Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 23 - The Electric Field - Exercises and Problems: 8

Answer

The electric field strength 5.0 cm from the wire is 4000 N/C

Work Step by Step

The strength of the electric field at a distance $r$ from a charged wire is $E = \frac{\lambda}{2\pi ~\epsilon_0~r}$ Let $r_1 = 10.0~cm$. We can write an expression for the electric field strength at a distance of $r_1$: $E_1 = \frac{\lambda}{2\pi ~\epsilon_0~r_1}$ Let $r_2 = 5.0~cm$. We can write an expression for the electric field strength at a distance of $r_2$: $E_2 = \frac{\lambda}{2\pi ~\epsilon_0~r_2}$ We can divide $E_1$ by $E_2$: $\frac{E_1}{E_2} = \frac{\frac{\lambda}{2\pi ~\epsilon_0~r_1}}{\frac{\lambda}{2\pi ~\epsilon_0~r_2}}$ $E_2 = \frac{r_1~E_1}{r_2}$ $E_2 = \frac{(10.0~cm)(2000~N/C)}{5.0~cm}$ $E_2 = 4000~N/C$ The electric field strength 5.0 cm from the wire is 4000 N/C.
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