Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 19 - Electric Charges, Forces, and Fields - Problems and Conceptual Exercises - Page 688: 98

Answer

a) $\sigma=-9.7\times 10^{-10}C/m^2$ b) $Q=-5.0\times 10^5C$ c) The electric field of the Moon will be greater than that of the Earth.

Work Step by Step

(a) We can find the charge density as $\phi=\frac{Q}{\epsilon_{\circ}}$ $\implies \phi=\frac{\sigma A}{\epsilon_{\circ}}$ This simplifies to: $\sigma=\epsilon_{\circ}E$ We plug in the known values to obtain: $\sigma=(8.85\times 10^{-12})(-110)$ $\sigma=-9.7\times 10^{-10}C/m^2$ (b) We can find the required electric charge as $Q=\sigma A$ $\implies Q=\epsilon_{\circ}E(4\pi R^2)$ $\implies Q=\frac{ER^2}{K}$ We plug in the known values to obtain: $Q=\frac{(-110)(6.38\times 10^6)^2}{8.99\times 10^9}$ $Q=-5.0\times 10^5C$ (c) We know that the surface area of the Moon is less than that of the Earth, so the surface charge density for the same amount of charge on the Moon will be larger than on the Earth. We also know that the electric field is directly proportional to the surface charge density and thus the electric field of the Moon will be greater than that of the Earth.
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