Charges of +2microC, +3microC and -8microC are placed in air at the vertices of an equilateral triangle of ide 10cm.What is the magnitude of the force acting on the -8microC due to the other two charges?

1 Answer
Mar 14, 2018

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Let charge 2 muC,3muC,-8 muC2μC,3μC,8μC are placed at point A,B,CA,B,C of the triangle shown.

So,net force on -8 muC8μC due to 2muC2μC will act along CACA

and the value is F_1=(9*10^9*(2*10^-6)*(-8)*10^-6)/(10/100)^2=-14.4NF1=9109(2106)(8)106(10100)2=14.4N

And due to 3muC3μC it will be along CBCB i.e F_2=(9*10^9*(3*10^-6)(-8)*10^-6)/(10/100)^2=-21.6NF2=9109(3106)(8)106(10100)2=21.6N

So,two forces of F_1F1 and F_2F2 are acting on the charge -8muC8μC with an angle of 60^@60 in between,so the nect force will be, F= sqrt(F_1^2 +F_2^2 + 2F_1 F_2 cos 60)=31.37NF=F21+F22+2F1F2cos60=31.37N

Making an angle of tan^-1((14.4 sin 60)/(21.6+14.4 cos 60))=29.4^@tan1(14.4sin6021.6+14.4cos60)=29.4 with F_2F2