JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 13)
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 = $${{\sqrt 3 } \over 2}L$$.
_26th_February_Morning_Shift_en_13_1.png)
$${Q \over {4\pi {\varepsilon _0}{L^2}}}$$
$${Q \over {3\pi {\varepsilon _0}{L^2}}}$$
$${Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}$$
$${{\sqrt 3 Q} \over {4\pi {\varepsilon _0}{L^2}}}$$
Explanation
_26th_February_Morning_Shift_en_13_2.png)
$$\tan \theta = {{L/2} \over {{{\sqrt 3 } \over 2}L}} \Rightarrow {1 \over {\sqrt 3 }}$$
$$\theta = 30^\circ $$
$${E_{net}} = {{K\lambda } \over {{{\sqrt 3 } \over 2}L}}(\sin 30^\circ + \sin 30^\circ ) = {{2KQ} \over {\sqrt 3 {L^2}}}\left( {{1 \over 2} + {1 \over 2}} \right)$$
$${E_{net}} = {1 \over {4\pi {\varepsilon _0}}}{{2Q} \over {\sqrt 3 {L^2}}}$$
$${E_{net}} = {Q \over {2\sqrt 3 \pi {\varepsilon _0}{L^2}}}$$
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