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Two identical pith balls each of mass .m. holding charge .q. each are suspended by silk threads of equal length from same point. They move apart due to repulsion. If the separation between the two balls is 2x and each string makes small angle theta to the vertical, |
Answer» Solution : From fig `T cos theta = mg, T sin theta = F` `(F)/(mg) = tan theta = (x)/(sqrt(1^(2)-x^(2)))` If `theta` is small `Tantheta = sin theta ~~ theta = (x)/(1), (F)/(mg) (x)/(1)` Where `F= (1)/(4pi in_(0))(q^(2))/(4x^(2)) rArr (q^(2))/(16pi in_(0)x^(2) mg) = (x)/(1)` `rArr x= ((q^(2)1)/(16pi in_(0)mg))^(1//3)` `rArr` Tension in the string `T= sqrt(F^(2) + (mg)^(2))` In the above CASE if the BALLS are suspended in a liquid of density `rho` and the distance between the balls remains the same, then `(F)/(mg) =(F^(1))/(mg^(1)) rArr (F)/(F^(1)) = (mg)/(mg^(1)) rArr (mg)/(mg(1-(rho)/(d)))( because k = (F_(a))/(F_(m)))` `rArr (1- (rho)/(d)) = (1)/(K) or K = (d)/((d-p))` (d is density of material of the ball) `rArr` In the above case if the charges on teh two balls are different and the angles made by those two STRINGS to the VERTICAL are `theta_(1) and theta_(2)` respectively, then `theta_(1) = theta_(2)` (as F, mg are same ) if `m_(1) gt m_(2) " then " theta_(1)lt theta_(2)` `rArr` In the above application if the whole setup is kept in an artifical satellite (in zero gravity region) the angle between the two strings is `180^(@)` and tension in each string is `(1)/(4pi in_(0))(q^(2))/(41^(2))` |
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