Answer
$5MV$
Work Step by Step
We know that
$\Delta V=\frac{mv^2}{2e}$
$\Delta V=\frac{m(0.1c)^2}{2e}$
$\Delta V=\frac{(1.67\times 10^{-27})(0.1)^2(3\times 10^8)^2}{2(1.6\times 10^{-19})}$
$\Delta V=5MV$
Physics Technology Update (4th Edition)
by Walker, James S.
Published by
Pearson
ISBN 10:
0-32190-308-0
ISBN 13:
978-0-32190-308-2
Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 718: 20
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