Question

Two small beads having opposite charges q1= 3q and q2=q are fixed at the opposite ends of a horizontal insulating rod of length d= 1.50 m. The bead with charge q1 is at the origin. A third small charged bead is free to slide on the rod. (A) at what position X is the third bead in equilibrium? (B) can the equilibrium be stable

Answer 1

Answer on Question #38199 - Physics - Other

Question: two small beads having charges $q_{1} = 3q$ and $q_{2} = q$ are fixed at the opposite ends of a horizontal insulating rod of length $d = 1,5m$ . The bead with charge $q_{1}$ is at the origin. A third small charged bead is free to slide on the rod.

a) At what position $x$ is the third bead in equilibrium?

b) Can the equilibrium be stable?

Solution: let us draw the simplified picture of this situation:

The third charge is interacting with first and second ones with forces $F_{1} = k\frac{3qk}{x^{2}}$ and $F_{2} = k\frac{qk}{(l - x)^{2}}$ , where $x$ is the position of the charge $Q$ . These forces have opposite directions. The total force acting on the third chart is

$$F _ {t o t} = k \frac {3 q k}{x ^ {2}} - k \frac {q k}{(l - x) ^ {2}}.$$

In the equilibrium state $F_{tot} = 0$ , so we become condition for the equilibrium position $x$ :

$$x ^ {2} - 3 l x + \frac {3}{2} l ^ {2} = 0 \rightarrow x _ {1, 2} = \frac {3 \pm \sqrt {3}}{2} \cdot l$$

We choose the solution for $x$ , which is smaller than $l$ . Finally, we obtain the equilibrium position

$$x _ {e q} = \frac {3 - \sqrt {3}}{2} \cdot l \cong 0, 6 4 \cdot l.$$

Let us consider now small deviations from the equilibrium position $x$ :

$$F _ {t o t} (x + d x) = F _ {t o t} (x) + F _ {t o t} ^ {\prime} (x) \cdot d x = F _ {t o t} ^ {\prime} (x) \cdot d x$$

Because in the equilibrium position $F_{tot}(x) = 0$ . The force now takes form

$$F _ {t o t} (x + d x) = - 2 \left(k \frac {3 q k}{x ^ {3}} - k \frac {q k}{(l - x) ^ {3}}\right) d x$$

In the equilibrium position $\frac{3qk}{x^2} = \frac{qk}{(l - x)^2}$ and the total force becomes after substitution

$$F _ {t o t} (x + d x) = - \frac {6 k l}{x ^ {3} (l - x)} \cdot q Q \cdot d x$$

By the stable equilibrium the force that acts on the body by small deviations from the equilibrium state should be directed to the equilibrium position. If we pull the bead rightwards, the force should be directed leftwards (force should be negative). Because in the expression of the

force $\frac{6kl}{x^3(l - x)} > 0$, $qQ$ should be also positive. That implies that $Q$ and $q_1, q_2$ should have the same sign. The same result we become when we try to pull the bead leftwards ($dx < 0$).

**Answer:**

a) The equilibrium position is

$$x_{eq} = \frac{3 - \sqrt{3}}{2} \cdot l \cong 0,95 \, \text{m}.$$

b) The equilibrium will be stable, if charges $q_3$ and $q_1, q_2$ will have the same sign.