Download Chapter 19: Electric Charges, Forces, and Fields

Ch. 20 Solution
PHY 205
Hayel Shehadeh
Chapter 19: Electric Charges, Forces, and Fields
Answers Conceptual Questions
4. The two like charges, if released, will move away from one another to
infinite separation, converting the positive electric potential energy into
kinetic energy. The two unlike charges, however, attract one another – if
their separation is to be increased, a positive work must be done. In fact, the
minimum amount of work that must be done to create an infinite separation
between the charges is equal to the magnitude of the original negative
electric potential energy.
6.
The electric field is either zero on this surface, or it is nonzero and
perpendicular to the surface. If the electric field had a nonzero component
parallel to the surface, the electric potential would decrease as one moved in
the direction of the parallel component.
10. Capacitors store electrical energy in the form of an electric field between
their plates. This stored energy can remain in a capacitor even when a device
such as a television is turned off. Accidentally touching the terminals can
cause the sudden and potentially dangerous release of this energy.
Solutions to Problems and Conceptual Exercises

5
2. A 4.5 µC charge moves in a uniform electric field E   4.110 N/C xˆ .
The change in electric potential energy of a charge that moves against an electric field
is given by equation 20-1, U  q0 Ed . If the charge moves in the same direction as
the field, the work done by the field is positive and U  W  Fd  q0 Ed . If the
charge moves in a direction perpendicular to the field, the work and change in
potential energy are both zero. The change in potential energy from point A to point
B is independent of the path the charge takes between those two points.
Solution: 1. (a) The
charge moves
perpendicular to the field:
U  0
2. (b) The charge moves
in the
same direction as the
field:
U  q0 Ed    4.5 10 6 C 4.1105 N/C  6.0m   11 J
1
Ch. 20 Solution
PHY 205
3. (c) The potential
energy changes by an
amount equal to the sum
of paths 1 and 2 above:
Hayel Shehadeh
U  0  11 J  11 J
Note that U changes only along the direction parallel to the field. This is consistent
with statement, “The electric field points downhill on the potential surface,” as
described by figure 20-3 and the accompanying text.
8. An ion loses electric potential energy as it is accelerated through a potential
difference.
The electric potential is the electric potential energy per charge (equation 20-2). To
find the charge we need only divide the electric potential energy by the electric
potential.
Solution: Solve equation 20-2
for q0 :
q0 
U 1.37 1015 J

  6.40 1019 C   4.00 e
V
2140 V
If the charge were positive, its electric potential energy would increase as it moves
from a lower to a higher potential. In this problem the negative charge loses electric
potential energy and gains kinetic energy as it accelerates.

12 A uniform electric field E   1200 N/C xˆ
creates a change in the electric potential along
the x̂ direction.
The change in electric potential is given by
equation 20-4, V  Es , where s is equal to
x because the field points in the x̂ direction.
There is no change in the electric potential
along the yˆ and zˆ directions. The change in
electric potential between any two points is
independent of the path taken between those
two points.
Solution: 1. (a) x  0 between points A
and B:
V  VB  VA   E x  0
2. (b) Solve equation 20-4 for V
between points B and C:
V  VB  VC   E x   E  xB  xC 
3. (c) Repeat for points C and A:
   1200 N/C  0.040 m   48 V
V  VC  VA   E x   E  xC  xA 
   1200 N/C  0.040 m   48 V
4. (d) No, we cannot determine the value of the electric potential at point A because
the location of zero potential has not been established. We could always choose
2
Ch. 20 Solution
PHY 205
Hayel Shehadeh
VA  0 if we wish!
Note that the potential changes only along the direction parallel to the field and is
higher at point C than at either point A or point B. This is consistent with statement,
“The electric field points downhill on the potential surface,” as described by figure
20-3 and the accompanying text.
16. The kinetic energy of an electron changes as it moves to locations of different electric
potential.
As the electron moves from place to place, its kinetic energy increases as it moves
toward higher potential (remember, it is negatively charged, so it is accelerated
opposite the field and toward positive charges) and decreases as it moves toward
lower potential. Use the conservation of energy to determine the potential at point C
and the kinetic energy at point A.
0  U A  KA  U C
Solution: 1. (a) Set Ki  Ui  Kf  Uf for
the first case:
0  U B  2KA  U C
2. Now set Ki  Ui  Kf  Uf for the
second case:
3. Solve the first equation for
U B  2 U A  U C   U C
KA  U A  U C
U C  2U A  U B
and substitute into the second, and
solve for U C :
 eVC  2   eVA     eVB 
4. Divide both sides by e to find VC :
VC  2VA  VB  2  332 V   149 V  515 V
5. (b) Substitute the result from part (a)
into the expression from step 1 to find
K A  U A  U C   e VA  VC 
  e  332  515 V   183 eV
KA :
In the first case the electron moves from 332 V (point A) to 515 V (point C), gaining
183 eV of kinetic energy, but in the second case the electron moves from 149 V (point
B) to 515 V (point C), gaining 366 eV of kinetic energy.
22. The speed of a proton is changed as it moves through a potential difference.
Use conservation of energy, with the potential energy of the proton given by
U  qV  eV , to determine the potential difference required to change the proton’s
speed and kinetic energy.
1
2
Solution: 1. (a) Set
Ki  Ui  Kf  Uf
and solve for V :
1
2
mvi2  eVi  12 mvf2  eVf
mvi2  12 mvf2  e Vf  Vi   eV
V 
m  vi 2  vf 2 
2e
1.673 10

27
kg  4.0  105 m/s   0
2
2 1.60 1019 C 
 840 V  0.84 kV
3
Ch. 20 Solution
2. (b) Find V for
when vf  12 vi :
PHY 205
V 

3. (c) Set
Ki  Ui  12 Ki  U f
and solve for
V :
Hayel Shehadeh
m  2 1 2  mvi2  1  3mvi2
vi   2 vi  
1

 2e  4 
2e 
8e
3 1.673 1027 kg  4.0 105 m/s 
8 1.60 1019 C 
2
 630 V  0.63 kV
Ki  eVi  12 Ki  eVf
1
2
Ki  eV
27
5
K
mv 2 1.673 10 kg  4.0 10 m/s 
V  i  i 
 0.42 kV
2e
4e
4 1.60 1019 C 
2
It requires a smaller potential difference to cut the kinetic energy in half than it does
to cut the speed in half because the kinetic energy is proportional to the speed
squared; cutting the speed in half reduces K by a factor of four.
30. Four different arrangements of point
charges are shown at right. In each case
the charges are the same distance from
the origin.
The electric potential at the origin is the
sum of the potentials from each
individual charge. The electric potential
at the origin is proportional to the sum
of the charges, because each charge is
the same distance from the origin. Use
this observation to determine the
ranking of the electric potentials at the
origin.
Solution: The sum of the charges in the
four cases are: A, 0; B, 0; C, −3q; and
D, +q. Therefore, the ranking of electric
potential at the origin is as follows: C <
A = B < D.
The electric potential a long distance
from the origin also follows the same
ranking, because from a long distance
the charges appear as a single charge
with a magnitude equal to the sum of
the charges.
4
Ch. 20 Solution
PHY 205
Hayel Shehadeh
37. Three charges are arranged at the corners of a
rectangle as indicated in the diagram at right.
The work required to move the 2.7 µC charge to
infinity is equal to the change in its electric
potential energy due to the other two charges.
Sum the electric potential energies due to the
other two charges to find the electric potential
energy at the corner of the rectangle. At infinity
the electric potential energies are zero. Use these
facts to find the work required to move the 2.7 µC
charge to infinity, and then repeat the procedure to
find the work required to move the − 6.1 µC
charge.
Solution: 1. (a) Find
U  U f  Ui  0  U1  U 2  :
  kq q kq q  
q q 
W  U  0   2 1  2 3    kq2  1  3 
r
r
23
 
 r21 r23 
  21
  6.1 10 6 C

N  m2 
3.3  10 6 C
6
   8.99  109

  2.7  10 C  
2
C 
 0.25 m
(0.25 m) 2  (0.16 m) 2




W  0.86 J
2. (b) The  6.1 C charge is repelled by the 3.3 C charge more than it is attracted
by the 2.7 C charge. The work required will be negative, which is less than the work
required in part (a).
3. (c) Repeat step 1 for
the − 6.1 µC charge:
  kq q kq q  
q q 
W  U  0   1 2  1 3    kq1  2  3 
r13  
  r12
 r12 r13 
 2.7  10 6 C 3.3 10 6 C 
   8.99  109 N  m 2 C2   6.110 6 C  


0.16 m 
 0.25 m
W   0.54 J
Note that the work done by the electric field is W  U (equation 20-1) but the work
you must do to move the charge against the field is W  U . The negative work of
part (c) means the charge would fly off on its own.
47. The figure at right shows a series of
equipotentials in a particular region of space,
and five different paths along which an
electron is moved.
Equipotential lines always intersect electric
field lines at right angles. Use this principle
together with the concepts of electric field
intensity and work done by an electric field to
answer the questions.
Solution: 1. (a) The value of the electric
potential decreases as we move in the direction
of the electric field (see, for example, figure
5
Ch. 20 Solution
PHY 205
Hayel Shehadeh
20-3.) This means that the electric field must
point to the left, corresponding to a
decrease in the electric potential from 60 V to –5 V.
2. (b) Recall that the force exerted on an electron is opposite in direction to the
direction of the electric field; therefore, the electric field does positive work on an
electron that moves opposite in direction to the electric field, because the force and
the direction of motion are the same. This means that any path in the diagram that
moves the electron toward the right corresponds to positive work done on the electron
by the electric field. It follows that the works done along the paths in the figure have
the following signs: A, positive; B, positive; C, positive; D, negative; E, negative.
3. (c) The work done by the electric field is proportional to the difference in electric
potential between the starting and ending points of the path. In fact, equation 20-2
says that W   q0 V    e  V  e V for these electrons. The V for each case is A,
+5 V; B, +40 V; C, +35 V; D, −15 V; E, −25 V. The ranking is thus: E < D < A < C <
B.
4. (d) The magnitude of the electric field is proportional to the magnitude of the
potential difference divided by the distance. The distance for paths A and E is the
same, but the magnitude of the potential difference is much greater for path E than for
path A. It follows that the electric field near path A is less than the electric field near
path E.
The work done by the electric field is independent of the particular path because the
electric force is a conservative force. The work depends only upon the endpoints of
the path.
58. A capacitor is formed with two parallel metal plates separated by a small distance.
Equation 20-12 gives the relationship between the capacitance of a parallel-plate
capacitor, the plate area A, and the plate separation d. Substitute this expression into
equation 20-9, C  Q V , and solve for Q. The air between the plates has a dielectric
constant (1.00059) that is close enough to one that we can approximate it as a
vacuum.
Solution: 1. (a)
Solve equation 20-9
for Q and substitute
equation 20-12 for
C:
Q  CV 
 0 AV
d

8.85 10
12
C2
N  m2
  0.0066 m  12 V   1.6 nC
2
0.45 103 m
2. (b) Because Q is inversely proportional to d, the answer to part (a) will decrease if
d is increased.
3. (c) Repeat part (a) with Q 
the new d:

1.00059 8.85 1012
C2
N  m2
 (0.0066 m )(12 V) 
0.90 103 m
2
7.8 1010 C
6
Ch. 20 Solution
PHY 205
Hayel Shehadeh
The battery maintains a constant potential difference V between the plates, so that E =
V / d must decrease if d is increased. In order to do that, the charge density  and
therefore charge Q must decrease ( E    0 , see active example 19-3).
82. Three charges are arranged as shown at right.
y
(− 4.50 m, 6.75 m) 8
The electric potential at the origin is the sum of
the potentials from each of the three charges.
Sum the three potentials and set the sum equal to
zero, then solve the resulting expression for q3 .
Solution: 1. Sum the
potentials:
q2
(4.40 m, 6.22 m)
4
2
−8 −6 −4 −2 O
q q q 
V  0  k 1  2  3 
 r1 r2 r3 
−4
2
q3
x
4 6 8
(2.23 m, −3.31 m)
−6
2. Solve for q3 :
q q 
q3  r3  1  2 
 r1 r2 
3. Find r1 , r2 , and r3 :
r1 
 4.40 m    6.22 m 
r3 
 2.23 m    3.31 m 
3. Insert the numerical
values:
q1
6
2
2
2
 7.62 m, r2 
2
 4.50 m    6.75 m 
2
2
 3.99 m
q q 
 24.5  C 11.2  C 
q3  r3  1  2     3.99 m  

  7.32  C
8.11 m 
 7.62 m
 r1 r2 
A third charge of relatively small magnitude is required because that charge is closer
to the origin than the other two charges.
7
 8.11 m