problemset3

# The dashed curves are the rst 2 and second 1 terms

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Unformatted text preview: Spring 2009 So σs = ε0 Ey (x, y = 0, z ) σs = −qh 3 2π [(x − U t)2 + h2 + z 2 ] 2 E � w � l −qh Q= 3 0 0 dxdz 2π [(x − U t)2 + h2 + z 2 ] 2 For w very small, σs does not change signiﬁcantly from z = 0 to z = w, so integral in z becomes just multiplication at z = 0. � Q= 0 l −qhw dx 2π [(x − U t)2 + h2 ] � Let x = x − U t ⇒ dx� = dx So: � l−U t −qhw � Q= 3 dx −U t 2π [((x� )2 + h2 ] 2 ⎡ ⎤ ⎢ ⎥ ⎥ qw ⎢ Ut ⎢� l − Ut ⎥ Q=− ⎢ (l − U t)2 + h2 + �(U t)2 + h2 ⎥ 2πh ⎣ ⎦ � �� �� �� � (2) (1) q (1) (2) l Ut (3) Figure 9: Representative shape of total charge Q on the electrode versus U t. The dashed curves are the ﬁrst (2) and second (1) terms for Q and (3) is the sum (1) + (2). (Image by MIT OpenCourseWare.) F dQ −qw i= = dt 2πh � −U h2 � U h2 3+ 3 [(l − U t)2 + h2 ] 2 [(U t)2 + h2 ] 2 � � qwR −U h2 U h2 V = −iR = + 3 2πh [(l − U t)2 + h2 ] 3 2 [(U t)2 + h2 ] 2 7 Problem Set 3 6.641, Spring 2009 v l/U t Figure 10: Voltage V versus time across small electrode resistance R (Image by MIT OpenCourseWare.) Problem 3.3 A 1 → − H= 4π → J (− � ) × ˆr� r � r i → −� − − → |2 dv |r r � y= a 2 1 → −...
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## This document was uploaded on 03/19/2014 for the course EECS 6.641 at MIT.

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