Two beads, each with charge q and mass m , are on a horizontal,...

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 [

\varepsilon_0

is the permittivity of free space.] Two beads, each with charge q and mass m , are on a horizontal, frictionless, non-conducting, circular hoop of radius R

Two beads, each with charge q and mass m , are on a horizontal, frictionless, non-conducting, circular hoop of radius R

Options

A $q^2 /\left(4 \pi \varepsilon_0 R^3 m\right)$
B $q^2 /\left(32 \pi \varepsilon_0 R^3 m\right)$
C $q^2 /\left(8 \pi \varepsilon_0 R^3 m\right)$
D $q^2 /\left(16 \pi \varepsilon_0 R^3 m\right)$

Correct answer

B. $q^2 /\left(32 \pi \varepsilon_0 R^3 m\right)$

Step-by-step solution

Restoring force

=\mathrm{qE} \sin \left(\frac{\theta}{2}\right)

\begin{aligned} & \therefore \tau=\mathrm{qE} \sin \left(\frac{\theta}{2}\right) \mathrm{R}=\mathrm{I} \alpha \\ & \mathrm{E}=\frac{\mathrm{Kq}}{\left(2 \mathrm{R} \cos \frac{\theta}{2}\right)^2}=\frac{1}{4 \pi \varepsilon_0} \frac{\mathrm{q}}{4 \mathrm{R}^2 \cos ^2\left(\frac{\theta}{2}\right)} \\ & \therefore \frac{1}{4 \pi \epsilon_0} \frac{\mathrm{qR}}{4 \mathrm{R}^2 \cos ^2\left(\frac{\theta}{2}\right)} \sin \left(\frac{\theta}{2}\right) \mathrm{q}=\mathrm{mR}^2 \alpha\end{aligned}

For $\theta$ very small,

\begin{aligned} & \frac{-\mathrm{q}^2}{32 \pi \varepsilon_0 \mathrm{R}^3 \mathrm{~m}} \theta=\alpha \\ & \therefore \omega^2=\frac{\mathrm{q}^2}{32 \pi \varepsilon_0 \mathrm{mR}^3}\end{aligned}$$ Hence option (2)

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