A charge (uniform linear density$\mathbf{=}\mathbf{}\mathbf{9}\mathbf{.0}\mathbf{}\mathbf{\text{nC/m}}$) lies on a string that is stretched along an xaxis from$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$to$\mathit{x}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{.0}\mathbf{}\mathbf{\text{m}}$. Determine the magnitude of the electric field at$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.0}\mathbf{}\mathbf{\text{m}}$on the xaxis.

1. Uniform charge density,$\lambda =9\text{nC/m}$
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: $d\overrightarrow{E}=\frac{dq}{4\pi {ϵ}_o|x-x_p{|}^2}$ (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,$\lambda dx$where.$\lambda =9.0x{10}^{–9}\text{C/m}$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:

$\overrightarrow{E}=\underset{0}{\overset{3.0m}{}}\frac{\lambda dx}{4\pi {ϵ}_o|x-x_p{|}^2}\widehat{i}$

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

$\begin{array}{c}\overrightarrow{E}=\frac{\lambda }{4\pi {ϵ}_o}\underset{-4.0m}{\overset{-1.0m}{}}\frac{du}{u^2}\widehat{i}\\ =\frac{\lambda }{4\pi {ϵ}_o}[\frac{-1}{-\text{1.0m}}-\frac{-1}{-\text{4.0m}}]\widehat{i}\\ =81[\frac{-1}{-\text{1.0m}}-\frac{-1}{-\text{4.0m}}]\widehat{i}\\ =\left(60.75\text{N/C}\right)\widehat{i}\\ \approx \left(61\text{N/C}\right)\widehat{i}\end{array}$

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