A small bead of mass m and charge q is free to move in a horizontal tube. The tube is placed in between two spheres with charges Q = − q. The spheres are separated by a distance 2a. What is the frequency of small oscillations around the equilibrium point of the bead? You can neglect any friction in the tube.
(Hint: When the bead is only slightly displaced, the force acting on it changes negligibly.)
To find the frequency of small oscillations around the equilibrium point of the bead, we need to determine the effective force acting on the bead. The equilibrium point is where the net force on the bead is zero.
Let's consider the forces acting on the bead:
1. Gravitational force (mg): This force acts vertically downwards and is constant.
2. Electric force (Fe): The bead has a charge q, and the spheres have charges Q = -q. The electric force between them follows Coulomb's law and depends on the distance between them. Since the bead is free to move in the horizontal tube, the electric force will also act horizontally. The direction of the electric force will depend on the charge of the bead and the direction of the electric field.
When the bead is in the equilibrium position, the electric force and the gravitational force balance each other out, resulting in no net force.
Now, let's examine the behavior of the bead when it is slightly displaced from the equilibrium position.
Since the forces change negligibly, we can approximate the electric force to be constant over the displacement range. Therefore, the electric force acting on the bead remains nearly the same throughout the small oscillations.
For small displacements, the force acting on the bead can be approximated as a linear spring force (F = -kx), where k is the effective spring constant and x is the displacement from the equilibrium position.
Therefore, the equation of motion for the bead is given by:
m * d²x/dt² = -kx
Since we know that F = -kx and Fe is the force acting on the bead due to electric forces, we can equate these two forces:
Fe = -kx
Now, we can set up the equation for the electric force acting on the bead:
Fe = qE
Here, E is the electric field created by the charged spheres. The electric field depends on the charges of the spheres and the separation distance between them.
Assuming the spheres are point charges, the electric field at a distance x from the center of one of the spheres is given by:
E = k * Q / (x + a)²
Here, k is the Coulomb constant.
Substituting E into our previous equation:
qE = -kx
q * k * Q / (x + a)² = -kx
q * Q / (x + a)² = -x
q * Q = -x(x + a)²
qx² + 2aqx + a²x + qx = 0
qx² + (3aq + a²)x = 0
This is a quadratic equation, and to find the displacement x, we can use the equilibrium position (x = 0) and the small displacement assumption to linearize it:
qx = -(3aq + a²)x
Simplifying,
q = -3aq - a²
3aq + a² = -q
3a + a²/q = -1
Solving for x, we get:
x = -a² / (3aq + a²)
Now, the frequency of small oscillations around the equilibrium point can be found using the formula:
ω = √(k / m)
Since we've already established that the restoring force acting on the bead is linear, we can write k = q / x:
k = q / (-a² / (3aq + a²))
k = -(3aq + a²) / a²
Substituting k and m into the frequency equation:
ω = √(k / m) = √((-3aq - a²) / (a²m))
Therefore, the frequency of small oscillations around the equilibrium point of the bead is ω = √((-3aq - a²) / (a²m)).