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1
Class examples for Chapter 22
8.
We place the origin of our coordinate system at point
P
and orient our
y
axis in the
direction of the
q
4
= –12
q
charge (passing through the
q
3
= +3
q
charge). The
x
axis is
perpendicular to the
y
axis, and thus passes through the identical
q
1
=
q
2
= +5
q
charges.
The individual magnitudes

, 
, 
,
E
E
E
1
2
3
and


E
4
are figured from Eq. 223, where the
absolute value signs for
q
1
,
q
2
, and
q
3
are unnecessary since those charges are positive
(assuming
q
> 0). We note that the contribution from
q
1
cancels that of
q
2
(that is,




E
E
1
2
), and the net field (if there is any) should be along the
y
axis, with magnitude
equal to
E
q
d
q
d
q
d
q
d
net
j
j
F
H
G
I
K
J
F
H
G
I
K
J
1
4
2
1
4
12
4
3
0
4
2
3
2
0
2
2
b
g
which is seen to be zero.
12.
The field of each charge has magnitude
19
9
2
2
6
2
22
1.60 10
C
(8.99 10 N m C )
3.6 10
N C.
(0.020 m)
0.020m
kq
e
Ek
r
The directions are indicated in standard format below. We use the magnitudeangle
notation (convenient if one is using a vectorcapable calculator in polar mode) and write
(starting with the proton on the left and moving around clockwise) the contributions to
E
net
as follows:
E
E
E
E
E
20
130
100
150
0
b
g
b
g
b
g
b
g
b
g
.
This yields
393
10
764
6
.
.
c
h
, with the N/C unit understood.
(a) The result above shows that the magnitude of the net electric field is
6
net

 3.93 10 N/C.
E
(b) Similarly, the direction of
E
net
is –76.4
from the
x
axis.
22.
(a) We use the usual notation for the linear charge density:
=
q/L
.
The arc length is
L = r
with
is expressed in radians.
Thus,
L
= (0.0400 m)(0.698 rad) = 0.0279 m.
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 Fall '08
 IASHVILI
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