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)

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