To determine the period of small amplitude oscillations of a charged rod along the axis of a charged ring, we need to analyze the forces acting on the rod due to the electric field created by the ring. This problem involves concepts from electrostatics and simple harmonic motion (SHM).
Understanding the System
We have a thin rod of mass m and uniform negative charge -q positioned symmetrically along the axis of a thin ring of radius r that carries a uniformly distributed positive charge Q. The length of the rod is 2r, meaning it extends from -r to r along the axis of the ring.
Electric Field Due to the Ring
First, we need to calculate the electric field produced by the ring at a distance x along its axis. The electric field E at a point along the axis of a charged ring is given by the formula:
E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Qx}{(x^2 + r^2)^{3/2}}
Here, \epsilon_0 is the permittivity of free space, and x is the distance from the center of the ring to the point where we want to calculate the electric field.
Force on the Rod
The force F acting on the rod due to the electric field can be calculated using the formula:
F = qE
Substituting the expression for the electric field, we have:
F = -q \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{Qx}{(x^2 + r^2)^{3/2}}
Small Oscillations Approximation
For small displacements x (where x is small compared to r), we can approximate the force as:
F \approx -k x
where k is the effective spring constant. To find k, we can differentiate the expression for force with respect to x and evaluate it at x = 0.
Calculating the Spring Constant
Taking the derivative of the force:
k = -\frac{dF}{dx} = -\frac{d}{dx} \left(-q \cdot \frac{1}{4\pi \epsilon_0} \cdot \frac{Qx}{(x^2 + r^2)^{3/2}}\right)
Evaluating this at x = 0 gives:
k = \frac{qQ}{4\pi \epsilon_0 r^2}
Period of Oscillation
The period T of small amplitude oscillations for a mass-spring system is given by:
T = 2\pi \sqrt{\frac{m}{k}}
Substituting our expression for k:
T = 2\pi \sqrt{\frac{m \cdot 4\pi \epsilon_0 r^2}{qQ}}
Final Result
Thus, the period of small amplitude oscillations of the rod along the axis of the ring is:
T = 2\pi \sqrt{\frac{4\pi \epsilon_0 m r^2}{qQ}}
This result shows how the mass of the rod, the charges involved, and the geometry of the system influence the oscillation period. The rod will oscillate back and forth around the equilibrium position due to the restoring electric force provided by the charged ring.
