18
CHAPTER 15
15.26
If the resultant field is to be zero, the contributions
of the two charges must be equal in magnitude
and must have opposite directions. This is only
possible at a point on the line between the two
negative charges.
Assume the point of interest is located on the
y
-axis at
. Then, for equal
magnitudes,
4.0 m
6.0 m
y
−<
<
12
22
ee
kq
rr
=
or
()
C
8.0 C
m
4.0 m
yy
9.0
6.0
µ
=
−+
Solving for
y
gives
8
6.0 m
9
+=
−
4.0 m
, or
0.85 m
y
=+
±
±
²³´µ¶
±
·
²
±
¸
³
±
³
¸
¹³´µ¶
´
±
µºµ»¼³´µ
±
½
´
¸
µºµ»¾³´µ
±
½
±²
15.27
If the resultant field is zero, the
contributions from the two charges must
be in opposite directions and also have
equal magnitudes. Choose the line
connecting the charges as the
x
-axis, with
the origin at the –2.5
C charge. Then, the
two contributions will have opposite

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