Chapter 26 Solutions
(solutions marked with * are graded problems)
26.3.
Model:
The electric field is that of the two 1 nC charges located on the
y
axis.
Visualize:
Please refer to Figure Ex26.3. We denote the top 1 nC charge by
q
1
and the bottom 1 nC charge by
q
2
. The
electric fields (
1
E
r
and
2
E
r
) of both the positive charges are directed away from their respective charges. With vector
addition, they yield the net electric field
net
E
r
at the point P indicated by the dot.
Solve:
The electric fields from
q
1
and
q
2
are
1
1
2
0
1
1
, along
axis
4
q
E
x
r
!"
#
$
=
+
%
%
’
(
r
(
)(
)
(
)
9
2
2
9
2
9.0 10 N m /C
1 10 C
ˆ
ˆ
3600 N/C
0.05 m
i
i
!
"
"
=
=
2
2
2
0
2
1
, above
axis
4
q
E
x
r
!
"#
$
%
=
+
’
’
(
)
r
Because
( )
1
tan
10 cm 5 cm,
tan
2
63.43 .
"
=
=
=
°
So,
(
)(
)
(
)
(
)
(
) (
)
9
2
2
9
2
2
2
9.0 10 N m /C
1 10 C
ˆ
ˆ
ˆ
ˆ
cos63.43
sin63.43
322
644 N/C
0.10 m
0.05 m
E
i
j
i
j
!
"
"
=
°
+
°
=
+
+
r
The net electric field is thus
(
)
net at P
1
2
ˆ
ˆ
3922
644 N/C
E
E
E
i
j
=
+
=
+
r
r
r
To find the angle this net vector makes with the
x
axis, we calculate
644 N/C
tan
3922 N/C
=
!
9.3
=
°
Thus, the strength of the electric field at P is
(
)
(
)
2
2
net
3922 N/C
644 N/C
E
=
+
=
3975 N/C
and
net
E
r
makes an angle of
9.3
°
above the
+
x
axis.
Assess:
Because of the inverse square dependence on distance,
2
1
E
E
<
. Additionally, because the point P has no special
symmetry relative to the charges, we expected the net field to be at an angle relative to 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
rod
F
qE
=
r
r
. From Example 26.3, the electric field a distance
r
from the center of a charged
rod is
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(
)(
)
(
) (
)
(
)
9
2
2
9
5
rod
2
2
5
2
0
ˆ
9.0 10 N m /C
50 10 C
1
ˆ
ˆ
1.757 10 N/C
4
/2
0.04 m
0.04 m
0.05 m
i
Q
E
i
i
r r
L
!"
#
$
$
=
=
=
$
+
+
r
Thus, the force is
(
)(
)
9
5
4
ˆ
ˆ
5 10 C 1.757 10 N/C
8.78 10
N
F
i
i
!
!
=
! "
"
=
!
"
r
More generally,
(
)
4
8.78 10 N, toward the rod
F
!
=
"
r
.
26.13.
Model:
Assume that the rings are thin and that the charge lies along circle of radius
R
.
Visualize:
Solve:
(a)
Let the rings be centered on the
z
axis. According to Example 26.5, the field of the left ring at
z
=
10 cm is
( )
(
)
(
)
(
)
(
)
(
)
(
)
9
2
2
9
4
1
1
3/2
3/2
2
2
2
2
0
9.0 10 N m /C
0.10 m 20 10 C
1.28 10 N/C
4
0.10 m
0.05 m
z
zQ
E
z
R
#
$
$
=
=
=
$
%
+
+
’
(
)
*
That is,
(
)
4
1
1.288 10 N/C, right
E
=
!
r
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 Spring '08
 ALL
 Charge, Electric charge, Enet

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