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: 6a

Answer

$V_B-V_A=2.46V$

Work Step by Step

We can find $V_B-V_A$ as: $V_B-V_A=\frac{W}{e}$ We plug in the known values of $W$ and $e$ in the formula above to obtain: $V_B-V_A=\frac{3.94\times 10^{-19}}{1.6\times 10^{-19}}$ $V_B-V_A=2.46V$
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