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Electric Charge and Electric Field
2119
The components of the two fields are shown in Figure 21.49g.
12
2
0
1
4
q
EE
r
π
==
P
9
92
2
1
2
6.00 10 C
(8.988 10 N m /C )
(0.250 m)
E
−
×
=×
⋅
862.8 N/C
=
=
Figure 21.49g
11
2 2
cos ,
cos
xx
θ
=−
=+
0
xxx
EEE
=+=
sin ,
sin
yy
( )( )
1
1
2
2
sin
2 862.8 N/C 0.800
1380 N/C
yyy
y
EE E E E
=
=
=
1380 N/C, in the
direction.
E
y
EVALUATE:
Point
a
is symmetrically placed between identical charges, so symmetry tells us the electric field
must be zero. Point
b
is to the right of both charges and both electric fields are in the +
x
direction and the resultant
field is in this direction. At point
c
both fields have a downward component and the field of
2
q
has a component to
the right, so the net
E
G
is in the 4th quadrant. At point
d
both fields have an upward component but by symmetry
they have equal and opposite
x
components so the net field is in the +
y
direction. We can use this sort of reasoning
to deduce the general direction of the net field before doing any calculations.
21.50.
IDENTIFY:
Apply Eq.(21.7) to calculate the field due to each charge and then calculate the vector sum of those fields.
SET UP:
The fields due to
1
q
and to
2
q
are sketched in Figure 21.50.
EXECUTE:
9
2
2
0
1
(6.00 10 C)
ˆˆ
() 1
5
0
N
/
C
4(
0
.
6
m
)
−
×
=
−
Ei
i
G
P
.
9
1
22
0
1
ˆ
ˆ
(4.00 10 C)
(0.600)
(0.800)
(21.6
28.8 )N C
1
.
0
0
m
)
(
1
.
0
0
m
)
−
⎛⎞
+
=
⎜⎟
⎝⎠
j
i
+
j
G
P
.
( 128.4 N/C)
(28.8 N/C)
E=E +E
i+
j
GGG
.
(128.4 N/C)
(28.8 N/C)
131.6 N/C
E
=
at
1
28.8
tan
12.6
128.4
−
°
above the
x
−
axis and therefore 196.2
°
counterclockwise from the +
x
axis.
EVALUATE:
1
E
G
is directed toward
1
q
because
1
q
is negative and
2
E
G
is directed away from
2
q
because
2
q
is positive.
Figure 21.50
21.51.
IDENTIFY:
The resultant electric field is the vector sum of the field
of
and
of .
qq
G
G
SET UP:
The placement of the charges is shown in Figure 21.51a.
Figure 21.51a
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Chapter 21
EXECUTE:
(a)
The directions of the two fields are shown in Figure 21.51b.
1
12
2
01
1
4
q
EE
r
π
==
P
9
92
2
1
2
6.00 10 C
(8.988 10 N m /C )
(0.150 m)
E
−
×
=×
⋅
2397 N/C
=
=
Figure 21.51b
11
2
x2
2397 N/C,
0
2397 N/C,
0
xy
y
=−
=
=
2( 2397 N/C)
4790 N/C
xxx
EE E
=+=
−
=
−
0
yyy
EEE
The resultant electric field at point a in the sketch has magnitude 4790 N/C and is in the
direction.
x
−
(b)
The directions of the two fields are shown in Figure 21.51c.
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.
 Spring '06
 Buchler
 Physics, Charge

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