Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 24 - Electric Potential - Problems - Page 713: 54a

Answer

The positron will emerge from the field at $~~x = 0$

Work Step by Step

We can find the change in electric potential energy if the positron moves from $V=0$ to $V = 500~V$: $\Delta U = q~\Delta V$ $\Delta U = (1.6\times 10^{-19}~C)~(500~V-0)$ $\Delta U = 8.00\times 10^{-17}~J$ In order to reach a location with an electric potential of $500~V$ the initial kinetic energy of the positron must be at least $8.00\times 10^{-17}~J$ We can find the initial kinetic energy: $K_i = \frac{1}{2}mv_i^2$ $K_i = \frac{1}{2}(9.109\times 10^{-31}~kg)(1.0\times 10^7~m/s)^2$ $K_i = 4.55\times 10^{-17}~J$ The positron does not have enough mechanical energy to reach a location with an electric potential of $500~V$ Therefore, the positron will emerge from the field at $~~x = 0$
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