00
22
0
1.2 J
44
f
qq
qq
KUU
x
yx
y
πε
′
′
+=
⇒
+
=
++
.
This yields
y
=
−
8.5 m.
59. We apply conservation of energy for particle 3 (with
q'
=
−
15
×
10
−
6
C):
K
0
+
U
0
=
K
f
+
U
f
where (letting
x
=
±
3 m and
q
1
=
q
2
= 50
×
10
−
6
C =
q
)
12
00 0
2
444
qq
qq
U
x
y
′′
′
=+=
+++
.
(a) We solve for
K
f
(with
y
0
= 4 m):
0
0
21
1
1.2 J
3.0 J
4

ff
qq
KK
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