JEE PYQ: Electrostatics Question 13
Question 13 - 2021 (26 Feb Shift 1)
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 $a = \frac{\sqrt{3}}{2}L$.
(1) $\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$
(2) $\frac{\sqrt{3}Q}{4\pi\varepsilon_0 L^2}$
(3) $\frac{Q}{3\pi\varepsilon_0 L^2}$
(4) $\frac{Q}{4\pi\varepsilon_0 L^2}$
Show Answer
Answer: (1)
Solution
$\tan\theta = \frac{L/2}{\frac{\sqrt{3}}{2}L} = \frac{1}{\sqrt{3}}$, so $\theta = 30°$. $E_{\text{net}} = \frac{K\lambda}{a}(2\sin 30°) = \frac{2KQ}{\sqrt{3}L^2}\left(\frac{1}{2} + \frac{1}{2}\right) = \frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$.
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