BW#2 Ch. 22: CQ7-9 P. 13, 14, 16, 23, 26, 30, 32, 34.
Q22.7. Reason: The wire with the greater resistivity will have a greater resistance ( R   L/A). The wire with the greater
resistance will require a greater electric potential difference to maintain the same current as the wire with less resistance
(V  IR). This larger electric potential difference will result in a larger electric field ( E  V /d ). This allows us to
conclude that the wire with the greater resistivity will require a greater electric field in order to sustain the same current as
the wire with the smaller resistivity. Since the field required for metal 1 is greater, the resistivity of metal 1 is greater.
Assess: In the electricity chapters it is important to know the basic concepts and how they are related.
Q22.8. Reason: The emf and potential difference created by a source of electricity are related by   V  Ir. In this
case it is the 70 mV potential difference is V of the source and is not an emf.
Assess: The output of the ion pumps is the equivalent of the terminal potential difference for a battery.
Q22.9. Reason: (a) The electric potential difference (3 V as produced by the battery) establishes an electric field from
the point of higher electric potential ( terminal of the battery at point 2) to the point of lower electric potential (– terminal
of the battery at point 3). The electrons travel in a direction opposite the direction of the electric field, hence they will
travel counterclockwise through the wire.
(b) The electron’s electric potential energy decreases as it moves through the wire. It must use some of its energy to
“work” its way through the wire and overcome the “atomic-level friction.”
(c) The electric potential energy of the electron decreases because it is working to make its way through the wire. As the
electron “works” its way through the wire, numerous collisions (these are long range collisions in that the electrons don’t
actually bounce off each other as billiard balls would) are occurring at the atomic level. This in effect acts as resistance to
the electron’s movement through the wire. We often refer to this in a global manner as “atomic-level friction.”
(d) We established in part (a) that the electrons travel through the external circuit from the negative terminal to the
positive terminal. As a result, in order to get back to the negative terminal (in order to make another trip), they must go
through the battery from the positive terminal to the negative terminal.
(e) The electron’s electric potential energy increases as it moves through the battery. The chemical reactions occurring in
the battery provide the energy needed to move the electron internally from the positive to the negative terminal. Since
work has been done on the electron (at the expense of chemical energy), its electric potential energy has increased.
(f ) The increase in the electron’s electric potential energy comes at the expense of chemical energy.
Assess: It is important to understand not only each concept dealt with in this question, but also the connections between
the concepts.
P22.13. Prepare: A battery is a charge escalator. When a wire is connected to a battery, there is a sustained motion of
positive ions. A current of 1.5 A means that a charge of Q  1.5 C flows per second through a cross section of the wire.
Because Q  Nee, the number of ions transported per second can thus be calculated.
Solve: The number of ions transported per second is
Q
1.5 C

 9.4  1018 ions
e 1.60 1019 C
The ions generally aren’t traveling very fast, but a lot of them go by to make a current of 1.5 A.
Ne 
Assess:
P22.14. Prepare: The work done in moving a positive charge from the negative terminal to the positive terminal is
exactly equal to the increase in the potential energy of the charge. We will use Equation 22.4.
Solve:
W  U  qV  q(Vf  Vi )  (1.0 106 C)(1.5 V)  1.5 106 J
Assess: The work done by the escalator on the charge is stored as electric potential energy of the charge.
P22.16. Prepare: Charge, electric current, and time are related by I  Q/t. Work, electric potential difference, and
charge are related by W  qV.
Solve:
The charge transferred is q  I t  (2.5 103 A)(5.0 hr)(3.6 103 s/hr)  45 C
The work done on these charges by the battery is W  qV  (45 C)(9.0 V)  4.1102 J.
Assess: These values are reasonable for this case.
P22.23. Prepare: From Table 22.1, copper and iron have resistivities of 1.7 108   m and 9.7 108   m. We will use
Equation 22.8, which connects resistance with resistivity.
Solve: (a) The resistance is
R
L

A
 L (1.7 108   m)(1 m)

 0.087 
 r2
 (2.5 104 m)2
(b) The resistance is
R
L
A

L
d2

(9.7  108   m)(0.1 m)
 0.0097 
(0.001 m)2
Assess: Small wires or pieces of metal such as copper and iron have low resistance as obtained previously.
P22.26. Prepare: The current I in a wire when a potential difference is applied to the ends of the wire can be obtained
from Equation 22.6, I  V /R, and Equation 22.8, R   L/A. The resistivity of nichrome from Table 22.1 is 1.5 106   m.
Solve:
 A  (3.0 V) (0.40 103 m) 2
I  V 
 2.0 A

6
  L  (1.5  10   m)(0.50 m)
Assess: The resistivity of nichrome is small, so a current of 2.0 A is reasonable.
P22.30. Prepare: The slope of the I versus V graph, according to Ohm’s law, is the reciprocal of the resistance R. The
resistance is directly proportional to the length of the resistor according to Equation 22.8.
Solve: (a) From the graph in Figure P22.30, the reciproical of the slope and hence the resistance of the resistor is
V
1
10 V
R

 2.0 
I
Slope 5.0 A
(b) Doubling the length of the wire will double the resistance, which means the slope will decrease. The current-versuspotential-difference graph is shown.
Assess: Since the resistance is the reciprocal of the slope, a smaller slope means a higher resistance.
P22.32. Prepare: We’ll assume the filament is ohmic so we can use Ohm’s law. Equation 22.8 will help us find the
resistance, and we need to recall E  V /d .
Solve: (a) E  V /d  120 V/(060 m)  200 V/m.
(b) All else being equal, the new field strength would be E  V /d   120 V/(120 m)  100 V/m.
(c) Assuming the filament is ohmic, the original current would be
I
V 120 V

 050 A
R
240 
We are reminded that the current is proportional to the electric field, so halving the electric field (from 200 V/m to 100
V/m) means we also halve the current from 0.50 A to 0.25 A.
(d) We can use Ohm’s law with the new current to calculate the new resistance.
R 
V 120 V

 480 
I
025 A
Assess: Another way to see the new resistance: Since R   L/A, then doubling the length doubles the resistance, and
R  2  240   480 . This method is consistent with the other one.
The power dissipated (the brightness of the bulb) will be reduced correspondingly because P  (V )2 /R. V is still 120 V
but R has doubled.
P22.34. Prepare: The electric field strength is related to the electric potential difference and the length of the wire by
E  V / L.
V
18V

 12 V/m
L (40ft)(1m / 3.281ft)
Assess: Given the electric potential difference and length of the wire, this is a reasonable value for the electric field
strength.
Solve: The electric field strength is E 