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thomq
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1. Two small beads having a positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to point X. This length is the distance (d). A third small charged bead (Q) is free to slide on the rod. At what position is the third bead in equilibrium?
2. Coulomb's Law is used for this problem. F = kqq/d
F[1][/SUB] = kQ(3q)/x[2][/SUP]
F[2][/SUB] = kQq/(d-x)[2][/SUP]
F[1][/SUB] = F[2][/SUB]
3. I am attempting to solve for x.
3 (d-x)[2][/SUP] = (x)[2][/SUP]
I know I could make both side a square root so I have x one side, but I need help beyond this.
I also need to show all steps. Thanks!
2. Coulomb's Law is used for this problem. F = kqq/d
F[1][/SUB] = kQ(3q)/x[2][/SUP]
F[2][/SUB] = kQq/(d-x)[2][/SUP]
F[1][/SUB] = F[2][/SUB]
3. I am attempting to solve for x.
3 (d-x)[2][/SUP] = (x)[2][/SUP]
I know I could make both side a square root so I have x one side, but I need help beyond this.
I also need to show all steps. Thanks!