Download Physics I - Rose

Physics II
Homework IX
CJ
Chapter 29; 3, 6, 10, 13, 20, 25, 30, 38, 59, 71
29.3. 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.
Solve: The proton loses potential energy and gains kinetic energy as it moves toward the negative plate. The potential
energy 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  K i  21 mvf2  21 mvi2  21 mvf2
Applying the law of conservation of energy K  Up  0 J, we have
1
2
mvf2   qEd   0 J  vf2 
2qEd
m
When the amount of charge on each plate is doubled, then the final velocity of the proton is
vf2 
2qEd
m
Dividing these equations,
vf2 E
 vf 

E
vf2
E
vf
E
For a parallel-plate capacitor E    0  Q A 0 . Therefore,
Q
vf  2  50,000 m/s   7.07  10 4 m/s
Q
vf 
Assess: The proton’s velocity is expected to increase because an increased charge on the capacitor plates leads to a higher
electric field between the plates and hence to an increased force on the proton.
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:
Uelec  
i,j
Kqi q j
rij
 U12  U13  U23  K
qq
qq
q1q2
K 1 3 K 2 3
r12
r13
r23
  2 nC  1.0 nC   2 nC  1.0 nC   1 nC  1.0 nC  
 9.0  109 N m2 /C2 



0.03 m
0.03 m
0.03 m




 0.60  10 6 J  0.60  10 6 J  0.30  10 6 J=  0.90  10 6 J
Assess: Note that U12  U21, U13  U31, and U23  U32.
29.10. Model: An external electric field supplies energy to a dipole.
Visualize: On an energy diagram, the oscillation occurs between the points where the potential-energy curve
crosses the total energy line.
Solve: (a) The potential energy of an electric dipole moment in a uniform electric field is given by Equation 29.23:
U   p  E   pE cos . This means
U 0   pE  2  J
U 60   pE cos60   21 pE   21  2J   1 J
The mechanical energy Emech  U  K. We know that at   60°, K  60  0 J. So,
Emech  U 60  K 60  1  J  0 J   1  J
(b) Conservation of mechanical energy gives
U 60  K 60  U 0  K 0  1  J  0 J  2  J  K 0  K 0  1  J
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 a He ion moving through a potential difference. The ion’s
initial speed is zero and its final speed is 1.0  106 m/s. A positive charge speeds up as it moves into a region of lower
potential (U  K).
Solve:
potential V, is
The potential energy of charge q is U  qV. Conservation of energy, expressed in terms of the electric
Kf  qVf  Ki  qVi  q(Vf – Vi)  Ki  Kf


4 1.67  1027 kg 1.0  106 m/s
K  Kf 0 J  21 mvf2
 V  i


q
q
2 1.60  1019 C



2
 2.09  104 V
Assess: This result implies that the helium ion moves from a higher potential toward a lower potential.
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



0.708  10 9 C 1.0  10 3 m
 
 Q A d  Qd 
VC  Ed    d 
 200 V
0
A 0
4.0  10 4 m2 8.85  10 12 C2 /N m2
 0 



(b) For d  2.0 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.
29.25. Model: Outside a charged sphere the electric potential is identical to that of a point charge at the center.
Visualize:
Solve:
For r  R, V  Q 4 0r . The potentials at the two points are
V2 mm 
 9.0  10
9

N m2 /C2 Q
1 mm  2 mm 

9.0  10
 V2 mm  V4 mm  500 V 
Q

9

N m2 /C2 Q
3
3.0  10 m
 9.0  10
9
V4 mm 
9.0  10
9

N m2 /C2 Q
3
5.0  10 m
1
1


N m2 /C2 Q 


3
3
 3.0  10 m 5.0  10 m 

 15.0  10 6 m2 
10

  4.17  10 C
3
9.0  109 N m2 /C2  2.0  10 m 
500 V

Assess: Do not forget to include the radius of the glass bead in r.
29.30. Model: The net potential is the sum of the potentials due to each charge.
Visualize: Please refer to Figure Ex29.30.
Solve: (a) Let q1  5 nC, q2  10 nC, and q3  10 nC. Also, r1  2 cm, r2  4 cm, and r3 
 2 cm    4 cm 
2
2
 4.47 cm . Each individual potential is simply that of a point charge, so
q3
q1
q2
1  q1 q2 q3 



 
 
4 0 r1 4 0 r2 4 0 r3 4 0  r1 r2 r3 
 5  10 9 C 10  10 9 C 10  10 9 C 
 9.0  109 N m2 /C2 


  2010 V
0.04 m
0.0447 m 
 0.02 m
VB  V1  V2  V3 


(b) The potential energy of an electron at point B is
Uelectron  qelectronVB  eVB  (1.6  1019 C)(2010 V)  3.22  1016 J
Assess: Note that the units in part (a) are N m/C. But Problem 29.17 showed that 1 N/C  1 V/m, so 1 N m/C  1 V.
29.38. Model: The charged beads are point charges.
Visualize:
Solve: Once the beads are released they will accelerate according to the force acting on them. The magnitude of
the force between the beads is given by Coulomb’s law:
FAB 
1
QA QB
4 0
2
rAB

 9.0  10 N m /C
9
2
2
 5.0  10


9
C 10.0  10 9 C
 0.12 m 
2
  3.125  10
5
N
The same force acts on bead A and bead B. Since F  ma,
aA 
3.125  105 N
 2.08  103 m/s2
0.015 kg
aB 
3.125  103 N
 1.25  103 m/s2
0.025 kg
where aA is directed toward the left and aB, is directed toward the right. To find the maximum speeds vfA and vfB it is easier to
use the conservation of momentum and conservation of energy equations than kinematics. The momentum conservation
equation along x-direction pafter  pbefore  0 kg m/s means
mAvfA  mBvfB  0 kg m/s  (0.015 kg)vfA  (0.025 kg)vfB  0 kg m/s  vfB  53 vfA
The conservation of energy conservation is Uf  Kf  Ui  Ki. Noting that Uf  0 J as the masses are infinitely separated and




9.0  109 N m2 /C2 5.0  10 9 C 10.0  10 9 C
1 QA QB
Ui 

 3.75  10 6 J
4 0 rAB
0.12 m
0J 

1
2

1
2

2
2
mA vfA
 21 mB vfB
 3.75  10 6 J  0 J
 0.015 kg  vfA2  21  0.025 kg   53 vfA 
2
2
 3.75  10 6 J  0.012vfA
 3.75  10 6 J
vfB 
Solving these equations yields vfA  1.77 cm/s and
3
5
1.77 cm/s  1.06 cm/s .
These are the maximum
speeds of the two beads and occur when they are infinitely separated from each other.
29.59. Solve: (a) Outside a uniformly charged sphere, the electric field is identical to that of a point charge Q at the
center. For r  R,
E
1
Q
4 0 r 2
Evaluating this at r  R and using Equation 29.36,
E0 
Q V0

4 0 R2
R
1
(b) The field strength is
E0 
500 V
 100,000 V/m
0.5  102 m
29.71. Model: Because the rod is thin, assume the charge lies along the semicircle of radius R.
Visualize: Please refer to Figure P29.71. The bent rod lies in the xy-plane with point P as the center of the
semicircle. Divide the semicircle into N small segments of length s and of charge Q   Q  R s , each of which can be
modeled as a point charge. The potential V at P is the sum of the potentials due to each segment of charge.
Solve:
The total potential is
V  Vi  
i
i
1 Q
1
1 Q
1 Q
 Q 

s 
R 

  .


2 
4 0 R
4 0 R i   R 
4 0  R i
4 0  R i
All of the terms come to the front of the summation because these quantities did not change as far as the summation is
concerned. The summation does not have to convert to an integral because the sum of all the  around the semicircle is .
Hence, the potential at the center of a charged semicircle is
Vcenter 
1 Q
1 Q

4 0  R 4 0 R