problemset4

# Taking the normal derivative and multiplying by d we

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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 &gt; 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 non-returning 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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