JEE Advanced 2024 Paper 1 Q03 Physics Electrostatics & Circuits Coulomb's Law Hard

JEE Advanced 2024 Paper 1 · Q03 · Coulomb's Law

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.]

  1. A. $\dfrac{q^2}{4\pi\varepsilon_0 R^3 m}$
  2. B. $\dfrac{q^2}{32\pi\varepsilon_0 R^3 m}$
  3. C. $\dfrac{q^2}{8\pi\varepsilon_0 R^3 m}$
  4. D. $\dfrac{q^2}{16\pi\varepsilon_0 R^3 m}$
Reveal answer + step-by-step solution

Correct answer:B

Solution

Equilibrium of the free bead on the hoop is the point diametrically opposite to the glued bead. Parameterise the free bead’s position by angle $\theta$ measured from the centre between the two beads, so the chord separation is $2R\sin(\theta/2)$. The Coulomb force along the chord is $F_C = \dfrac{kq^2}{[2R\sin(\theta/2)]^2}$ where $k = 1/(4\pi\varepsilon_0)$. The tangential component along the hoop (the direction of motion) is $F_C\cos(\theta/2)$, hence the tangential restoring acceleration about $\theta=\pi$ becomes, after writing $\theta = \pi - \phi$ for small $\phi$, $a_t = -\dfrac{kq^2}{4R^2 m\cos^2(\phi/2)}\sin(\phi/2) \approx -\dfrac{kq^2}{4R^2 m}\cdot \dfrac{\phi}{2}$ for small $\phi$. Arc displacement $s = R\phi$, so $\ddot s = -\dfrac{kq^2}{8R^3 m}\,s$. Therefore $\omega^2 = \dfrac{kq^2}{8mR^3} = \dfrac{q^2}{32\pi\varepsilon_0 m R^3}$.

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