2114
Chapter 21
EVALUATE:
The magnitude of the field approaches infinity at the location of one of the point charges.
Figure 21.43
21.44.
IDENTIFY:
For a point charge,
2
q
E
k
r
=
. For the net electric field to be zero,
1
E
G
and
2
E
G
must have equal
magnitudes and opposite directions.
SET UP:
Let
1
0.500 nC
q
=+
and
2
8.00 nC.
q
E
G
is toward a negative charge and away from a positive charge.
EXECUTE:
The two charges and the directions of their electric fields in three regions are shown in Figure 21.44.
Only in region II are the two electric fields in opposite directions. Consider a point a distance
x
from
1
q
so a
distance 1.20 m
x
−
from
2
q
.
12
E
E
=
gives
22
0.500 nC
8.00 nC
(1.20
)
kk
x
x
=
−
.
16
(1.20 m
)
x
x
=−
. 4
(1.20 m
)
x
x
=±
−
and
0.24 m
x
=
is the positive solution. The electric field is zero at a point between the two charges, 0.24 m from
the 0.500 nC charge and 0.96 m from the 8.00 nC charge.
EVALUATE:
There is only one point along the line connecting the two charges where the net electric field is zero.
This point is closer to the charge that has the smaller magnitude.
Figure 21.44
21.45.
IDENTIFY:
Eq.(21.7) gives the electric field of each point charge. Use the principle of superposition and add the
electric field vectors. In part (b) use Eq.(21.3) to calculate the force, using the electric field calculated in part (a).
(a) SET UP:
The placement of charges is sketched in Figure 21.45a.
Figure 21.45a
The electric field of a point charge is directed away from the point charge if the charge is positive and toward the
point charge if the charge is negative. The magnitude of the electric field is
2
0
1
,
4
q
E
r
π
=
P
where
r
is the distance
between the point where the field is calculated and the point charge.
(i) At point a the fields
11
2 2
of
and
of
qq
EE
GG
are directed as shown in Figure 21.45b.
Figure 21.45b
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentElectric Charge and Electric Field
2115
EXECUTE:
9
1
92
2
1
22
01
12
.
0
0
1
0
C
(8.988 10 N m /C )
449.4 N/C
4
(0.200 m)
q
E
r
π
−
×
==
×
⋅
=
P
9
2
2
2
02
15
.
0
0
1
0
C
124.8 N/C
4
(0.600 m)
q
E
r
−
×
×
⋅
=
P
11
449.4 N/C,
0
xy
EE
124.8 N/C,
0
449.4 N/C 124.8 N/C
574.2 N/C
xxx
EE E
=+=
+
+
=
+
0
yyy
EEE
The resultant field at point a has magnitude 574 N/C and is in the +
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics, Charge, Electric charge, N/C

Click to edit the document details