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 Pmakes 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.)

Short 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-by-step solution

The given data

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

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 distancefrom 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:

localid="1661918090783" $\theta ={\tan }^{-1}\left(\frac{E_y}{E_x}\right)$ (ii)

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.