Question

A ring of radius r carriges a charge Q uniformly distributed over its length. A charge q placed at its centre will experience a force is equal to

Answer 1

let point of intersection of ring & sphere are P,Q ...let center of ring is O & center of sphere is O1 ...

OO1 bisects PQ in two equal parts , if S is the point of intersection of OO1 & PQ then

in triangle right angled POS

PO = R , OS = R/2 so angle POS = @

using trigonometry , cos@ = OS/PS = 1/2

@ = pi/3

total angle substended by arc is 2pi/3...this is the arc of ring which lies inside sphere & substends

2pi/3 angle at the center of ring...





for 2pi radian charge = q

for unit radian = q/2pi

for 2pi/3 radian q1 = (q/2pi)(2pi/3) = q/3

total flux through sphere = total charge enclosed/ebsilen = q/3ebsilen

this is the required flux