Question

Two small beads having positive charges 112q and q2-q are fixed at the opposite ends of a horizontal insulating rod of length d1.50 m. The bead with charge q1 is at the origin. As shown in the figure below, a third small, charged bead is free to slide on the rod 91 92 (a) At what position x is the third bead in equilibrium? (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. O No

Answer 1

(a) Let us assume that the bead has charge Q.

The bead will be in equilibrium when the forces are exerted by the other two charges are equal and opposite, The force exerted by the first charge is,

$F_1 = \frac{1}{4 \pi \epsilon_0} \frac{q_1Q}{r^2} \Rightarrow F_1 = \frac{1}{4 \pi \epsilon_0} \frac{12qQ}{x^2}$

The force exerted by the second charge is,

$F_2 = \frac{1}{4 \pi \epsilon_0} \frac{q_2Q}{r^2} \Rightarrow F_2= \frac{1}{4 \pi \epsilon_0} \frac{qQ}{(d-x)^2}$

Now, for equilibrium, these forces will be equal. This gives,

$\Rightarrow F_1 = F_2$

$\Rightarrow \frac{1}{4 \pi \epsilon_0} \frac{12qQ}{x^2} = \frac{1}{4 \pi \epsilon_0} \frac{qQ}{(d-x)^2}$

$\Rightarrow \frac{12}{x^2} = \frac{1}{(d-x)^2} \Rightarrow \frac{12}{x^2} = \frac{1}{(1.5-x)^2}$

$\Rightarrow \frac{12}{x^2} = \frac{1}{2.25 + x^2 - 3x}$

$\Rightarrow 27 + 12x^2 - 36x = x^2$

$\Rightarrow 11x^2 - 36x + 27 = 0$

Solving the quadratic equation, we get two solutions 1.16 m and 2.11 m, Since 2.11 m is greater than d, it is not a possible solution, this gives,

​​​​​​$\Rightarrow x = 1.16 \, m$

(b) The equilibrium will be stable only if the bead has a positive charge because if the bead is displaced from the equilibrium position it will return back to its stable position due to the repulsive forces from the other two charges.

But if the bead has a negative charge, then it will lead to an unstable equilibrium because a small shift from the stable position will make the bead to go towards one of the charges due to their attractive force on the bead (Positive and negative charges will attract each other).

Hence the correct option is 1. Yes, if the third bead has a positive charge.