Question

Find the direction $\theta$ of $\overrightarrow{E}$ at point $\text{P}$ due to uniformly charged finite rod will be at an angle

• A. ${30}^{\circ }$ from $\text{x-}$axis
• B. ${45}^{\circ }$ from $\text{x-}$axis
• C. ${60}^{\circ }$ from $\text{x-}$axis
  
• D. none of these

Answer 1

The correct option is A ${30}^{\circ }$ from $\text{x-}$axis

Electric field parallel to the wire of charge per unit length $\lambda$ at perpendicular distance $d$ from the wire is, $$E_y=E_{\text{par}}=\frac{K\lambda }{d}[\cos {\theta }_2-\cos {\theta }_1]..\left(1\right)$$ Where, the directions of ${\theta }_1$ and ${\theta }_2$ are given below,


Comparing the given diagram with the above diagram, we get:
${\theta }_1={60}^{\circ }$
${\theta }_2=0^{\circ }$

Substituting the values in equation $\left(1\right)$, we get

$E_y=\frac{K\lambda }{d}[\cos 0^{\circ }-\cos {60}^{\circ }]$

$\Rightarrow E_y=\frac{K\lambda }{d}[1-1/2]$

$\Rightarrow E_y=\frac{K\lambda }{2d}$

Now,
Electric field perpendicular to the wire of charge per unit length $\lambda$ at perpendicular distance $d$ from the wire is, $$E_x=E_{\text{Perp}}=\frac{K\lambda }{d}[\sin {\theta }_2+\sin {\theta }_1]$$ Substituting the values, we get

$\Rightarrow E_x=\frac{K\lambda }{d}[\sin 0^{\circ }+\sin {60}^{\circ }]$

$\Rightarrow E_x=\frac{\sqrt{3}K\lambda }{2d}$


Now, the direction $\theta$ of net electric field $\overrightarrow{E}$ with $\text{x}-$axis is given by,

$\tan \theta =\frac{E_y}{E_x}$

$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={30}^{\circ }$

Hence, option (a) is the correct answer.