quiz1 - 6.641 Electromagnetic Fields Forces and Motion...

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6.641 — Electromagnetic Fields, Forces, and Motion Spring 2006 Quiz 1 - Solutions Prof. Markus Zahn MIT OpenCourseWare Problem 1 z z z q -q σ = ∞ ε 0 image charge Figure 1: Charge q of a mass m above a perfectly conducting ground plane and its image charge q . (Image by MIT OpenCourseWare.) A Question: What is the velocity of the charge as a function of position z ? Solution: Force = ma dv m dt = F due to image charge dv q ( q ) dv z q 2 m = m = dt 4 πε 0 (2 z ) 2 dt 4 πε 0 4 z 2 dv z q 2 dv z dv z dz dv z d p v 2 P dt = 16 πε 0 mz Use of chain rule: dt = dz dt = dz v z = dz 2 z 2 i z 1
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r i R ± ± ² R ± bB³´ R b ´ R ± Spring 2006 Quiz 1 6.641, Spring 2005 d p v z 2 P q 2 i p v z 2 P i q 2 2 2 dz 2 = 16 πε 0 mz d 2 = 16 πε 0 mz dz 2 2 q v = + + C z 8 πε 0 mz Use I.C. v z ( z = d ) = 0 2 8 πε 0 md C = q q 2 p 1 1 P 2 v = z 8 πε 0 m z d ( 1 1 ) z 2 ( z ) = q v z 8 πε 0 m d since particle is moving towards the conducting ground plane v = v z i z ( v z has minus sign) B Question:
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quiz1 - 6.641 Electromagnetic Fields Forces and Motion...

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