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QJEEM21Ph4946: JEE Main 2021 Physics - Class 12 JEEM21Ph4946 Solution: Find the electric field at point P (as shown in figure) on the…

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 = 32L{{\sqrt 3 } \over 2}L .

JEE Main 2021 (Online) 26th February Morning Shift Physics - Electrostatics Question 135 English

Multiple Choice Options

  • A Q4πε0L2{Q \over {4\pi {\varepsilon _0}{L^2}}}
  • B Q3πε0L2{Q \over {3\pi {\varepsilon _0}{L^2}}}
  • C Q23πε0L2{Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}
  • D 3Q4πε0L2{{\sqrt 3 Q} \over {4\pi {\varepsilon _0}{L^2}}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 26th February Morning Shift Physics - Electrostatics Question 135 English Explanation
tan⁡θ=L/232L⇒13\tan \theta = {{L/2} \over {{{\sqrt 3 } \over 2}L}} \Rightarrow {1 \over {\sqrt 3 }}

θ=30∘\theta = 30^\circ

Enet=Kλ32L(sin⁡30∘+sin⁡30∘)=2KQ3L2(12+12){E_{net}} = {{K\lambda } \over {{{\sqrt 3 } \over 2}L}}(\sin 30^\circ + \sin 30^\circ ) = {{2KQ} \over {\sqrt 3 {L^2}}}\left( {{1 \over 2} + {1 \over 2}} \right)

Enet=14πε02Q3L2{E_{net}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over {\sqrt 3 {L^2}}}

Enet=Q23πε0L2{E_{net}} = {Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}

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