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: 9a

Answer

$W=1.87\times 10^{-21}J$

Work Step by Step

We know that $W=Fd$. This equation can be written as: $W=qEd$.....................eq(1) However, we know that; $E=\frac{\sigma}{2{\epsilon_{\circ}}}$ Substituting values into this equation and solving: $E=\frac{5.8\times 10^{-12}}{2\times 8.85\times 10^{-12}}$ $E=0.32768\frac{N}{C}$ We plug in the known values in eq(1) to obtain: $W=1.6\times 10^{-19}\times 0.32768\times 3.56\times 10^{-2}$ $W=1.87\times 10^{-21}J$
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