Question

A uniformly charged finite rod is placed along $x-$axis, as shown in the figure. At point $P$, find the angle made by $\overrightarrow{E}$ with $x-$direction.

• A. ${45}^{\circ }$
• B. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
• D. None of these
  

Answer 1

The correct option is A ${45}^{\circ }$


We know,
$E_{||}=\frac{k\lambda }{d}(\cos \beta -\cos \alpha )$
$\Rightarrow E_{||}=\frac{k\lambda }{d}(\cos {30}^{\circ }-\cos {60}^{\circ })$

$[\because \alpha =+{60}^{\circ }$ and $\beta =-{30}^{\circ },\cos (-{30}^{\circ })=\cos {30}^{\circ }]$

and,

$E_{\perp }=\frac{k\lambda }{d}(\sin \alpha +\sin \beta )$
$\Rightarrow E_{\perp }=\frac{k\lambda }{d}(\sin {60}^{\circ }-\sin {30}^{\circ })$

$[\because \sin (-{30}^{\circ })=-\sin {30}^{\circ }]$

So, $\tan \theta =\frac{E_{\perp }}{E_{||}}=\frac{\sin {60}^{\circ }-\sin {30}^{\circ }}{\cos {30}^{\circ }-\cos {60}^{\circ }}$
$\Rightarrow \tan \theta =\frac{\frac{\sqrt{3}}{2}-\frac{1}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=1$
$\Rightarrow \theta ={45}^{\circ }$