Question

A semi-infinite insulating rod has linear charge density $\lambda$ . The electric field at the point P shown in figure is :-

(A) ${\displaystyle \frac{2{\lambda }^2}{{\left(4\pi {\epsilon }_{0}r\right)}^2}}$ at ${45}^{\circ }$ with AB
(B) ${\displaystyle \frac{\sqrt{2}\lambda }{\left(4\pi {\epsilon }_0{r}^2\right)}}$ at ${45}^{\circ }$ with AB
(C) ${\displaystyle \frac{\sqrt{2}\lambda }{\left(4\pi {\epsilon }_{0}r\right)}}$ at ${45}^{\circ }$ with AB
(D) ${\displaystyle \frac{\sqrt{2}\lambda }{\left(4\pi {\epsilon }_{0}r\right)}}$ at perpendicular with AB

Answer 1

Hint: We will calculate the x-component and the y-component of the electric field at the point P, as shown in the figure, due to the semi-infinite insulating rod carrying uniform linear charge of density $\lambda$ .We integrate the electric field $d{E}_x$ over the length of the “semi-infinite” rod that is with respect to the variable $x$ from 0 to $\mathrm{\infty }$ .

Complete step by step answer:
It has been given that a semi-infinite insulating rod has linear charge density $\lambda$ .

Consider an infinitesimal charge element of length $dx$ at distance $x$ from the end of the rod as shown in the figure.
 $\cos \theta ={\displaystyle \frac{x}{\sqrt{r^2+x^2}}}$ by Pythagoras Theorem, which states that the square of the hypotenuse is equal to the square of the other two sides of a right angled-triangle.
As the direction of electric field due to a charge element is along the line joining the element to the point where the field is to be calculated.
From the figure, the x-component of the field will be,
 $d{E}_x=dE\cos \theta$
 $\Rightarrow d{E}_x={\displaystyle \frac{\lambda dx}{\left(r^2+x^2\right)}}\times {\displaystyle \frac{x}{\sqrt{r^2+x^2}}}$
Integrating $d{E}_x$ over the length of the “semi-infinite” rod that is with respect to the variable $x$ from 0 to $\mathrm{\infty }$ , we get
 $\int d{E}_x=k\lambda \int {\displaystyle \frac{xdx}{{\left(r^2+x^2\right)}^{3/\phantom{32}2}}}$
For simplicity we take $t=r^2+x^2$ .
Thus, $dxdx=dt$ .
On simplifying the equation, we can get,
 $\int d{E}_x=k\lambda \int {\displaystyle \frac{dt}{{\left(t\right)}^{3/\phantom{32}2}}}$ .
By rules of integration,
 $\int dE={\displaystyle \frac{k\lambda }{2}}\times {\displaystyle \frac{-3}{2}}{t}^{-1/\phantom{12}2}$ .
Substituting, the value of $t=r^2+x^2$ , and setting limits of integration at 0 and $\mathrm{\infty }$ ,
 $\int d{E}_x=k\lambda {\left[{\displaystyle \frac{1}{\sqrt{r^2+x^2}}}\right]}_0^{\mathrm{\infty }}$
 $\Rightarrow E_x={\displaystyle \frac{k\lambda }{r}}$
We note that $\left|E_x\right|=\left|E_y\right|$ for all $r$ .
Therefore, the angle that the electric field vector makes at P is
 $\tan \theta ={\displaystyle \frac{\left|E_x\right|}{\left|E_y\right|}}$ .
Since $\left|E_x\right|=\left|E_y\right|$ , $\tan \theta =1$ .
It implies that $\theta ={45}^{\circ }$ and is independent of $r$ , that is the distance of the point P from the edge of the “semi-infinite” rod.
Now, the net electric field is given by, $E_{net}=\sqrt{{E_X}^2+{E_Y}^2}=\sqrt{2}{\displaystyle \frac{k\lambda }{r}}$ where $k={\displaystyle \frac{1}{4\pi {\epsilon }_0}}$ is a constant.
Hence the net electric field is ${\displaystyle \frac{\sqrt{2}\lambda }{\left(4\pi {\epsilon }_{0}r\right)}}$ at ${45}^{\circ }$ with AB.
The correct answer is Option C.

Note:
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge.