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30.
(a) The two charges are
q
=
αQ
(where
α
is a pure number presumably less than 1 and greater than
zero) and
Q
−
q
=(1
−
α
)
Q
. Thus, Eq. 224 gives
F
=
1
4
πε
0
(
αQ
)((1
−
α
)
Q
)
d
2
=
Q
2
α
(1
−
α
)
4
πε
0
d
2
.
(b) The graph below, of
F
versus
α
, has been scaled so that the maximum is 1. In actuality, the
maximum value of the force is
F
max
=
Q
2
/
16
πε
0
d
2
.
0
1
Force
1
alpha
(c) It is clear that
α
=
1
2
gives the maximum value of
F
.
(d) Seeking the halfheight points on the graph is diﬃcult without grid lines or some of the special
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Charge

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