Please and thank you for your help! Two small beads having positive charges q,19q and92= q are fixed at the opposite ends of a horizontal insulating rod of length d = 1.50 m. The bead with charge q, is at the origin. As shown inthe figure below, a third small, charged bead is free to slide on the rod.92(a) At what position x is the third bead in equilibrium?X =(b) Can the equilibrium be stable?O Yes, if the third bead has a positive charge.O Yes, if the third bead has a negative charge.No

(a) At equilibrium the net force is zero F13=F23kq1qx2=kq2qd-x2q1x2=q2d-x2d-x2x2=q2q1d-xx=q2q11.5…

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College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Please and thank you for your help!

Two small beads having positive charges q,
19q and
92
= q are fixed at the opposite ends of a horizontal insulating rod of length d = 1.50 m. The bead with charge q, is at the origin. As shown in
the figure below, a third small, charged bead is free to slide on the rod.
92
(a) At what position x is the third bead in equilibrium?
X =
(b) Can the equilibrium be stable?
O Yes, if the third bead has a positive charge.
O Yes, if the third bead has a negative charge.
No

Expert Solution

Step 1

(a)

At equilibrium the net force is zero

$F_{13} = F_{23} \frac{k q_{1} q}{x^{2}} = \frac{k q_{2} q}{{(d - x)}^{2}} \frac{q_{1}}{x^{2}} = \frac{q_{2}}{{(d - x)}^{2}} \frac{{(d - x)}^{2}}{x^{2}} = \frac{q_{2}}{q_{1}} \frac{(d - x)}{x} = \sqrt{\frac{q_{2}}{q_{1}}} \frac{(1.5 m - x)}{x} = \sqrt{\frac{q}{19 q}} 1.5 m - x = 0.2294 x 1.2294 x = 1.5 m x = \frac{1.5 m}{1.2294} = 1.22 m$

Therefore, the third bead will be at equilibrium when x=1.22 m

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