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Transcript
Kapittel 29
Løsning på alle oppgavene finnes på CD. Lånes ut hvis behov.
29.1. Model: The mechanical energy of the proton is conserved. A parallel-plate capacitor has a uniform electric
field.
Visualize:
The figure shows the before-and-after pictorial representation. The proton has an initial speed vi  0 m/s and a final
speed vf after traveling a distance d  2.0 mm.
Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The
potential energy U is defined as U  U0  qEx, where x is the distance from the negative plate and U0 is the potential
energy at the negative plate (at x  0 m). Thus, the change in the potential energy of the proton is
Up  Uf – Ui  (U0  0 J) – (U0  qEd)  qEd
The change in the kinetic energy of the proton is
K  K f  Ki  12 mvf2  12 mvi2  12 mvf2
The law of conservation of energy is K  Up  0 J. This means
1
2
 vf 
mvf2   qEd   0 J
2  1.60 1019 C   50,000 N/C   2.0 103 m 
2qEd

 1.38 105 m/s
27
m
1.67

10
kg


Assess: As described in Section 29.1, the potential energy for a charge q in an electric field E is U  U0  qEx, where
x is the distance measured from the negative plate. Having U  U0 at the negative plate (with x  0 m) is completely
arbitrary. We could have taken it to be zero. Note that only U, and not U, has physical consequences.
29.2. Model: The mechanical energy of the electron is conserved. A parallel-plate capacitor has a uniform electric
field.
Visualize:
The figure shows the before-and-after pictorial representation. The electron has an initial speed vi  0 m/s and a final
speed vf after traveling a distance d  1.0 mm.
Solve: The electron loses potential energy and gains kinetic energy as it moves toward the positive plate. The
potential energy U is defined as U  U0 qEx, where x is the distance from the negative plate and U0 is the potential
energy at the negative plate (at x  0 m). Thus, the change in the potential energy of the electron is
Ue  Uf – Ui  (U0  qEd) – (U0  0 J)  qEd
The change in the kinetic energy of the electron is
K  K f  Ki  12 mvf2  12 mvi2  12 mvf2
Now, the law of conservation of mechanical energy gives K  U  0 J. This means
1
2
2qEd

m
 vf 
mvf2  qEd  0 J
 2  1.60 1019 C   20,000 N/C  1.0 103 m 
9.111031 kg
 2.7 106 m/s
Assess: Note that Ue  qEd is the change in the potential energy of the electron. It is negative because q   e for
the electron. Thus, the potential energy becomes more negative as d increases, that is, the potential energy of the
electron decreases with an increase in d (or x).
29.5. Model: The charges are point charges.
Visualize: Please refer to Figure EX29.5.
Solve: The electric potential energy of the electron is
U electron  U13  U 23

19
19
 1.60 10 C  1.60 10 C 
  9.0 10 N m /C  

2
2
9
9
  2.0 10 m    0.50 10 m 
 1.12 1019 J  1.12 1019 J  2.24 1019 J
9
2
2
1.60 10
 2.0 10

C  1.60 10 19 C  
2
2 
m    0.50 10 9 m  

19
9
29.6. Model: The charges are point charges.
Visualize: Please refer to Figure EX29.6.
Solve: For a system of point charges, the potential energy is the sum of the potential energies due to all pairs of
charges:
U elec  
i,j
Kqi q j
rij
 U12  U13  U 23  K
q1q2
qq
qq
K 1 3 K 2 3
r12
r13
r23
  2.0 nC  1.0 nC   2.0 nC  1.0 nC   1.0 nC  1.0 nC  
  9.0  109 N m 2 /C2  



0.030 m
0.030 m
0.030 m


6
6
6
6
 0.60  10 J  0.60  10 J  0.30  10 J  0.90  10 J
Assess: Note that U12  U21, U13  U31, and U23  U32.
29.13. Model: Energy is conserved. The potential energy is determined by the electric potential.
Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference.
Because the negative electron gains speed as it travels, it moves into a region of higher potential (U  K).
Solve: The potential energy of charge q is U  qV. Using the conservation of energy equation,
Kf  qVf  Ki  qVi  Vf  Vi  V 
1
1
 Ki  Kf  
 0 J  12 mvf2 
q
 e 
31
6
mv 2  9.1110 kg  2.0 10 m/s 
 V  f 
 11.4 V
2e
2 1.60 1019 C 
2
Assess: A positive value of V shows that the electron moved from a region of lower potential to a region of higher
potential.
29.14. Model: Energy is conserved. The potential energy is determined by the electric potential.
Visualize:
The figure shows a before-and-after pictorial representation of an electron moving through a potential difference.
Solve: (a) Because the electron is a negative charge and it slows down as it travels, it must be moving from a region
of higher potential to a region of lower potential.
(b) Using the conservation of energy equation,
Kf  Uf  Ki  Ui  Kf  qVf  Ki  qVi
 Vf  Vi 
1
1 1 2
 Ki  K f  
 mv  0 J 
q
 e  2 i
 9.111031 kg   500,000 m/s   0.712 V
mvi2

2e
2 1.60 1019 C 
2
 V  
Assess:
region.
The negative sign with V verifies that the electron moves from a higher potential region to a lower potential
29.19. Model: The electric potential difference between the plates is determined by the uniform electric field in
the parallel-plate capacitor.
Solve: (a) The potential difference VC across a capacitor of spacing d is related to the electric field inside by
E
VC
 VC  Ed  1.0 105 V/m   0.0020 m   200 V
d
(b) The electric field of a capacitor is related to the surface charge density by
E
 Q A

0
0
 Q   0 AE  8.85 1012 C2 /N m 2  4.0 104 m 2 1.0 105 V/m   3.5 10 10 C
29.20. Model: The electric potential between the plates of a parallel-plate capacitor is determined by the uniform
electric field between the plates.
Solve: (a) The potential difference across the plates of a capacitor is
 
 Q A d  Qd   0.708 109 C1.00 103 m   200 V
VC  Ed    d 
0
A 0  4.00 104 m2 8.85 1012 C2 /N m2 
 0 
(b) For d  2.00 mm, VC  400 V.
Assess: Note that the units in part (a) are N m/C. But Exercise 29.17 showed that 1 N/C  1 V/m, so 1 N m/C  1 V.
We also see that the potential difference across a parallel-plate capacitor is directly proportional to the plate separation.
29.23. Model: The charge is a point charge.
Visualize: Please refer to Figure EX29.23.
Solve: (a) The electric potential of the point charge q is
V
 2.0 109 C  18.0 N m 2 /C
1 q
  9.0 109 N m 2 /C2  

4 0 r
r
r


For points A and B, r  0.010 m. Thus,
VA  VB 
18.0 N m2 /C
Nm
V
 1800
 1800   m  1.80 kV
0.010 m
C
m
For point C, r  0.020 m and VC  900 V.
(b) The potential differences are
VAB  VB  VA  1.80 kV 1.80 kV  0 V
VBC  VC  VB  0.90 kV  1.80 kV  0.90 kV
29.26. Model: The net potential is the sum of the potentials due to each charge.
Visualize: Please refer to Figure EX29.26.
Solve: The potential at the dot is
V
1 q1
1 q2
1 q3


4 0 r1 4 0 r2 4 0 r3
 2.0 109 C 2.0 109 C 2.0 109 C 
3
  9.0 109 N m2 /C2  


   1.4110 V
0.040
m
0.050
m
0.030
m


Assess: Potential is a scalar quantity, so we found the net potential by adding three scalar quantities.
29.29. Model: The net potential is the sum of the scalar potentials due to each charge.
Visualize:
Solve: Let the point on the x-axis where the electric potential is zero be at a distance x from the origin. At this point,
V1  V2  0 V. This means
 3.0 109 C 1.0 109 C 

1 


  0 V   x  3 x  4.0 cm  0 cm
4 0 
x
x

4.0
cm



Either  x  3 x  4.0 cm   0 cm, or  x  3 4.0 cm  x   0 cm. In the first case, x  6.0 cm and, in the second case, x

3 cm. That is, we have two points on the x-axis where the potential is zero.
29.37. Model: While the net potential is the sum of the potentials due to each charge, the net electric field is the
vector sum of the electric fields.
Visualize:
The charge Q1  20.0 nC is at the origin. The charge Q2  10.0 nC is 15.0 cm to the right of the charge Q1 on the xaxis.
Solve: (a) As the pictorial representation shows, the point P on the x-axis where the electric field is zero can only be
on the right side of the charge Q2, that is, at x  15.0 cm. At this point E1  E2 , so we have
1 20.0 109 C
1 10.0 10 9 C

 x2  2(x – 15.0 cm)2
2
4 0
x
4 0  x  15.0 cm 2
 x2  (60.0 cm)x 450 cm2  0  x 
 60.0 cm  
3600 cm2  1800 cm 2
2
 x  51.2 cm and 8.8 cm.
The root x  8.8 cm is not possible physically. So, the electric fields cancel out at x  51.2 cm. The electric potential at
this point is
V
 20.0 109 C
1 Q1
1 Q2
10.0 109 C 

  9.0 109 N m2 /C2  

  103 V
4 0 r1 4 0 r2
 0.512 m  0.150 m  
 0.512 m
(b) The point on the x-axis where the potential is zero can be obtained from the condition V1  V2  0 V, which is
1 Q1
1 Q2
20.0 109 C 10.0 109 C

0 V

0
4 0 r1 4 0 r2
x
15.0 cm  x 
 2(15.0 cm – x) – x  0  x  10.0 cm
The electric field 10.0 cm away from charge Q1 is
9
1 20.0 109 C ˆ
1 10.0 10 C  ˆ
Enet  E1  E2 
i

i  5.40 104iˆ N/C
4 0  0.100 m 2
4 0  0.050 m 2
29.42. Model: Energy is conserved. The potential energy is determined by the electric potential.
Visualize: Please refer to Figure P29.42.
Solve: The proton at point A is at a potential of 30 V and its speed is 50,000 m/s. At point B, the proton is at a
potential of 10 V and we are asked to find its speed. Clearly, the proton moves into a lower potential region, so its
speed will increase. The conservation of energy equation Kf  Uf  Ki  Ui is
1
2
 vf 
mvf2   e  10 V   12 m  50,000 m/s    e  30 V 
2
 50,000 m/s  
2
2 1.60 1019 C   40 V 
1.67 1027 kg
 1.01105 m/s
Assess: The speed of the proton is higher, as expected.
29.45. Model: Energy is conserved. The proton’s potential energy inside the capacitor can be found from the
capacitor’s potential difference.
Visualize: Please refer to Figure P29.45.
Solve: (a) The electric potential at the midpoint of the capacitor is 250 V. This is because the potential inside a
parallel-plate capacitor is V  Es where s is the distance from the negative electron. The proton has charge q  e and its
potential energy at a point where the capacitor’s potential is V is U  eV. The proton will gain potential energy
U  eV  e(250 V)  1.60  1019 C (250 V)  4.00  1017 J
if it moves all the way to the positive plate. This increase in potential energy comes at the expense of kinetic energy
which is
K  12 mv 2 
1
2
1.67 10
27
kg   200,000 m/s   3.34 1017 J
2
This available kinetic energy is not enough to provide for the increase in potential energy if the proton is to reach the
positive plate. Thus the proton does not reach the plate because K < U.
(b) The energy-conservation equation Kf  Uf  Ki  Ui is
1
2
 vf  vi2 
mvf2  qVf  12 mvi2  qVi  12 mvf2  12 mvi2  q Vi  Vf 
2q
Vi  Vf  
m
29.46. Model: Energy is conserved.
Visualize:
 2.0 105 m/s  
2
2 1.60 1019 C   250 V  0 V 
1.67 1027 kg
 2.96 105 m/s
Solve: (a) The electric field inside a parallel-plate capacitor is constant with strength
E
3
V  25 10 V 

 2.1106 V/m.
d
0.012 m
(b) Assuming the initial velocity is zero, energy conservation yields
Ui  Kf  U f
1
0  mevf 2   e  Ed 
2
 vf 
2 1.60 1019 C  2.1106 V/m   0.012 m 
9.111031 kg
 9.4 107 m/s
Assess: This speed is about 31% the speed of light. At that speed, relativity must be taken into account.
29.47. Model: Mechanical energy is conserved.
Visualize:
Solve: (a) Both gravitational and electric potential energy are involved. The electric potential between the plates is V
= Es, where s is measured from the more negative plate. The electric field between the plates is
E
V 3.0 106 V

 1.00 106 V/m.
d
3.0 m
Conservation of energy is used to find the height.
Ki  U gi  U ei  Kf  U gf  U ef
1
mvi 2  0 J  q  0 V   0 J  mgh  q  Eh 
2
1
1
mvi 2
1.0 103 kg  5.0 m/s 2
2
2
h

 0.85 m
mg  qE 1.0 103 kg  9.8 m/s 2    4.0 109 C 1.00 106 V/m 
(b) The top plate is at a negative potential, so the charge is attracted to it. The electric potential is now zero at the top
plate and positive at the bottom plate. Does the charge make it all the way to the top plate? If we assume it does not
and obtain a height more than 3.0 m then we will know that it does.
Conservation of energy for the charge requires
1
1
mvi 2  0 J  qE  3.0 m   mgh  qE  3.0 m  h   mvi 2   mg  qE 
2
2
1
mvi 2
2
h
 2.6 m
mg  qE
Assess: Since h < 3.0 m the charge did not hit the top plate.
29.56. Model: Energy is conserved.
Visualize:
The alpha particle is initially at rest (vi alpha  0 m/s) at the surface of the thorium nucleus. The potential energy of the
alpha particle is U i alpha . After the decay, the alpha particle is far away from the thorium nucleus, Uf alpha  0 J, and
moving with speed vf alpha .
Solve: Initially, the alpha particle has potential energy and no kinetic energy. As the alpha particle is detected in the
laboratory, the alpha particle has kinetic energy but no potential energy. Energy is conserved, so Kf alpha  Uf alpha 
Ki alpha  Ui alpha. This equation is
1
2
 vf alpha 
1
4 0
mvf2alpha  0 J  0 J 
1
 2e  90e 
4 0
ri
360e   9.0 10 N m /C  360 1.60 10 C 
mr
4 1.67 10 kg  7.5 10 m 
2
9
2
19
2
27
15
2
 4.1107 m/s
i
29.78. Model: Energy and momentum are conserved.
Visualize: Please refer to Figure CP29.78. Let the two spheres have masses mA and mB, and speeds vA and vB when
they are very far apart.
Solve: The energy-conservation equation Kf  Uf  Ki  Ui is

1
2
mA vA2  12 mBvB2   0 J  0 J 
1 qA qB
4 0 rAB
 12  0.002 kg  vA2  12  0.001 kg  vB2  9.0 109 N m 2/C 2 
 2.0 10
9
C 1.0 109 C 
2.0 103 m
 vA2  0.5 vB2  0.0090 m 2 /s 2
To solve for vA and vB, we need another equation relating vA and vB. From the momentum conservation equation Pafter 
Pbefore we get
mAvA  mBvB  0 kg m/s  (2.0 g)vA  (1.0 g)vB  0 kg m/s  vB  2vA
Substituting this expression into the energy-conservation equation,
vA2  0.5 2vA   0.009 m2 /s2  3vA2  0.009 m 2 /s 2
2
Solving these equations, we get vA  –0.0548 m/s and vB  0.110 m/s. Thus the speeds are 0.055 m/s for A and 0.110
m/s for B.