Question

A small bead with a positive charge q is free to slide on a horizontal wire of length 4.5 cm . At the left end of the wire is a fixed charge q, and at the right end is a fixed charge 4q. How far from the left end of the wire does the bead come to rest?

Answer 1

Answer:

1.5cm

Explanation:

The place where the bead will come to rest is the place where the force between the left charge and the bead is the same, but in opposite direction, as he force between the right charge and the bead.

$F_{left} = F_{right}\\K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}$

Also:

$r_{left}+r_{right}=0.045m\\r_{right} = 0.045m-r_{left}$

$q_{left}= q\\q_{right}=4q$

So, we replace and simplify. Notice that the charges and the Coulomb constant will cancel:

$K\frac{q_{left}*q_b}{r_{left}^2} =K\frac{q_{right}*q_b}{r_{right}^2}\\\frac{q*q}{r_{left}^2} = \frac{4q*q}{(0.045m-r_{left})^2}\\(0.045m-r_{left})^2 =4r_{left}^2\\0.002025m - 2*0.045r_{left} + r_{left}^2 = 4r_{left}^2\\3r_{left}^2 + 0.09r_{left} - 0.002025m = 0$

$r_{left} = \frac{-(0.09)+-\sqrt{(0.09)^2-4*(3)(-0.002025)}}{2(3)}\\ r_{left} = 0.015m| -0.045m$

But the only solution that would place the bead on the wire is 0.015m or 1.5cm, so this is our answer.