Two uniformly charged semi circular rings of radius R and linear charge density λ, are placed in y-z and x-z plane as shown. Electric field intensity at point P is λnε0R. Where point P is intersection point of axis of both rings and at distance R from center of both rings. Find value of n.

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Two uniformly charged semi circular rings of radius R and linear charge density $λ,$ are placed in y-z and x-z plane as shown. Electric field intensity at point P is $\frac{λ}{n ε_{0} R} .$ Where point P is intersection point of axis of both rings and at distance R from center of both rings. Find value of n.

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

answer is 8.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At P, the axial component of electric field of both half rings will be at $90 °$ and components to the axis will cancel each other, being at $180 °$
$E_{n e t} = E \sqrt{2}$
Where E is field due to one half ring.
$= [\frac{k λ (π R)}{{(R^{2} + R^{2})}^{3 / 2}} . R] \sqrt{2}$