Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 27 - Gauss's Law - Exercises and Problems - Page 806: 15

Answer

$3.5\times 10^{-4}\;\rm N\cdot m^2/C$

Work Step by Step

We know that the electric flux is given by $$ \Phi=\vec E\cdot \vec A $$ Hence, $$\Phi=(350x+150)\;\hat i\cdot \vec A$$ The box has 6 faces but there are only two faces that the electric field enters and comes out from them since the given electric field formula is on the $x$-direction only. The other 4 faces the angle between the electric field and the area vector is 90$^\circ$ which means that $\cos\theta=0$ and hence the flux is zero as well on all of them. The flux enters on the face where $x=0$ cm and comes out from the face where $x=1$ cm since the dimensions of the box are 1 cm $\times$ 1 cm $\times$ 1 cm. $$\Phi=(350x+150)\;\hat i\cdot \vec A_1+(350x+150)\;\hat i\cdot \vec A_2$$ $$\Phi=(350[0]+150)\;\hat i(0.01^2)\cos180^\circ\;\hat i+(350[0.01]+150)(0.01^2)\cos0^\circ\;\hat i$$ where at $x=0$ cm, $\theta=180^\circ$ and at $x=1$ cm, $\theta=0^\circ$ $$\Phi=\color{red}{\bf 3.5\times 10^{-4}}\;\rm N\cdot m^2/C$$
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