Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 24 - Gauss's Law - Conceptual Questions - Page 681: 6


$\phi_{A}=\frac{4q}{\epsilon_0}$ $\phi_{B}=-\frac{4q}{\epsilon_0}$ $\phi_{C}=0$ $\phi_{D}=\frac{3q}{\epsilon_0}$ $\phi_{E}=0$

Work Step by Step

Flux through a body due to charges present outside is 0. So, there will be flux only due to charges present inside the body. For surface A, charge inside is $ 4q$ while for B, the charge inside is $-4q$. For C, it is $0 $ and for surface D it is $3q$. For E, it is $0$. 1.Charge in surface A $=4q$ so, electric flux through $\phi_{A}=\frac{4q}{\epsilon_0}$ 2. Charge in surface B $=-4q$ so, electric flux through $\phi_{B}=-\frac{4q}{\epsilon_0}$ 3. Charge in surface C $=0$ so, electric flux through $\phi_{C}=0$ 4.Charge in surface D $=3q$ so, electric flux through $\phi_{D}=\frac{3q}{\epsilon_0}$ 5.Charge in surface E $=0$ so, electric flux through $\phi_{E}=0$ Hence, $\phi_{A}=\frac{4q}{\epsilon_0}$ $\phi_{B}=-\frac{4q}{\epsilon_0}$ $\phi_{C}=0$ $\phi_{D}=\frac{3q}{\epsilon_0}$ $\phi_{E}=0$
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