Two beads, each with charge q and mass m, are on a horizontal, frictionless, non-conducting, circular hoop of radius R. One of the beads is glued to the hoop at some point, while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [E0 is the permittivity of free space.]

To determine the square of the angular frequency ${\omega }^2$ for small oscillations of a charged bead on a circular hoop, we analyze the electrostatic restoring force and model the system as a harmonic oscillator.

Key Steps:

1. Electrostatic Force:
   The Coulomb force between the two beads separated by distance $d$ is:

   $$F=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{d^2}\mathrm{.}$$

   At equilibrium ($d=2R$), the beads are diametrically opposed. For a small displacement $\alpha$, the separation becomes $d\approx 2R\cos (\alpha \mathrm{/}2)\approx 2R(1-\frac{{\alpha }^2}{8})$.

2. Tangential Restoring Force:
   The tangential component of the force drives oscillations. For small $\alpha$, the angle between the Coulomb force and the tangent is $\alpha \mathrm{/}2$. The tangential force is:

   $$F_{\text{tangent}}\approx \frac{q^2}{32\pi {ϵ}_0{R}^2}\alpha \mathrm{.}$$

3. Effective Spring Constant:
   Relating displacement $s=R\alpha$, the force becomes:

   $$F_{\text{tangent}}=-\frac{q^2}{32\pi {ϵ}_0{R}^3}s=-ks,$$

   giving $k=\frac{q^2}{32\pi {ϵ}_0{R}^3}$.

4. Angular Frequency:
   Using ${\omega }^2=\frac{k}{m}$:

   $${\omega }^2=\frac{q^2}{32\pi {ϵ}_0{R}^{3}m}\mathrm{.}$$

Final Answer:

$$\boxed{{\displaystyle \frac{q^2}{32\pi {ϵ}_0{R}^{3}m}}}$$

Correct Option: (B)