The figure is below. All the forces represented with thin lines are equal in module. The electric field lines exits the positive charges and enters the negative charge. Thus the direction of forces are (see the figure)
`F_(12) =F_(21)= k_e*q^2/R^2=9*10^9*3.92^2*10^-12/0.0556= 2.4874 N`
away from charges 1 and 2, on the...
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The figure is below. All the forces represented with thin lines are equal in module. The electric field lines exits the positive charges and enters the negative charge. Thus the direction of forces are (see the figure)
`F_(12) =F_(21)= k_e*q^2/R^2=9*10^9*3.92^2*10^-12/0.0556= 2.4874 N`
away from charges 1 and 2, on the 12 line
`F_(13)=F_(23) = k_e*q^2/R^2 =2.4874 N` ,
away from charges 1 and 2 towards 3
`F_31 =F_32=k_e*q^2/R^2 =2.4874 N` ,
away from charge 3 towards 1 and 2
Now the resultant forces on charges 1, 2 and 3 are (see the law of cosine in a triangle)
`F_1 = sqrt(F_12^2+F_13^2-2*F_12*F_13*cos(60)) =2.4874 N`
`F2 =F1 =2.4874 N` by symmetry arguments
`F_3 =sqrt(F_(31)^2 +F_(32)^2-2*F_31*F_32*cos(120)) =4.3083 N`
`F_1` and `F_2` make an angle of 60 degree with the horizontal and `F3` is downwards.
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