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67. (a) Since the two charges in question are of the same sign, the point
x
= 2.0 mm
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
=⇒ =
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This note was uploaded on 07/10/2010 for the course PHYS 241 taught by Professor Fr during the Fall '10 term at The Petroleum Institute.
 Fall '10
 FR
 Charge

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