Q33P - Fundamentals Of Physics

Question

In Fig. 22-56, a “semi-infinite” non-conducting rod (that is, infinite in one direction only) has uniform linear charge density l. Show that the electric field   $\overrightarrow{{\mathbf{E}}_{\mathbf{p}}}$at point P makes an angle of  $\mathbf{45}\mathbf{°}$with the rod and that this result is independent of the distance R. (Hint: Separately find the component of  $\overrightarrow{{\mathbf{E}}_{\mathbf{p}}}$parallel to the rod and the component perpendicular to the rod.)

Step-by-Step Solution

Answer

The electric field$\overrightarrow{E_p}$  at point P makes an angle with the rod and it is independent of the distance R.

Step 1: The given data

A semi-infinite non-conducting rod (infinite in one direction only) has uniform charge density, $\lambda =l$.

 

Step 2: Understanding the concept of electric field

Using the concept of the electric field at an axial point, we can find the net electric field of the point at a distance from the rod and extending to infinity from one direction only.

Consider an infinitesimal section of the rod of length, a distance  from the left end, as shown in the following diagram. It contains charge, $dq=\lambda dx$ 

 

Formula:

The magnitude of the electric field due to the rod at a point,  $dE=\frac{1}{4\pi {\epsilon }_o}\frac{\lambda dx}{r^2}\widehat{r}$       (i)

Where,$\lambda$ is the linear charge density of the charge distribution,

r is the distance of the point from the small charge element. 

The angle between two components of the vectors can be given as:

     $\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$                                                                                                      (ii)

Step 3: Calculation of the angle made by the electric field

The magnitude of x and the y components of the electric field for that small charge are given using equation (i) as follows:

$d{E}_x=-\frac{1}{4\pi {\epsilon }_o}\frac{\lambda dx}{r^2}sin\theta$

and $d{E}_y=-\frac{1}{4\pi {\epsilon }_o}\frac{x}{r^2}cos\theta$

We use $\theta$ as the variable of integration and now substituting, 

,$$\begin{gathered} r=R/cos\theta \\ \begin{array}{c}x=Rtan\theta \\ dx=(R/co{s}^2\theta )d\theta \end{array} \end{gathered}$$

The limits of integration are . $0and\pi /2rad$

Thus, the x-component of the electric field can be given using the above substituted values as follows:

$\begin{array}{c}{E}_x=-\frac{\lambda }{4\pi {\epsilon }_{o}R}\underset{0}{\overset{\frac{\pi }{2}}{}}\text{sin}\theta d\theta \\ =\frac{\lambda }{4\pi {\epsilon }_{o}R}{\left[\text{cos}\theta \right]}_0^{\frac{\pi }{2}}\end{array}$. 

$\Rightarrow E_x=-\frac{\lambda }{4\pi {\epsilon }_{o}R}$  ……. (a)

Now, the y-component of the electric field is given using above values as follows:

$\begin{array}{c}{E}_y=-\frac{\lambda }{4\pi {\epsilon }_{o}R}\underset{0}{\overset{\frac{\pi }{2}}{}}\text{cos}\theta d\theta \\ =-\frac{\lambda }{4\pi {\epsilon }_{o}R}{\left[\text{sin}\theta \right]}_0^{\frac{\pi }{2}}\end{array}$ 

$\Rightarrow E_y=-\frac{\lambda }{4\pi {\epsilon }_{o}R}$  ……. (b)

Now, the angle of the electric field with the rod is given using equations (a) and (b) in equation (ii) as:

$\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{-\frac{\lambda }{4\pi {\epsilon }_{o}R}}{-\frac{\lambda }{4\pi {\epsilon }_{o}R}}\right)\\ ={45}^0\end{array}$

Hence, the value of the required angle is ${45}^0$and from the data given, we can say that for equal value of x and y components of electric field, the field makes ${45}^0$angle with all points of the rod.