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C
HAPTER
16
Quick Quizzes
1.
(b). The field exerts a force on the electron, causing it to accelerate in the direction opposite
to that of the field. In this process, electrical potential energy is converted into kinetic
energy of the electron. Note that the electron moves to a region of higher potential, but
because the electron has negative charge this corresponds to a decrease in the potential
energy of the electron.
2.
Either (a) or
(b), but not both. The absence of any electrical charges within a finite distance
from the point would produce an electric potential of zero at the point. Thus, (a) could be
a true statement. If electrical charges exist at finite distances from the point, then (b) must
be true. Both positive and negative charges must be present in the vicinity so their
contributions to the electrical potential at the observation point may cancel each other.
3.
(a) and (b). Both the electric potential and the magnitude of the electric field decrease as
the distance from the charged particle increases.
4.
(c). The battery moves negative charge from one plate and puts it on the other. The first
plate is left with excess positive charge whose magnitude equals that of the negative
charge moved to the other plate.
5.
(a)
C
decreases.
(b)
Q
stays the same.
(c)
E
stays the same
(d)
∆
V
increases.
(e)
The energy stored increases.
Because the capacitor is removed from the battery, charges on the plates have nowhere to
go. Thus, the charge on the capacitor plates remains the same as the plates are pulled
apart. Because
00
QA
E
σ
==
∈∈
, the electric field is constant as the plates are separated.
Because
∆
V
=
Ed
and
E
does not change,
∆
V
increases as
d
increases. Because the same
charge is stored at a higher potential difference, the capacitance has decreased. Because
2
Energy stored
=2
QC
and
Q
stays the same while
C
decreases, the energy stored
increases.
The extra energy must have been transferred from somewhere, so work was
done. This is consistent with the fact that the plates attract one another, so work must be
done to pull them apart.
6.
(a)
C
increases.
(b)
Q
increases.
(c)
E
stays the same.
(d)
∆
V
remains the same.
(e)
The energy stored increases.
31
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View Full Document CHAPTER 16
Problem Solutions
16.1
(a) The work done is
( )
cos
WF
s
q
Es
θ
=⋅
=
⋅
, or
( )
( )( )
19
2
1.60
10
C2
0
0
NC2
.
00
mc
o
s
0
W
−−
=×
×
°
=
6.40
J
−
×
(b) The change in the electrical potential energy is
e
PE
W
∆=
−=
J
−
−×
(c) The change in the electrical potential is
19
J
C
e
PE
V
q
−
∆
=
=
×
4.00 V
−
16.2
(a) We follow the path from (0,0) to (20 cm,0) to (20 cm,50 cm). The work done on the
charge by the field is
( ) ( )
()
() ()
[]
( )
12
1 1
2 2
64
0.20
o
s
0
0
.50
o
s
9
0
C
250 V m
m06
.
01
0 J
WWW q
q
qE
θθ
=+= ⋅
+ ⋅
=°
+
°
+=×
Thus,
e
PE
W
4
6.
−
(b)
4
6
50 J C
10 C
e
PE
V
q
−
∆
=
=
−
=
×
50 V
−
16.3
The work done by the agent moving the charge out of the cell is
( )
input
fiel
de
WW
P
E
q
=−
=− −∆
=+
∆
V
( )
3
J
C9
0
1
0
C
⎛⎞
+
×
=
20
1.
41
0
J
−
×
⎜⎟
⎝⎠
16.4
e
KE
W
PE
q V
∆==
−
∆ =∆
, so
17
7.37
J
115 J C
q
V
−
∆×
=
==
∆
6.41
C
−
×
32
16.5
2
25 000 J C
51
0 m
V
E
d
−
∆
=
×
6
71
NC
×
CHAPTER 16
16.6
Since potential difference is work per unit charge
W
V
q
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This note was uploaded on 03/21/2008 for the course PY 211 / 212 taught by Professor Chilton during the Spring '08 term at N.C. State.
 Spring '08
 Chilton

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