91Ó°ÊÓ

1. Uniform charge density,$\mathrm{λ}=9\text{nC/m}$
2. Stretch along the x-axis is from $x=0.0\text{m}$to.$x=3.0\text{m}$

Using the concept of the electric field, we can get the required magnitude of the electric field by calculating the charge from linear charge density at the given distance.

Formula:

The electric field of a particle at a given distance within the given range of stretch is given by: $d\stackrel{→}{E}=\frac{dq}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o|x−x_p{|}^2}$ (i)

where,$x_p$= The distance of field point

$dq$= small charge of the particle

A small section of the distribution that has charge$dq$is,$\mathrm{λ}dx$where.$\mathrm{λ}=9.0x{10}^{–9}\text{C/m}$Its contribution to the field at$x_P=4.0\text{m}$is given by integrating equation (ii) within the given range stretch pointing in the +x direction as follows:

$\stackrel{→}{E}=\underset{0}{\overset{3.0m}{}}\frac{\mathrm{λ}dx}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o|x−x_p{|}^2}\widehat{i}$

Using the substitution, $u=x–x_P$we get the above equation as:

$\begin{array}{c}\stackrel{→}{E}=\frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}\underset{−4.0m}{\overset{−1.0m}{}}\frac{du}{u^2}\widehat{i}\\ =\frac{\mathrm{λ}}{4\mathrm{Ï€}{\mathrm{Ï\mu }}_o}[\frac{−1}{−\text{1.0m}}−\frac{−1}{−\text{4.0m}}]\widehat{i}\\ =81[\frac{−1}{−\text{1.0m}}−\frac{−1}{−\text{4.0m}}]\widehat{i}\\ =\left(60.75\text{N/C}\right)\widehat{i}\\ ≈\left(61\text{N/C}\right)\widehat{i}\end{array}$

Hence, the value of the magnitude of the electric field is.$61\text{N/C}$