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 center of the rod is $a = \frac{\sqrt{3}}{2} L .$

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$\frac{Q}{4 π ε_{0} L^{2}}$

$\frac{Q}{2 \sqrt{3} π ε_{0} L^{2}}$

$\frac{\sqrt{3} Q}{4 π ε_{0} L^{2}}$

$\frac{Q}{3 π ε_{0} L^{2}}$

answer is B.

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Detailed Solution

$\begin{array}{l} E_{n e t} = \frac{λ}{4 π ε_{0} r} (\sin α + \sin β) \\ h e r e \\ λ = \frac{Q}{L}, r = \frac{\sqrt{3}}{2} L, \sin α = \sin β = \frac{1}{2} \\ \Rightarrow E_{n e t} = \frac{2 Q}{4 π ε_{0} \sqrt{3} L^{2}} (\frac{1}{2} + \frac{1}{2}) \\ \Rightarrow E_{n e t} = \frac{Q}{2 \sqrt{3} π ε_{0} L^{2}} \end{array}$