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114. (a) The net potential is
V
=
V
1
+
V
2
=
q
1
4
πε
o
r
1
+
q
2
4
πε
o
r
2
where
r
1
=
x
2
+
y
2
and
r
2
=
(
x
−
d
)
2
+
y
2
.
The distance
d
is 8.6 nm. To find the locus
of points resulting in
V
= 0, we set
V
1
equal to the (absolute value of)
V
2
and square both
sides.
After simplifying and rearranging we arrive at an equation for a circle:
y
2
+
⎝
⎜
⎛
⎠
⎟
⎞
x
+
9
d
16
2
=
225
256
d
2
.
From this form, we recognize that the center of the circle is –9
d
/16 =
– 4.8 nm.
(b) Also from this form, we identify the radius as the square root of the righthand side:
R
=
15
d
/16 = 8.1 nm.
(c) If one uses a graphing program with “implicitplot” features, it is certainly possible to
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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