Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 16 - Electric Charge and Electric Field - General Problems - Page 471: 56

Answer

$Q=8.0\times 10^{-9}C$

Work Step by Step

According to given condition: $mg=K\frac{Q^2}{d^2}$ As $d=2r$, $mg=\frac{KQ^2}{(2r)^2}$.......eq(1) We also know that: $m=V\rho $ As the volume for sphere is $V=\frac{4}{3}\pi r^3$; $\implies m=\frac{4}{3}\pi r^3 \rho$ Thus equation(1) becomes $\frac{4}{3}\pi r^3 \rho g=\frac{KQ^2}{(2r)^2}$ This simplifies to $Q=\sqrt{\frac{16\rho \pi g r^5}{3K}}$ We plug in the known values to obtain: $Q=\sqrt{\frac{16\times 35\times 3.1416\times 9.8(1\times 10^{-2})^5}{3\times 8.99\times 10^9}}=8.0\times 10^{-9}C$
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