P24_031

# P24_031 - v is the nal velocity, and x is the distance...

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31. The charge on the metal plate, which is negative, exerts a force of repulsion on the electron and stops it. First ±nd an expression for the acceleration of the electron, then use kinematics to ±nd the stopping distance. We take the initial direction of motion of the electron to be positive. Then, the electric ±eld is given by E = σ/ε 0 ,where σ is the surface charge density on the plate. The force on the electron is F = eE = eσ/ε 0 and the acceleration is a = F m = ε 0 m where m is the mass of the electron. The force is constant, so we use constant acceleration kinematics. If v 0 is the initial velocity of the electron,
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Unformatted text preview: v is the nal velocity, and x is the distance traveled between the initial and nal positions, then v 2 v 2 = 2 ax . Set v = 0 and replace a with e/ m , then solve for x . We nd x = v 2 2 a = mv 2 2 e . Now 1 2 mv 2 is the initial kinetic energy K , so x = K e . We convert the given value of K to Joules. Since 1 . 00 eV = 1 . 60 10 19 J, 100 eV = 1 . 60 10 17 J. Thus, x = (8 . 85 10 12 C 2 / N m 2 )(1 . 60 10 17 J) (1 . 60 10 19 C)(2 . 10 6 C / m 2 ) = 4 . 4 10 4 m ....
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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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