Three charges +q each are kept at the vertices A, B and C of an equilateral triangle of side L. The electric field intensity at the mid point of BC is *
Answer (1)
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)
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