Q67P - Fundamentals Of Physics

Question

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 x axis 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 x axis.

Step-by-Step Solution

Answer

The magnitude of the electric field at$x=4.0\text{m}$ on the x-axis is.$61\text{ N/C}$

Step 1: The given data

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

Step 2: Understanding the concept of the electric field

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

Step 3: Calculation of the electric field at x   =   4 .0   m on the x-axis

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.$61\text{ N/C}$