P25_088

# P25_088 - kinetic term: v = r 2(2 U ) m = 2 | q | r k m d ....

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88. (Second problem of Cluster ) (a) We argue by symmetry that of the total potential energy in the initial conFguration, a third converts into the kinetic energy of each of the particles. And, because the total potential energy consists of three equal contributions U = 1 4 πε 0 q 2 d then any of the particle’s Fnal kinetic energy is equal to this U . Therefore, using k for1 / 4 πε 0 ,we obtain v = r 2 U m = | q | r 2 k md . (b) In this case, two of the U contributions to the total potential energy are converted into a single
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Unformatted text preview: kinetic term: v = r 2(2 U ) m = 2 | q | r k m d . (c) Now it is clear that the one remaining U contribution is converted into a particle’s kinetic energy: v = r 2 U m = | q | r 2 k m d . (d) This leaves no potential energy to convert into kinetic for the last particle that is released. It maintains zero speed....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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