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?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{2 R}{π}$

$\frac{R}{2 π}$

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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