for equilibrium net, the electric field on third must be zero
suppose net field is zero from +3q at x distance so
E_net = 0
field due to +3q = field due to +q
3q/x^2 = q/ ( d-x)^2
3/x^2 = 1/(d-x)^2
x^2 / (d-x)^2 = 1/3
x = 0.6339*d
is the only solution
so this is the distance from +3 q where samll bead willl be at equblirium
b)
electric potential
= 3q/0.6339*d +q/ ( d-0.6339*d )
= q/d * ( 3/0.6339 + 1/(1-0.6339) )
= 7.464*q/d
C)
if you take -3q and -q only potential will change but the equilibrium position will be same as before
new potential = - 7.464*q/d
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