Two small beads having positive charges 3q and q are fixed...

Two small beads having positive charges 3$q$ and $q$ are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to the point, $x=d$ . As shown in Figure $\mathrm{P} 23.10$ , a third small charged bead is free to slide on the rod. At what position is the third bead in equilibrium? Can it be in stable equilibrium?

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding the forces between charged particles. It states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance separating them. This law forms the basis for calculating individual forces in systems with multiple charges.

Principle of Superposition

The Principle of Superposition allows us to determine the net force acting on a charge by adding the individual forces contributed by each source charge. In systems with multiple charges, this concept is crucial because it enables the analysis of the net electric force or field by treating each interaction independently before combining them.

Electrostatic Equilibrium

Electrostatic Equilibrium occurs when the net force on a charged particle is zero, meaning all the forces acting on it cancel each other out. Finding the position of equilibrium in a multi-charge system involves setting the sum of all forces equal to zero, which often leads to solving algebraic equations to determine the equilibrium position.

Stability of Equilibrium

The stability of an equilibrium position refers to whether a small displacement of the charged particle results in forces that restore it to equilibrium (stable) or push it further away (unstable). Analyzing the stability involves examining how the net force changes with small perturbations around the equilibrium point, helping to predict the particle's behavior under slight disturbances.

Transcript

00:01 In the given problem, there is a light rod kept along x -axis, starting from the origin like this.

00:18 Then there are two small beats having the positive charges 3q and q, and which are fixed at the ends of this light rod.

00:31 Here this charge is 3 q and this charge is q.

00:36 This is point o and let the other point to be a.

00:43 Then we have to find the position where we should put a third charge so that that third charge may remain in equilibrium.

00:57 The total distance between these two charges means the length of this light rod is given as d.

01:03 Suppose we put this third charge whose magnitude is q at a distance x from origin o, then its distance from the other charge will be d minus x.

01:19 Now, as this middle charge is in equilibrium, in equally so, we can say the two forces acting on it due to the two charges, 3q and q should be equal and opposite in direction.

01:55 Means f at b due to o should be equal to f at b due to a.

02:02 Hence using coulin's law for f at b due to o, this is k, q into 3q by x square, is equal to k, q again, but for the other charge at b due to o, this is k, and, for the other charge, a this is small q and the distance between them is d minus x to the whole square now canceling the similar terms on both the sides k with the k q with the q and small q with the small q with the small q so finally it comes out to be three by x square is equal to one by d minus x to the whole square now making a perfect square of both the sides for three this is is root 3 by x to the whole square is equal to 1 by d minus x to the whole square.

02:59 Then taking square root of both sides, it remains root 3 by x is equal to 1 by d minus x.

03:11 Now, when making a cross multiplication, root 3d minus root 3x is equal to x.

03:19 So if we shift this root 3x towards right hand side, root 3d will be equal to 1 plus root 3x.

03:30 So finally, the answer for this x here will become root 3d by 1 plus root root 3.

03:39 Or we can say this is 1 .732d divided by 2 .732.

03:49 So finally this x here comes out to be 0 .634 times of d...

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