(b) In this case, the same procedure yields these two equipotential lines:
(c) One way to search for the “crossover” case (from a single equipotential line, to two) is
to “solve” for a point on the
y
axis (chosen here to be an absolute distance
ξ
below
q
1
–
that is, the point is at a negative value of
y
, specifically at
y
=
−ξ
) in terms of
V
(or more
conveniently, in terms of the parameter
η
= 4
πε
o
V
x
10
10
).
Thus, the above expression
for
V
becomes simply
115. The (implicit) equation for the pair (
x,y
) in terms of a specific
V
is
V
=
q
1
4
πε
o
x
2
+
y
2
+
q
2
4
πε
o
x
2
+ (
y
−
d
)
2
where
d
= 0.50 m.
The values of
q
1
and
q
2
are given in the problem.
(a)
We set
V
= 5.0 V and plotted (using MAPLE’s implicit plotting routine) those points
in the
xy
plane which (when plugged into the above expression for
V
) yield 5.0 volts.
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 Spring '08
 Any
 Physics

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