Answered: 1. A point charge of q = 70.0 µC, a…

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

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Question

1. A point charge of q = 70.0 µC, a line charge segment of A
second line charge segment of 2 = 13.0 µC/m are arranged as shown in the
Diagram.
17.0 µC/m ând a
%3D
%3D
b) What is the electric field at the origin of the system Due to only the line charge
segment, 21? Express this in unit vector notation.
c) What is the electric field at the origin of the system Due to only the line charge
segment, 2? Express this in unit vector notation.

Expert Solution

Step 1

To calculate the electric field at the origin of the system Due to only the line charge segment, the line charge density given in part A equal to the 17 microC/m.

A point P(0, 0, 0) on the y axis at which to determine the field.
This is a perfectly general point in view of the lack of variation of the field with φ
and z. To find the incremental field at P due to the incremental charge
d Q = ρL dy'

$d E = \frac{ρ_{L} d x' (r - r')}{4 {πε}_{o} {|r - r'|}^{3}} r = - 0.65 a_{x} r' = 0.25 a_{x} d E = \frac{17 \times 10^{- 6} d x' (0.40 a_{x})}{4 {πε}_{o} {(0.65)}^{\frac{3}{2}}} d E = 0.032 \times 10^{3} \int_{- 0.25}^{- 0.9} d x E = 0.032 \times 10^{3} (- 0.9 + 0.250) a_{x} E = - 20.8 a_{x}$

Step 2

Part (C):-

To determine the electric field at the origin of the system Due to only the line charge segment, As the line segment is symmetrical about y axis,

so determine the for the one portion and then multiply with 2,

$d E = \frac{ρ_{L} d x' (r - r')}{4 {πε}_{o} {|r - r'|}^{3}} r = - 0.3 a_{y} r' = 0.5 a_{x} d E = - \frac{13 \times 10^{- 6} \times 0.3 d y' a_{y} (- 0.3 a_{y} - 0.5 a_{x})}{4 {πε}_{o} {(0 . 3^{2} + 0 . 5^{2})}^{\frac{3}{2}}} d E = \frac{10.53}{0.445} d y d E = 23.65 \times 10^{3} \int_{0}^{- 0.3} d y E = - 7.09 \times 10^{3} a_{y}$

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