Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 18 - Electric Forces and Electric Fields - Problems - Page 509: 11

Answer

(a)Magnitude of force on each sphere is $F=0.8285N$ (b) Attractive

Work Step by Step

Distance between the spheres is $r=0.50m$ $1$ electron $= 1.6\times10^{-19}C $ charge so$ 3.0\times10^{13}$ electrons is equal to $ 3.0\times10^{13}\times1.6\times10^{-19}C=4.8\times10^{-6}C $ charge so sphere from where electron is removed will get positively charged due to electron deficiency let us assume it $q_{1}=+4.8\times10^{-6}C $ similarly second sphere where these electron are placed will get negatively charged due to excessive electrons let us assume it $q_{2}=-4.8\times10^{-6}C $ (a) from coulombs law we know that magnitude of force between two charged objects is given by $F=k\frac{|q_{1}|\times|q_{2}| }{r^2}$ in our problem it is given that magnitude of force $|q_{1}|=4.8\times10^{-6} C$ $ |q_{2}| =4.8\times10^{-6} C$ $k=8.99\times10^{9}N.m^2/C^2$ $r=0.5m$ so magnitude of force will be $F=8.99\times10^{9}N.m^2/C^2\times\frac{4.8\times10^{-6} C|\times4.8\times10^{-6} C }{(0.5m)^2}$ $F=828.518\times10^{-3}N$=$0.8285N$ magnitude of force on each sphere is $F=0.8285N$ (b) due to opposite polarity on spheres force will be attractive in nature.
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