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Unformatted text preview: phere at r = R, we must have
q
v
�
= −�
2 sin2 θ − (r − R cos θ )2
2 sin2 θ − (R cos θ − s)2
R
R
0
q 2 (R2 sin2 θ − (R cos θ − s)2 ) = v 2 (R2 sin2 θ − (r0 − R cos θ)2 )
2
q 2 (R2 − s2 − 2Rs cos θ) = v 2 (R2 − r0 − 2Rr0 cos θ). M. Z. Bazant – 18.366 Random Walks and Diﬀusion – Problem Set 4 Solutions 6 To be valid for all θ, we must have q 2 s = v 2 r0 , and
2
q 2 (R2 − s2 ) = v 2 (R2 − r0 )
R2
2
(R − s2 ) = (R2 − r0 )
s
� 2�
R
= 0.
(s − r0 ) s
r0 Thus s = R2 /r0 , which is inside of the sphere, since r0 > R. The magnitude of the charge is given
by
� 2�
2
2R
v r0 = q
r0
−qR
v=
.
r0
Thus the electric potential is
q
qR
Φ(�) = �
r
−�
2
r2 sin θ − (r0 − r cos θ)2 r0 r2 sin2 θ − (r cos θ − R2 2
r0 ) . Taking the normal derivative and multiplying by −D, we ﬁnd that the PDF of absorption at a
position (R, θ) on the sphere is
1
P (R, θ) =
4πRr �
1− 1−
2R
r0 R2
2
r0 cos θ + R2
2
r0 �3/2 . The ratio between the probability of hitting at the nearest point on the sphere and the farthest is
P (R, 0)
=
P (R, π ) 4 � 2
1 + 2R/r0 + R2 /r0
2
1 − 2R/r0 + R2 /r0 �d/2 �
= 1 + R/r0
1 − R/r0 �d
. The Ballot Problem We deﬁne Pi and Qi be the partial scores for the two candidates after i votes have been counted,
and let Ri = Pi − Qi be the diﬀerence between the two. At each step, Ri can either increase or
decrease by one, and it is therefore a Bernoulli pathway on the integers, as discussed in lecture 14.
We know that Pp+q = p and Qp+q = q , so Rp+q = p − q . In terms of the quantities introduced
in lecture, we know that the number of possible ways to count the votes is therefore N (p − q , p + q ),
and each of these paths is equally likely.
If the ﬁrst candidate always has more votes than the second, we know that the Ri trace out a
nonreturning path to (p−q, p+q ), and as shown in the lecture there are (p−q )N (p−q, p+q...
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 Fall '06
 MartinBazant

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