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Taking the (positive) square root and solving, we obtain
x
= 0.683 m. If one takes the
negative root and ‘solves’, one finds the location where the net force
would
be zero
if
q
1
and
q
2
were of like sign (which is not the case here).
(d) From the above, we see that
y
= 0.
64. Charge
q
1
= –80
×
10
–6
C is at the origin, and charge
q
2
= +40
×
10
–6
C is at
x
= 0.20
m. The force on
q
3
= +20
×
10
–6
C is due to the attractive and repulsive forces from
q
1
and
q
2
, respectively. In symbols,
G
G
G
FF
F
33
1
3
2
net
=+
, where
31
3
2
31
32
22
32

, 
.
qq
q
q
Fk
F
k
rr
==
GG
(a) In this case
r
31
= 0.40 m and
r
32
= 0.20 m, with
F
G
directed towards –
x
and
F
G
directed in the +
x
direction. Using the value of
k
in Eq. 215, we obtain
3
2
12
3net
3
2
2
66
92
2
6
ˆˆ
ˆ
ˆ
i 
i
i
i
80 10 C
40 10 C
ˆ
(8.99 10 N m C )(20 10 C)
i
(0.40m)
(0.20m)
ˆ
(89.9 N)i .
q
q
F
k
k
k
q
r
r
−−
−
⎛⎞
⎛
⎞
=−
+
= −
+
=
−
+
⎜⎟
⎜
⎟
⎜
⎟
⎝⎠
⎝
⎠
−×
+×
=×⋅
×
+
=
G
(b) In this case
r
31
= 0.80 m and
r
32
= 0.60 m, with
31
F
G
directed towards –
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 Spring '08
 Any
 Physics, Charge, Force

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