P25_015.PDF
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Pobierz
Chapter 25 - 25.15
15. First, we observe that
V
(
x
) cannot be equal to zero for
x>d
.Infact
V
(
x
) is always negative for
x>d
.
Now we consider the two remaining regions on the
x
axis:
x<
0and0
<x<d
.For
x<
0 the separation
between
q
1
andapointonthe
x
axis whose coordinate is
x
is given by
d
1
=
−
x
; while the corresponding
separation for
q
2
is
d
2
=
d
x
.Weset
V
(
x
)=
k
q
1
d
1
+
q
2
=
q
4
πε
0
1
−
x
+
−
3
=0
d
2
d
−
x
to obtain
x
=
−
d/
2. Similarly, for 0
<x<d
we have
d
1
=
x
and
d
2
=
d
−
x
.Let
V
(
x
)=
k
q
1
d
1
+
q
2
=
q
4
πε
0
x
+
−
3
=0
d
2
d
−
x
and solve:
x
=
d/
4.
−
1
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Inne pliki z tego folderu:
P25_086.PDF
(80 KB)
P25_025.PDF
(59 KB)
P25_001.PDF
(58 KB)
P25_002.PDF
(53 KB)
P25_003.PDF
(64 KB)
Inne foldery tego chomika:
chap01
chap02
chap03
chap04
chap05
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