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Vx k
q
d
q
d
q
xdx
()
=+
F
H
G
I
K
J
−
−
F
H
G
I
K
J
=
1
1
2
20
4
13
0
π
ε
and solve:
x = d
/4. With
d
= 24.0 cm, we have
x
= 6.00 cm.
(b) Similarly, for
x
< 0 the separation between
q
1
and a point on the
x
axis whose
coordinate is
x
is given by
d
1
= –
x
; while the corresponding separation for
q
2
is
d
2
=
d – x
.
We set
Vx
k
q
d
q
d
q
F
H
G
I
K
J
=
−
+
−
−
F
H
G
I
K
J
=
1
1
2
4
0
π
to obtain
x = –d
/2. With
d
= 24.0 cm, we have
x
=
–
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This note was uploaded on 10/28/2010 for the course PHYS 219 taught by Professor Gf during the Fall '10 term at Hudson Valley Community College.
 Fall '10
 GF

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