Two idential pith balls. Each of mass 'm' and charges 'q' hang from noncoducting threads of length I as shown in figure , `theta ` is very small and hence `tan theta ~~sin theta . `Show that , for equilibrium ` x=[( q^(2) l)/( 2pi in _0mg) ]^((1)/(3)) `

Explanation

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) `