C 3 + C1 C2   C2 + C3 C1 

−1

One capacitor in parallel with a series combination of the other two:  CC   CC   CC  Ceq =  1 2  + C3 , Ceq =  3 1  + C2 , C eq =  2 3  + C1 C C C C  1+ 2  C2 + C3   3+ 1  10.

Nothing happens to the charge if the wires are disconnected. If the wires are connected to each other, the charge rapidly recombines, leaving the capacitor uncharged.

12.

All connections of capacitors are not simple combinations of series and parallel circuits. As an example of such a complex circuit, consider the network of five capacitors C1, C2, C3, C4, and C5 shown below. C1

C2

C5

C3

C4

This combination cannot be reduced to a simple equivalent by the techniques of combining series and parallel capacitors.

38

CHAPTER 16

14.

The material of the dielectric may be able to withstand a larger electric field than air can withstand before breaking down to pass a spark between the capacitor plates.

16.

(a) i (b) ii

18.

(a) The equation is only valid when the points A and B are located in a region where the electric field is uniform (that is, constant in both magnitude and direction). (b) No. The field due to a point charge is not a uniform field. (c) Yes. The field in the region between a pair of parallel plates is uniform.

Electrical Energy and Capacitance

Answers to Even Numbered Problems − 6.0 × 10 −4 J

2.

(a)

4.

− 3.20× 10− 19 C

6.

4.3 × 106 J

8.

(a)

(b)

1.52 × 105 m s

–50 V

(b)

6.49 × 10 6 m s

10.

40.2 kV

12.

2.2 × 102 V

14.

-9.08 J

16.

8.09 × 10 −7 J

18.

7.25 × 10 m s

22.

(a)

48.0 µ C

(b)

6.00 µ C

24.

(a)

800 V

(b)

Q f = Qi 2

26.

31.0 Å

28.

1.23 kV

30.

(a)

(b)

1.78 µF

32.

3.00 pF and 6.00 pF

34.

(a)

(b)

Q4 =144 µ C, Q 2 =72.0 µ C, Q 24 =Q 8 =216 µC

36.

Yes. Connect a parallel combination of two capacitors in series with another parallel combination of two capacitors. ∆ V = 45.0 V .

38.

30.0 µ F

40.

6.04 µ F

42.

12.9 µ F

44.

(a)

46.

9.79 kg

6

18.0 µ F

12.0 µ F

0.150 J

(b)

268 V

39

40

CHAPTER 16

48.

(a)

50.

1.04 m

54.

1 1 2 1 1 2 C1 = Cp ± Cp − Cp Cs , C2 = Cp ∓ Cp − Cp Cs 2 4 2 4

3.00 × 10 3 V m

(b)

(a) (c)

1.8 × 104 V −1.8 × 10 4 V

58.

(a)

C=

60.

κ = 2.33

62.

1.8 × 102 µ C on C1 , 89 µ C on C2

64.

121 V

66.

(a)

56.

42.5 nC

(b)

−3.6 × 10 4 V − 5.4 × 10 −2 J

(b)

at x = 4.4 mm

(d)

ab k e ( b − a)

0.11 m

(c)

5.31 pC

Electrical Energy and Capacitance

Problem Solutions 16.1

(a) The work done is W = F ⋅ s cos θ = ( qE) ⋅ s cos θ , or W = ( 1.60 × 10 −19 C ) ( 200 N C) ( 2.00 × 10 −2 m )cos 0° = 6.40 × 10 −19 J

(b) The change in the electrical potential energy is ∆PEe = − W = −6.40 ×10 −19 J

(c) The change in the electrical potential is ∆V =

16.2

∆PEe −6.40 × 10 −19 J = = − 4.00 V 1.60 × 10 -19 C q

(a) We follow the path from (0,0) to (20 cm,0) to (20 cm,50 cm). The work done on the charge by the field is W = W1 + W2 = ( qE) ⋅ s1 cosθ1 + ( qE) ⋅ s2 cosθ 2 = ( qE )  ( 0.20 m ) cos0° + ( 0.50 m ) cos90° = ( 12× 10 −6 C ) ( 250 V m)  ( 0.20 m )+ 0  = 6.0× 10 −4 J

Thus, (b) ∆V =

16.3

∆PE e = −W = − 6.0 × 10−4 J ∆ PE e − 6.0 × 10 − 4 J = = − 50 J C = − 50 V 12 × 10 -6 C q

The work done by the agent moving the charge out of the cell is Winput = − Wfield = − ( −∆ PEe ) = + q( ∆ V) J   = ( 1.60 × 10 −19 C ) + 90 × 10 −3  = 1.4 × 10− 20 J C 

16.4

(

)

∆PE e = q ( ∆V ) = q V f − Vi , so q =

∆PE e −1.92 × 10 −17 J = = − 3.20 × 10 −19 C +60.0 J C Vf − Vi

41

42

CHAPTER 16

∆V 25 000 J C = = 1.7× 106 N C 1.5 × 10 −2 m d

16.5

E=

16.6

Since potential difference is work per unit charge ∆V =

W , the work done is q

W = q( ∆V ) = ( 3.6 × 10 5 C )( +12 J C ) = 4.3 × 10 6 J

16.7

(a)

E=

∆V d

=

600 J C = 1.13 × 105 N C 5.33 × 10 −3 m

(b) F = q E = ( 1.60 × 10 −19 C )(1.13 × 10 5 N C) = 1.80 × 10 −14 N

(c)

W = F ⋅ s cos θ = ( 1.80 × 10−14 N )  (5.33 − 2.90 )× 10−3 m  cos 0° = 4.38 × 10 −17 J

16.8

1 From conservation of energy, mv 2f − 0 = q ( ∆V ) 2 (a) For the proton,

vf =

(b) For the electron, v f =

or v f =

2 ( 1.60 × 10 −19 C ) ( −120 V ) 1.67 × 10 −27 kg

m

= 1.52 × 10 5 m s

2 ( − 1.60 × 10 − 19 C ) (+ 120 V ) 9.11 ×10 − 31 kg

2 q( ∆V)

= 6.49 × 10 6 m s

43

Electrical Energy and Capacitance

16.9

(a) Use conservation of energy

Q k

ur E

( KE + PEs + PEe ) f = ( KE + PEs + PEe )i or ∆ ( KE ) + ∆ (PE s ) + ∆ ( PE e ) = 0

x=0

∆ ( KE ) = 0 since the block is at rest at both beginning and end.

1 ∆ ( PEs ) = kx 2max − 0 , 2 where xmax is the maximum stretch of the spring. ∆ ( PE e ) = −W = − (QE ) x max 1 Thus, 0 + kx2max −( QE) xmax = 0 , giving 2 −6 5 2 QE 2 ( 50.0 × 10 C )( 5.00 × 10 V m ) x max = = 0.500 m = 100 N m k

(b) At equilibrium, ΣF = − Fs + Fe = 0, or Therefore,

x eq =

− kx eq + QE = 0

QE 1 = x max = 0.250 m k 2

Note that when the block is released from rest, it overshoots the equilibrium position and oscillates with simple harmonic motion in the electric field.

16.10

Using ∆y = v0y t +

0 = v0 y t +

1 2 a t for the full flight gives 2 y − 2 v0 y 1 2 ay t , or a y = 2 t

Then, using vy2 = v02y + 2 ay ( ∆y ) for the upward part of the flight gives

( ∆y) max =

0 − v20 y 2 ay

=

(

− v20 y

)

2 − 2 v0 y t

=

v0 y t 4

=

(20.1 m s ) ( 4.10 s ) 4

= 20.6 m

44

CHAPTER 16

From Newton’s second law, ay =

Σ Fy m

=

− mg − qE qE  = −  g +  . Equating m m 

qE  − 2 v 0y  this to the earlier result gives ay = −  g + , so the electric field strength is = m t    m   2 v0 y   2.00 kg   2 ( 20.1 m s ) E =   − g =  − 9.80 m s2  = 1.95 × 103 N C   −6    5.00 ×10 C   4.10 s  q  t Thus,

16.11

(a)

( ∆V) max = ( ∆ymax ) E = ( 20.6 m ) ( 1.95 ×103 N C ) = 4.02 ×104 V = 40.2 kV

V=

− 19 9 2 2 ke q ( 8.99 × 10 N ⋅ m C )( 1.60 × 10 C ) = = 1.44 × 10 −7 V r 1.00× 10-2 m

(b) ∆V = V2 − V1 =

1 1 ke q ke q − = ( ke q )  −  r2 r1  r2 r1 

  1 1 = ( 8.99 × 10 9 N ⋅ m 2 C 2)( 1.60 × 10 − 19 C)  −   0.020 0 m 0.010 0 m  = −7.19 × 10 −8 V

16.12

q q  V = V1 + V2 = ke  1 + 2  where r1 = 0.60 m − 0 = 0.60 m , and  r1 r2  r2 = 0.60 m − 0.30 m = 0.30 m. Thus,

 N ⋅ m2   3.0 ×10 −9 C 6.0 ×10 −9 C V =  8.99 × 10 9 +  2 0.30 m C   0. 60 m 

 2  = 2.2 × 10 V 

Electrical Energy and Capacitance

16.13

(a) Calling the 2.00 µ C charge q3 , V=∑ i

q q q k eq i = ke  1 + 2 + 2 3 2  r2 ri r r1 + r2  1

   

  N⋅ m2   8.00 × 10− 6 C 4.00 ×10 − 6 C =  8.99× 109 + +  2 C   0.060 0 m 0.030 0 m  

2.00 × 10 − 6 C

( 0.060 0 )2 + ( 0.030 0 )2

   m 

V = 2.67 × 106 V (b) Replacing 2.00 ×10 − 6 C by −2.00 ×10 −6 C in part (a) yields V = 2.13 × 106 V

16.14

(

)

W = q( ∆V) = q Vf − Vi , and Vf = 0 since the 8.00 µ C is infinite distance from other charges. q q   N ⋅m2 Vi = k e  1 + 2  =  8.99× 10 9 2 C  r1 r2  

   2.00 × 10−6 C +    0.030 0 m 

4.00 × 10 − 6 C

= 1.135× 106 V Thus, W = ( 8.00 × 10 −6 C )(0 − 1.1 35 × 10 6 V ) = − 9.08 J

16.15

(a)

V=∑ i

2

(0.030 0 ) + (0.060 0 )

k e qi ri

 N⋅ m2  5.00× 10− 9 C 3.00× 10− 9 C =  8.99 × 10 9 −   = 103 V C2  0.175 m 0.175 m  

2

   m 

45

46

CHAPTER 16

(b) PE =

ke qi q2 r12

2 (5.00 ×10 −9 C )( −3.00 ×10 − 9 C )  9 N ⋅m  =  8.99× 10 = − 3.85× 10− 7 J C2  0.350 m 

The negative sign means that positive work must be done to separate the charges (that is, bring them up to a state of zero potential energy).

16.16

9 The potential at distance r = 0.300 m from a charge Q = + 9.00 × 10− C is

V=

ke Q r

=

(8.99 × 10

9

N ⋅ m2 C2 )( 9.00 × 10− 9 C ) 0.300 m

= + 270 V

Thus, the work required to carry a charge q = 3.00 × 10−9 C from infinity to this location is W = qV = (3.00 × 10 −9 C )( +270 V ) = 8.09 × 10 −7 J

16.17

The Pythagorean theorem gives the distance from the midpoint of the base to the charge at the apex of the triangle as r3 =

2 2 ( 4.00 cm ) − (1.00 cm ) = 15 cm = 15 × 10 −2 m

Then, the potential at the midpoint of the base is V = ∑ k eq i ri , or i

2  −7.00 ×10 −9 C ) (−7.00 × 10 −9 C ) ( +7.00 ×10 − 9 C )  (  9 N ⋅m   V =  8.99× 10 + +  − 0.010 0 m C 2   0.010 0 m 15× 10 2 m  

= − 1.10× 104 V = − 11.0 kV

Electrical Energy and Capacitance

16.18

47

Outside the spherical charge distribution, the potential is the same as for a point charge at the center of the sphere, V = k eQ r , where Q =1.00 ×10 −9 C 1 1  Thus, ∆( PEe ) = q( ∆V) = − ek eQ  −   rf ri    and from conservation of energy ∆ (KE ) = −∆ ( PEe ) ,

or

  1 1  1 me v 2 − 0 = − − eke Q  −   This gives v =  rf ri   2   

2 ke Qe  1 1   −  , or me  rf ri 

 N ⋅m 2  −9 −19 2 8.99 × 109  ( 1.00 × 10 C )(1.60 × 10 C )  2  C 1 1  − v=    − 31 9.11 × 10 kg  0.020 0 m 0.030 0 m  v = 7.25 × 10 6 m s

16.19

From conservation of energy, ( KE + PE e ) f = ( KE + PE e ) i , which gives 0+

2 k Qq 2 k e ( 79 e )( 2 e ) k eQq 1 2 = mα vi + 0 or r f = e 2 = 2 mα v i mα v i2 rf

2  N ⋅ m2  −19 2  8.99 × 10 9  (158 )( 1.60 × 10 C ) 2 C  rf =  = 2.74× 10 −14 m 2 − 27 7 (6.64 × 10 kg )( 2.00 × 10 m s)

16.20

By definition, the work required to move a charge from one point to any other point on an equipotential surface is zero. From the definition of work, W = ( F cos θ ) ⋅ s , the work is zero only if s = 0 or F cosθ = 0 . The displacement s cannot be assumed to be zero in all cases. Thus, one must require that F cosθ = 0 . The force F is given by F = qE and neither the charge q nor the field strength E can be assumed to be zero in all cases. Therefore, the only way the work can be zero in all cases is if cosθ = 0 . But if cosθ = 0 , then θ = 90° or the force (and hence the electric field) must be perpendicular to the displacement s (which is tangent to the surface). That is, the field must be perpendicular to the equipotential surface at all points on that surface.

48

CHAPTER 16

V=

16.21

r=

ke Q r

so

−9 9 2 2 ke Q (8.99 × 10 N ⋅ m C )( 8.00 × 10 C ) 71.9 V ⋅ m = = V V V

For V = 100 V, 50.0 V, and 25.0 V,

r = 0.719 m, 1.44 m, and 2.88 m

The radii are inversely proportional to the potential.

16.22

16.23

(a)

Q = C( ∆ V) = ( 4.00× 10 −6 F ) ( 12.0 V ) = 48.0× 10 −6 C = 48.0 µ C

(b)

Q = C( ∆ V) = ( 4.00 × 10 −6 F) ( 1.50 V) = 6.00× 10 −6 C = 6.00 µ C

(a) (b)

C = ∈0

6 2 C 2  ( 1.0 × 10 m ) A  =  8.85 × 10 −12 = 1.1 × 10 −8 F  d  N ⋅ m2  ( 800 m )

Q max = C (∆V

)max = C (Emaxd )

= ( 1.11× 10−8 F )( 3.0 × 106 N C) ( 800 m ) = 27 C

16.24

For a parallel plate capacitor, ∆V =

Qd Q Q = = . C ∈0 ( A d ) ∈0 A

(a) Doubling d while holding Q and A constant doubles ∆ V to 800 V .

(∈0 A) ∆ V

Thus, doubling d while holding ∆V and A constant will cut the d charge in half, or Q f = Q i 2

(b) Q =

16.25

(a)

∆V 20.0 V = = 1.11× 104 V m = 11.1 kV m directed toward the negative d 1.80 × 10-3 m plate E=

(b) C =

∈0 A ( 8.85 ×10 = d

−12

C 2 N ⋅m 2 )(7.60 ×10 −4 m 2 -3

1.80× 10 m

= 3.74 × 10 −12 F = 3.74 pF