Help with solving the problem charge problem

[thomq]

1. Two small beads having a positive charges 3q and q are fixed at the opposite ends of a horizontal, insulating rod, extending from the origin to point X. This length is the distance (d). A third small charged bead (Q) is free to slide on the rod. At what position is the third bead in equilibrium?

2. Coulomb's Law is used for this problem. F = kqq/d

F[1][/SUB] = kQ(3q)/x[2][/SUP]
F[2][/SUB] = kQq/(d-x)[2][/SUP]

F[1][/SUB] = F[2][/SUB]

3. I am attempting to solve for x.

3 (d-x)[2][/SUP] = (x)[2][/SUP]

I know I could make both side a square root so I have x one side, but I need help beyond this.

I also need to show all steps. Thanks!

 

[Delphi51]

Welcome to PF, thom.
Looks good so far!
Recommend you expand the left side, collect like terms to get a quadratic equation.

I'm passing on a tip I find really helpful: don't bother with those sub, sup codes. Copy the ² symbol from the bottom of my post, or the complete set here:
https://www.physicsforums.com/blog.php?b=346

 

[susskind_leon]

$$\frac{3}{x^2}=\frac{1}{(d-x)^2} \Rightarrow 3x^2-6dx+3d^2=x^2\Rightarrow 2x^2-6dx+3d^2=0\Rightarrow x^2-3dx+3/2d^2=0$$
$$x_{1,2}=\frac{3d}{2}\pm\sqrt{\frac{9d^2}{4}-\frac{3d^2}{2}}=\frac{3d}{2}\pm\sqrt{\frac{3d^2}{4}}=\frac{3\pm\sqrt{3}}{2}d$$
I hope I haven't made a mistake...

 

[Delphi51]

That checks out when using the + sign. The other solution is larger than d, so it has to be extraneous.

 

[rwisz]

To solve this problem, we can use the principle of equilibrium, which states that the net force on an object must be equal to zero for it to be in equilibrium. In this case, the third bead will be in equilibrium when the net force acting on it is equal to zero.

Using Coulomb's Law, we can calculate the forces acting on the third bead from the two fixed beads. The force F[1] acting on the third bead from the bead with charge 3q is given by F[1] = kQ(3q)/x^2, where k is the Coulomb's constant and x is the distance from the third bead to the bead with charge 3q. Similarly, the force F[2] acting on the third bead from the bead with charge q is given by F[2] = kQq/(d-x)^2, where d is the distance between the two fixed beads and (d-x) is the distance from the third bead to the bead with charge q.

To find the equilibrium position, we need to set F[1] = F[2] and solve for x. This will give us the position where the net force on the third bead is zero. Using the equation F[1] = F[2], we get:

kQ(3q)/x^2 = kQq/(d-x)^2

Simplifying this equation, we get:

3q/(x^2) = q/((d-x)^2)

Cross-multiplying and simplifying, we get:

3q(d-x)^2 = qx^2

Expanding the brackets, we get:

3qd^2 - 6qdx + 3qx^2 = qx^2

Simplifying further, we get:

3qd^2 - 6qdx + 2qx^2 = 0

This is a quadratic equation in terms of x, and we can solve it using the quadratic formula. The solutions to this equation will give us the two possible equilibrium positions for the third bead. However, since the third bead is free to slide on the rod, it can only be at one position at a time. Therefore, we need to choose the positive solution for x, which will give us the unique equilibrium position for the third bead.

Finally, we can substitute the value of x into the equation d-x to find the distance from the origin to the equilibrium position of the