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 470: 34

Answer

$ \frac{Q_1}{Q_2}=\frac{1}{4}$

Work Step by Step

The charges are either both positive or both negative so that the electric fields are oppositely directed and add to zero. Their magnitudes must be equal. $$E_1=E_2$$ $$k\frac{Q_1}{(\mathcal{l}/3)^2}= k\frac{Q_2}{(2\mathcal{l}/3)^2}$$ $$ 9Q_1=\frac{9Q_2}{4}$$ $$ \frac{Q_1}{Q_2}=\frac{1}{4}$$
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