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: 15

Answer

(a) Proton will move from lower potential to higher potential against the electric field. (b) $3340$ volt

Work Step by Step

Given: Proton initial speed = 8,00,000 m/s Proton final speed =0 (a) Proton will move from lower potential to higher potential against the electric field. (b) By conservation of energy Loss in kinetic energy = Gain in potential energy $e\Delta V=\frac{1}{2}m_{p}v_{p}^{2}-0$ $e\Delta V=\frac{1}{2}m_{p}v_{p}^{2}$ $\Delta V=\frac{{\frac{1}{2}m_{p}v_{p}^{2}}}{1.6 \times 10^{-19}}$ $=\frac{1}{2}\times \frac{{ 1.67 \times 10^{-27} \times(8\times 10^{5})^{2}}}{1.6 \times 10^{-19}}$ $=3340$ volt
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