kaneez fathima 9th Jul, 2020

Dear Chirag,

solution to this problem is:

F1=force at A due to charge at B

F1=1/4piE X q^2/l^2 along BA

F2=Force at A due to charge C

=I/4 pi E X q^2/l^2 along CA

F1+F2=sqrt(F1*^2+F2^2+2f1f2cos 60)

=F1*(sqrt(3))=sqrt(3)*q^2/4piEl^2 along GA

Force at A due to charge Q at centroid G=1/4 pi e*Qq/AG^2=Qq/4 pi E (l/sqrt(3))^2=3Qq/4 pi E l^2

This is equal and opposite to F1+F2

3Qq/4piEl^2=-sqrt(3)q^2/4 pi E l^2

Q=q/sqrt(3)