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 = √3/2 L.

Answer 1

The correct option is (3).

\(E_{net} = \frac \lambda{4\pi \varepsilon_0r} (\sin \alpha + \sin \beta)\)

Here,

\(\lambda = \frac QL\), \(r = \frac{\sqrt 3}2 L\), \(\sin \alpha = \sin \beta = \frac 12\)

⇒ \(E_{net } = \frac{2Q}{4\pi \varepsilon_0\sqrt 3L^2} (\frac 12 + \frac 12)\)

⇒ \(E_{net } = \frac{Q}{2\sqrt 3\pi \varepsilon_0L^2}\)

Answer 2

Correct option is