Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 25 - The Electric Potential - Exercises and Problems: 13

Answer

$\Delta V=11.38 V\approx 11.4 V$

Work Step by Step

Initial speed of an electron = 0 Final speed of an electron = $2\times 10^{6} m/s$ By the conservation of energy; Loss in energy = Gain in energy $q \Delta V=\frac{1}{2}m_{e}v_{e}^{2}-0$ $\Delta V =\frac{\frac{1}{2}m_{e}v_{e}^{2}}{q}$ $q \Delta V=\frac{1}{2}\times 9.11\times 10^{-31}\times (2\times 10^{6} )^{2}$ $\Delta V=\frac{\frac{1}{2}\times 9.11\times 10^{-31}\times (2\times 10^{6} )^{2}}{1.6 \times 10^{-19}}$ $\Delta V=11.38 V\approx 11.4 V$
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