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# Good evening sir.This is Gauss law application.Explain it sir.

Arun
25763 Points
2 years ago
Dear Raju

let point of intersection of ring & sphere are P,Q ...let center of ring is C & center of sphere is C1 ...

CC1 bisects PQ in two equal parts , if S is the point of intersection of CC1 & PQ then

in triangle right angled PCS

PC = R , CS = R/2 so angle PCS = @

using trigonometry , cos@ = CS/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 2pi/3 radian q1 = (q/2pi)(2pi/3) = q/3

total flux through sphere = total charge enclosed/$\LARGE \varepsilon$o = q/3$\LARGE \varepsilon$o

this is the required flux