Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 23 - Gauss' Law - Problems: 4

Answer

$\phi^{\prime}_E=-1.1\times 10^{-4}\frac{N.m^2}{C}$

Work Step by Step

The flux through the rim of radius $a=0.11m$ can be calculated as: $\phi_E=EA$ $\implies \phi_E=E\pi a^2$ We plug in the known values to obtain: $\phi_E=(3.0\times 10^{-3})(3.1416)(0.11)^2=1.1\times 10^{-4}\frac{N.m^2}{C}$ The flux through the net $(\phi^{\prime}_E)$ is opposite to that of rim: $\implies \phi^{\prime}_E=-\phi_E=-1.1\times 10^{-4}\frac{N.m^2}{C}$
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