College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 17 - Problems - Page 654: 44


$\Delta V = 150~J/C$

Work Step by Step

By conservation of energy, the increase in kinetic energy is equal in magnitude to the decrease in electric potential energy. We can use conservation of energy to find the potential difference: $K_2+U_2 = K_1+U_1$ $U_2-U_1 = K_1-K_2$ $(\Delta V)~q = 0-\frac{1}{2}m~v_2^2$ $\Delta V = \frac{-\frac{1}{2}m~v_2^2}{q}$ $\Delta V = \frac{-\frac{1}{2}(9.1\times 10^{-31}~kg)(7.26\times 10^6~m/s)^2}{-1.6\times 10^{-19}~C}$ $\Delta V = 150~J/C$
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