A semicircular ring of radius R carries a uniform linear charge density of $λ$ . P is a point in the plane of the ring at a distance R from centre O. OP is perpendicular to AB. Then the electric field intensity at point P is $(K = \frac{1}{4 π ε_{0}})$

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$K \frac{λ}{2 R}$

$K \frac{λ \sqrt{2}}{R}$

$K \frac{λ}{R} \ln (\sqrt{2} + 1)$

$K \frac{λ}{\sqrt{2} R}$

answer is C.

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Detailed Solution

Consider an element of angular width $d ϕ$ as shown in figure. Change on the element $d q = λ R d ϕ$
Distance of the element from point $p = 2 R \cos θ$
Field at P due to the element is $d E = K \frac{d q}{r^{2}} = K \frac{λ R d ϕ}{{(2 R \cos θ)}^{2}}$
Field $= 2 \frac{K λ}{2 R} \int_{0}^{\frac{π}{4}} \sec θ d θ$
$= \frac{K λ}{2 R} [\ln (\sqrt{2} + 1)]$