Consider an infinitely thin charged rod of length L with uniform linear chargedensity i lying along the x-axis. The ends are at x = +L/2 and the center is at theorigin.a. Calculate the electric field (magnitude and direction) for x > L/2.b. Calculate the force on a test particle with charge qo<0 located at someposition x> L/2.c. Show that the force on the test particle will simplify to Coulomb’s Lawwhen it is sufficiently far away from the rod, i.e. when x >> L. Use Q =AL for the total charge on the rod.

Answered: Consider an infinitely thin charged rod…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

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Consider an infinitely thin charged rod of length L with uniform linear charge
density i lying along the x-axis. The ends are at x = +L/2 and the center is at the
origin.
a. Calculate the electric field (magnitude and direction) for x > L/2.
b. Calculate the force on a test particle with charge qo<0 located at some
position x> L/2.
c. Show that the force on the test particle will simplify to Coulomb’s Law
when it is sufficiently far away from the rod, i.e. when x >> L. Use Q =
AL for the total charge on the rod.

Expert Solution

Step 1

$A . dE = \frac{1}{4 {πε}_{0}} \frac{dq}{x^{2}} [dq = λdx] E_{p} = \int dE E_{p} = \frac{1}{4 {πε}_{0}} \int_{b}^{L + b} \frac{dq}{x^{2}} E_{p} = \frac{λ}{4 {πε}_{0}} \int_{b}^{L + b} \frac{d x}{x^{2}} E_{p} = \frac{λ}{4 {πε}_{0}} (\frac{- 1}{x}) \begin{matrix} L + b \\ b \end{matrix} E_{p} = \frac{λ}{4 {πε}_{0}} (\frac{1}{b} - \frac{1}{L + b}) if, λ > 0; then, E_{p} will be point away from the rod .$

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