zhou (ejz224) – Homework 4 – shih – (57480)
1
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001
10.0points
The current
I
=
a t
2

b t
+
c
in a section of a conductor depends on time.
What quantity of charge moves across the
section of the conductor from
t
= 0 s to
t
=
t
1
?
1.
q
= 2
a t
2
1

b t
1
2.
q
=
a
3
t
3
1

b
2
t
2
1
+
c t
1
correct
3.
q
= 2
a t
1

b
4.
q
=
a t
3
1

b t
2
+
c t
1
5.
q
= 2
a t
2
1

3
b t
1
+
c t
1
6.
q
= 3
a t
2
1

b
+
c
7.
q
=
a t
3
1

b
2
t
2
1
+
ct
1
8.
q
=
a
3
t
3
1

b
2
t
2
1
+
c
9.
q
= 3
a t
2
1

2
b t
1
+
c
10.
q
=
a t
2
1

b t
1
+
c
Explanation:
The unit of current is Coulomb per second
I
=
d q
dt
or
dq
=
I dt
.
To find the total charge in coulombs that
passes through the conductor, one must inte
grate the current over the time interval.
q
=
integraldisplay
t
1
0
dq
=
integraldisplay
t
1
0
I dt
=
integraldisplay
t
1
0
(
a t
2

b t
+
c
)
dt
=
bracketleftbigg
a
3
t
3
1

b
2
t
2
1
+
c t
bracketrightbigg vextendsingle
vextendsingle
vextendsingle
vextendsingle
t
1
0
=
a
3
t
3
1

b
2
t
2
1
+
c t
1
.
002
10.0points
In a fluorescent tube of diameter 4
.
6 cm, 1
.
3
×
10
18
1
/
s electrons and 2
.
3
×
10
18
1
/
s positive
ions (with a charge of +
e
) flow through a
crosssectional each second.
What is the current in the tube?
Correct answer: 0
.
576 A.
Explanation:
Let :
n
electron
= 1
.
3
×
10
18
1
/
s
,
n
ion
= 2
.
3
×
10
18
1
/
s
,
q
electron
= 1
.
6
×
10
−
19
C
,
and
q
ion
= 1
.
6
×
10
−
19
C
.
Note that, while the positive and negative
charges flow in opposite directions, the total
current is their sum:
I
=
I
electron
+
I
ion
=
n
electron
q
electron
+
n
ion
q
ion
= (1
.
3
×
10
18
1
/
s) (1
.
6
×
10
−
19
C)
+ (2
.
3
×
10
18
1
/
s) (1
.
6
×
10
−
19
C)
= 0
.
208 A + 0
.
368 A
=
0
.
576 A
.
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 Spring '08
 morrison
 Physics, Current, Work, Electric charge, Electrical resistance

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