q=1 C.
a=10 cm=0.1 m
F_12=F_13=F=(kq^2)/a^2
F_1x=F+F cos60=F+1/2 F=3/2 F
F_1y=F sin60=√3/2 F
Thus,
F_1=√((3/2 F)^2+(√3/2 F)^2 )=√3 F
F_1=√3 (kq^2)/a^2
F_1=√3 (9∙〖10〗^9 ) (1/0.1)^2=1.56∙〖10〗^12 N.
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