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 - Problems - Page 469: 14

Answer

Magnitude: $65.2 N$ Direction depends on the position of the charge Top left: North West Top right: North East Bottom left: South West Bottom right: South East

Work Step by Step

The charges can be treated as point charges. Because each charge has the same magnitude and is placed at different corners of a square, all charges will feel a force of equal magnitude. Use Coulomb's Law to find the magnitude. $F=\frac{kQ^{2}}{r^{2}}$ First, lets find the magnitude of the force of the top right and bottom left charges on the top left charge $F=\frac{(8.99\times10^{9}\frac{N\times m^{2}}{C})(6.15\times10^{-6}C)^{2}}{(0.100m)^{2}}=34.0N$ Now lets find the magnitude of the force of the bottom right charge on the top left charge. Because the two charges are on the opposite vertices of the square, the distance is $\sqrt (0.100m^{2}+0.100m^{2})=0.141m$ $F=\frac{(8.99\times10^{9}\frac{N\times m^{2}}{C})(6.15\times10^{-6}C)^{2}}{(0.141m)^{2}}=17.1N$ Now we need to find the total magnitude. $F_{net}=\sqrt (34.0N^{2}+34.0N^{2})+17.1N=65.2N$ Because all the charges are positive, the direction of this force is North West, away from the other charges. The direction of the force on the top right charge will be North East, the direction of the force on the bottom left charge will be South West, and lastly, the direction of the force on the bottom right charge will be South East.
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