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Two small beads having positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to the point x=d as shown in fig. A third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium ?

Text Solution

Verified by Experts

The correct Answer is:
`(sqrt(3)d)/(sqrt(3)+1)`


For Q to be in equilibrium
`(k3qQ)/(x^(2)) =(kQq)/((d-x)^(2))`
or , `x=(sqrt(3)d)/(sqrt(3)+1)`
The bead can be in stable equilibrium, if it has positive charges.
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