A semicircular wire with radius R has uniform charge density $- λ$ . At all points along the “axis” of the semicircle (the line through the centre, perpendicular to the plane of the semicircle, as shown in Fig.), the vectors of the electric field all point towards a common point in the plane of the semi circle. What is the distance between the point of concurrency of the fields and centre of the ring?

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$\frac{2 R}{π}$

$\frac{R}{2 π}$

$\frac{R}{\sqrt{2}}$

$\frac{R}{\sqrt{3}}$

answer is A.

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Detailed Solution

$E_{y} = k \frac{2 λ R z d θ}{{(Z^{2} + R^{2})}^{\frac{3}{2}}} E_{y} = k \frac{2 λ R z d θ . z}{{(Z^{2} + R^{2})}^{\frac{3}{2}}} E_{y} = \frac{2 k λ R d θ}{{(Z^{2} + R^{2})}^{\frac{3}{2}}} (\frac{π}{2}) = \frac{K λ R π z}{{(z^{2} + R^{2})}^{\frac{3}{2}}} = \frac{λ R π z}{(4 π ε_{0}) {(z^{2} + R^{2})}^{\frac{3}{2}}} = \frac{λ R z}{(4 ε_{0}) {(z^{2} + R^{2})}^{\frac{3}{2}}} 2 (\frac{k λ R d θ}{(z^{2} + R^{2})} \frac{R}{{(z^{2} + R^{2})}^{\frac{3}{2}}}) \cos θ \sin θ \int_{0}^{\frac{π}{2}} \frac{2 k λ R^{2}}{(4 π ε_{0}) {(z^{2} + R^{2})}^{\frac{3}{2}}} = \frac{λ R^{2}}{2 π ε_{0} {(z^{2} + R^{2})}^{\frac{3}{2}}} \tan γ = \frac{\frac{λ R z}{4 ε_{0} {(z^{2} + R^{2})}^{\frac{3}{2}}}}{\frac{λ R^{2}}{2 π ε_{0} {(z^{2} + R^{2})}^{\frac{3}{2}}}} \frac{z 2 π}{(4) R} = \frac{z}{(\frac{2 R}{π})}$