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:
$V_B-V_A=\frac{3.94\times 10^{-19}}{1.6\times 10^{-19}}$
$V_B-V_A=2.46V$
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 24 - Electric Potential - Problems - Page 710: 6a
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