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![[Post New]](/templates/default/images/icon_minipost_new.gif) 19 Jun 2008 12:03:28 IST
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the third charge is kept towards left of q2 .....so tat the force on q2 becomes zero.....now find unit vector along line joining q1 and q2.....from q1 to q2....now in equillibrium forces r equal....(fq2q3 and fq1q2)....so from this find distance b/w q2 and q3...(mgnitude),,by applying simple formul.....and now u have to find position vector of q3....so multiply the magnitude of distance which u have find above by formula....to unit vector which u have found above,,,this give vector bc....means position vector of line joining q2q3...,now suppose the position vector of third charge is rc and tt of q2 is rb....so rcb = vector u have find above already...so from this u will find rc.....
if any doubt nusdge me,,,
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