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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 |
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Answer» 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 , [email protected] = 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 |
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