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Two idential pith balls. Each of mass 'm' and charges 'q' hang from noncoducting threads of length I as shown in figure , thetais very small and hence tan theta ~~sin theta . Show that , for equilibriumx=[( q^(2) l)/( 2pi in _0mg) ]^((1)/(3)) |
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Answer» Solution :For equilibrium `T sin theta =F_e= ( 1)/( 4pi in _0) .( q^(2))/( x^(2))` ` "" T cos theta =MG ` ` "" therefore tan theta =sin theta =(x)/((2)/(l) =(x)/( 2L ) ` `(x)/(2l) =( q^(2))/( 4pi in _0 x^(2)mg ) ` ` x^(2) =( 2Q^(2) l)/( 4pi in_0mg ) "" therefore x= [(q^(2) l)/( 2PI in _0mg ) ]^(l) ` 
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