1) Let the small bead be placed at a distance x from +3q and in between both the charges
the third charge will be in equilibrium if the net force on it zero
so Force on third bead by +3q = force on third bead by +q
kQ*3q / x^2 = k*Qq / (d - x)^2
3(x^2 + d^2 - 2xd) = x^2
2x^2 - 6xd + 3d^2
x = +6d - sqrt( (6d)^2 - 4*2*3d^2)) / 2*2
= 0.64 d
so the bead will be positioned at 0.64d from the +3q charge
+ Q1. Two small beads having positive charges 39 and q are fixed at the opposite...
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