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PHYS 133, WEEK 2
CHAPTER 26
26.4.
Model:
The electric field is that of the two charges located on the
y
axis.
Visualize:
Please refer to Figure Ex26.4. We denote the positive charge by
q
1
and the negative charge by
q
2
. The electric
field
r
E
1
of the positive charge
q
1
is directed away from
q
1
, but the field
r
E
2
is toward the negative charge
q
2
. We will add
r
E
1
and
r
E
2
vectorially to find the strength and the direction of the net electric field vector.
Solve:
The electric fields from
q
1
and
q
2
are
r
E
1
=
1
4
!"
0
q
1
r
1
2
, away from
q
1
along
+
x
axis
#
$
%
&
’
(
=
9.0
)
10
9
N m
2
/ C
2
( )
1
)
10
*
9
C
( )
0.05 m
( )
2
ˆ
i
=
3600
ˆ
i
N / C
r
E
2
=
1
4
0
q
2
r
2
2
,
#
below
$
x
axis
%
&
’
(
)
*
From the geometry of the figure,
tan
!
=
10 cm
5 cm
"
=
63.43
°
!
r
E
2
=
9.0
"
10
9
N m
2
/ C
2
(
)
1
"
10
#
9
C
(
)
0.10 m
(
)
2
+
0.05 m
(
)
2
#
cos63.43
°
ˆ
i
#
sin63.43
°
ˆ
j
(
) =
#
322
ˆ
i
+
644
ˆ
j
(
)
N / C
!
r
E
net
=
r
E
1
+
r
E
2
=
3278
ˆ
i
"
644
ˆ
j
( )
N / C
!
E
net
=
3278 N / C
( )
2
+
"
644 N / C
( )
2
=
3340 N/C
To find the angle this net vector makes with the horizontal, we calculate
tan
=
E
net
(
)
y
E
net
x
(
)
=
644 N / C
3278 N / C
"
=
11.1
°
Thus, the strength of the net electric field at P is 3341 N/C and
r
E
net
makes an angle of
11.1
°
below the
+
x
axis.
26.9.
Model:
The rod is thin, so assume the charge lies along a
line
.
Visualize:
Solve:
The force on charge
q
is
r
F
=
q
r
E
rod
. From Example 26.3, the electric field a distance
r
from the center of a charged
rod is
r
E
rod
=
1
4
0
Q
r
r
2
+
L
/ 2
( )
2
ˆ
i
=
9.0
#
10
9
N m
2
/ C
2
( )
50
#
10
$
9
C
( )
ˆ
i
0.04 m
( )
0.04 m
( )
2
+
0.05 m
( )
5
=
1.757
#
10
5
ˆ
i
N / C
Thus, the force is
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F
=
!
5
"
10
!
9
C
( )
1.757
"
10
5
N / C
( )
ˆ
i
=
!
8.78
"
10
!
4
ˆ
i
N
More generally,
r
F
=
8.78
!
10
"
4
N, toward the rod
(
)
.
26.14.
Model:
Each disk is a uniformly charged disk. When the disk is charged negatively, the onaxis electric field of
the disk points toward the disk. The electric field points away from the disk for a positively charged disk.
Visualize:
Solve:
(a)
The surface charge density on the disk is
!
=
Q
A
=
Q
"
R
2
=
50
#
10
$
9
C
0.05 m
( )
2
=
6.366
#
10
$
6
C / m
2
From Equation 26.22, the electric field of the left disk at
z
=
0.10 m is
E
1
(
)
z
=
2
0
1
#
1
1
+
R
2
z
2
$
%
&
&
’
(
)
)
=
#
6.366
*
10
#
6
C / m
2
2 8.85
*
10
#
12
C
2
/ N m
2
(
)
1
#
1
1
+
0.05 m 0.10 m
(
)
2
$
%
&
&
&
’
(
)
)
)
=
#
38,000 N / C
In other words,
r
E
1
=
38,000 N / C, left
(
)
. Similarly, the electric field of the right disk at
z
=
0.10 m (to its left) is
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 Spring '04
 Saunders
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