Electrostatics

JEE Main level, single correct, medium

105 of 311

Question 105 of 311

Find the electric field at point PP (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length LL carrying a charge QQ. The distance of the point PP from the centre of the rod is a=32La=\dfrac{\sqrt{3}}{2}L

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Answer: (B) Q23πϵ0L2\dfrac{Q}{2\sqrt{3}\pi {\epsilon }_{0}{L}^{2}}

E=kλa(sin⁡θ1+sin⁡θ2)E=\dfrac{k\lambda }{a}\left(\sin {\theta }_{1}+\sin {\theta }_{2}\right)
E=14πϵ0×QL×1(3L2)×(2sin⁡θ)E=\dfrac{1}{4\pi {\epsilon }_{0}}\times \dfrac{Q}{L}\times \dfrac{1}{\left(\dfrac{\sqrt{3}L}{2}\right)}\times \left(2\sin \theta \right)
tan⁡θ=L/23L2=13\tan \theta =\dfrac{L/2}{\dfrac{\sqrt{3}L}{2}}=\dfrac{1}{\sqrt{3}}
sin⁡θ=12\sin \theta =\dfrac{1}{2}
E=14πϵ0×2Q3L2×(2×12)E=\dfrac{1}{4\pi {\epsilon }_{0}}\times \dfrac{2Q}{\sqrt{3}{L}^{2}}\times \left(2\times \dfrac{1}{2}\right)
E=Q23πϵ0L2E=\dfrac{Q}{2\sqrt{3}\pi {\epsilon }_{0}{L}^{2}}

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