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should be located in between them (so that the field vectors point in the opposite
direction). Let the coordinate of the second particle be
x'
(
x'
> 0). Then, the magnitude of
the field due to the charge –
q
1
evaluated at
x
is given by
E
=
q
1
/4
πε
0
x
2
, while that due to
the second charge –4
q
1
is
E'
= 4
q
1
/4
πε
0
(
x'
–
x
)
2
. We set the net field equal to zero:
G
EE
E
net
=
¡
=
′
0
so that
q
x
q
xx
1
0
2
1
0
2
4
4
4
π
π
ε
=
′ −
b
g
.
Thus, we obtain
x'
= 3
x
= 3(2.0 mm) = 6.0 mm.
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Fall '10 term at Kansas State University.
 Fall '10
 Rebello,NobelS
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