Question

Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q$. The distance of the point $P$ from the centre of the rod is $a=\frac{\sqrt{3}}{2}L$

• A. $\frac{\sqrt{3}Q}{4\pi {\epsilon }_0{L}^2}$
• B. $\frac{Q}{3\pi {\epsilon }_0{L}^2}$
• C. $\frac{Q}{2\sqrt{3}\pi {\epsilon }_0{L}^2}$
• D. $\frac{Q}{4\pi {\epsilon }_0{L}^2}$

Answer 1

Answer: (C)

$E=\frac{k\lambda }{a}\left(\sin {\theta }_1+\sin {\theta }_2\right)$
$E=\frac{1}{4\pi {\epsilon }_0}\times \frac{Q}{L}\times \frac{1}{\left(\frac{\sqrt{3}L}{2}\right)}\times (2\sin \theta )$
$\tan \theta =\frac{L/2}{\frac{\sqrt{3}L}{2}}=\frac{1}{\sqrt{3}}$
$\sin \theta =\frac{1}{2}$
$E=\frac{1}{4\pi {\epsilon }_0}\times \frac{2Q}{\sqrt{3}{L}^2}\times \left(2\times \frac{1}{2}\right)$
$E=\frac{Q}{2\sqrt{3}\pi {\epsilon }_0{L}^2}$