91影视

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

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,$dq=位dx$

Formula:

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

Where,$位$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" $胃={\tan }^{鈭�1}\left(\frac{E_y}{E_x}\right)$ (ii)

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

$d{E}_x=鈭�\frac{1}{4蟺{蔚}_o}\frac{位dx}{r^2}sin胃$

and$d{E}_y=鈭�\frac{1}{4蟺{蔚}_o}\frac{x}{r^2}cos胃$

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

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

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

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

$\begin{array}{c}{E}_x=鈭�\frac{位}{4蟺{蔚}_{o}R}\underset{0}{\overset{\frac{蟺}{2}}{}}\text{sin}胃d胃\\ =\frac{位}{4蟺{蔚}_{o}R}{\left[\text{cos}胃\right]}_0^{\frac{蟺}{2}}\end{array}$.

$鈬�E_x=鈭�\frac{位}{4蟺{蔚}_{o}R}$ 鈥︹��. (a)

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

$\begin{array}{c}{E}_y=鈭�\frac{位}{4蟺{蔚}_{o}R}\underset{0}{\overset{\frac{蟺}{2}}{}}\text{cos}胃d胃\\ =鈭�\frac{位}{4蟺{蔚}_{o}R}{\left[\text{sin}胃\right]}_0^{\frac{蟺}{2}}\end{array}$

$鈬�E_y=鈭�\frac{位}{4蟺{蔚}_{o}R}$ 鈥︹��. (b)

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

$\begin{array}{c}胃={\tan }^{鈭�1}\left(\frac{鈭�\frac{位}{4蟺{蔚}_{o}R}}{鈭�\frac{位}{4蟺{蔚}_{o}R}}\right)\\ ={45}^0\end{array}$

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.