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tiL12214_appe_643-698.indd Page 659 09/10/10 8:37 AM user-f463
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v _
charge
5.19. (a) v = f λ ∴ λ = _
electric

f 6.4. current = time
ft _ or
1,125
_
s
= 1 _ q _

440 I =
t
s
6.00 C
1,125 ft _ _ s = _
_

×
=
440 s 1 2.00 s
C _
_ = 3.00
ft·s
= 2.56 s
s
= 2.6 ft = 3.00 A
v _
(b) v = f λ ∴ λ = 6.5. A current of 1.00 amp is defined as 1.00 coulomb/s. Since the

f –19
fundamental charge of the electron is 1.60 × 10 C/electron,
5,020 ft _ _ s
_
×
= C _
1.00
440 s 1 __
s

–19 _
= 11.4 ft = 11 ft 1.60 × 10 C

electron
18 C _ _
CHAPTER 6 = 6.25 × 10 × electron

s C/
6.1. First, recall that a negative charge means an excess of electrons. 18 electrons
_

Second, the relationship between the total charge (q), the number = 6.25 × 10 s
of electrons (n), and the charge of a single electron (e) is q = ne. V _
6.6. R =

The fundamental charge of a single (n = 1) electron (e) is I
–19
1.60 × 10 C. Th us, _
120.0 V
=
q
_ 4.00 A

q = ne ∴ n =
e
V _

–14
__ = 30.0
1.00 × 10 C

= C A
–19 _
1.60 × 10 = 30.0 Ω

electron
–14
V _
1.00 × 10 _ _
__ C electron R = ∴ V _

= × 6.7. I =
1.60 × 10 –19 1 C I R
_
120.0 V
4 C ⋅ electron _

= 6.25 × 10 =
V _
C 60.0

4
= 6.25 × 10 electrons A
_ A _
120.0

= V ×
6.2. (a) Both balloons have negative charges so the force is 60.0 V
repulsive, pushing the balloons away from each other.
= 2.00 A
(b) The magnitude of the force can be found from Coulomb’s

V _
law: 6.8. (a) R = ∴ V = IR

I
V _
(
)
_ = (1.20 A) 10.0

kq 1 q 2
F = A
d 2
–12
2
9
2
–14
(9.00 × 10 N·m /C )(3.00 × 10
_____ = 12.0 V
C)(2.00 × 10
C)

=
–2
(2.00 × 10 m) 2 (b) Power = (current)(potential diff erence)
_ 2 × C × C or
N·m

–12

9
–14
(9.00 × 10 )(3.00 × 10 )(2.00 × 10 )
__
____ C 2 P = IV

=
–4
4.00 × 10 m 2 C _ _
s (
J
)

= (1.20 ) 12.0
–16
5.40 × 10 _ 2 2 _ C
1
__ N·m

= × C ×

/

J
4.00 × 10 –4 C / 2 m 2 C _ _
= (1.20)(12.0) ×

s
–12
= 1.35 × 10 N C
_ _ J
potential
work

6.3. = 14.4

=
s
diff erence charge
or = 14.4 W
W _
V = 6.9. Note that there are two separate electrical units that are rates:

q
(1) the amp (coulomb/s) and (2) the watt (joule/s). Th e
_
7.50 J
= question asked for a rate of using energy. Energy is measured in
5.00 C
_ joules, so you are looking for the power of the radio in watts. To
J
= 1.50 find watts (P = IV), you will need to calculate the current (I)

C
= 1.50 V
E-17 APPENDIX E Solutions for Group A Parallel Exercises 659