Question

The direction $(\theta )$ of $\vec{E}$ at point $P$ due to uniformly charged finite rod will be

• A. At angle $30^{\circ }$ from $X$ -axis
• B. At angle $45^{\circ }$ from X-axis
• C. At angle $60^{\circ }$ from X-axis
• D. None of these

Answer 1

Answer: (A)

$E_x=\frac{\lambda }{4\pi {\epsilon }_{0}x}[\sin \alpha +\sin \beta ]$
$E_x=\frac{\lambda }{4\pi {\epsilon }_{0}d}[0+\sin 60^{\circ }$
$E_x=\frac{\lambda }{4\pi {\epsilon }_{0}d}\left[0+\frac{\sqrt{3}}{2}\right]=\frac{\sqrt{3}\lambda }{8\pi {\epsilon }_{0}d}$
Similarly, $E_y=\frac{x}{4\pi {\epsilon }_{0}d}\left[\cos 0^{\circ }+\cos 60^{\circ }\right]$
$E_y=\frac{\lambda }{4\pi {\epsilon }_{0}d}\left[1+\frac{1}{2}\right]$
$E_x=\frac{\sqrt{3}\lambda }{8\pi {\epsilon }_{0}d},E_y=\frac{3\lambda }{8\pi {\epsilon }_{0}d}$
$\tan \theta =\frac{E_y}{E_x}=\frac{1}{\sqrt{3}}$
$\Rightarrow \theta =30^{\circ }$