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

q^2 \over 32\pi \varepsilon _0a^3m

q^2 \over 4\pi \varepsilon _0a^3m

q^2 \over 8\pi \varepsilon _0a^3m

q^2 \over 16\pi \varepsilon _0a^3m

Electrostatics PYQ — JEE Advanced 2024 · Shift 1 · Q37 with Solution

View written solutionFree

Correct answer: B

Set up the geometry

Let the fixed bead be at angle $0$ on the circular hoop of radius $a$ , and let the მოძრable bead be at angular position $\theta$ .

The distance between the two beads is the chord length: $r=2a\sin\frac{\theta}{2}$

So the electrostatic potential energy of interaction is

=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a\sin(\theta/2)}$$ 2. **Find the equilibrium position** The movable bead is constrained to move on the hoop, so equilibrium occurs where $U(\theta)$ is minimum. Since $$U(\theta)\propto \frac{1}{\sin(\theta/2)}$$ this is minimum when $\sin(\theta/2)$ is maximum, i.e. $$\theta=\pi$$ So the movable bead oscillates about the diametrically opposite point. 3. **Expand about equilibrium** Let $$\theta=\pi+\phi$$ where $\phi$ is small. Then $$r=2a\sin\left(\frac{\pi+\phi}{2}\right)=2a\sin\left(\frac{\pi}{2}+\frac{\phi}{2}\right)=2a\cos\frac{\phi}{2}$$ Hence $$U(\phi)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a\cos(\phi/2)}$$ For small $\phi$, $$\cos\frac{\phi}{2}\approx 1-\frac{\phi^2}{8}$$ so $$\frac{1}{\cos(\phi/2)}\approx 1+\frac{\phi^2}{8}$$ Therefore, $$U(\phi)\approx \frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a}\left(1+\frac{\phi^2}{8}\right)$$ So the variable part of potential energy is $$\Delta U=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16a}\phi^2$$ 4. **Compare with SHM form** The kinetic energy of the bead moving on the hoop is $$T=\frac12 m(a\dot\phi)^2=\frac12 ma^2\dot\phi^2$$ For angular coordinate $\phi$, SHM potential is $$\Delta U=\frac12 k_\phi \phi^2$$ Thus $$\frac12 k_\phi=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16a}$$ which gives $$k_\phi=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{8a}$$ Equation of motion: $$ma^2\ddot\phi+k_\phi\phi=0$$ So $$\omega^2=\frac{k_\phi}{ma^2} =\frac{1}{4\pi\varepsilon_0}\frac{q^2}{8a}\cdot \frac{1}{ma^2}$$ Hence $$\boxed{\omega^2=\frac{q^2}{32\pi\varepsilon_0 a^3 m}}$$ 5. **Match with options** This corresponds to **Option B**.

More from Electrostatics

A small electric dipole $p_{0}$ , having a moment of inertia $I$ about its center, is kept at a distance $r$ from the center of a spherical shell of radius $R$ . The surface charge density $σ$ is uniformly distributed on the… Includes diagram2024 · Multiple correct
A charge is kept at the central point $P$ of a cylindrical region. The two edges subtend a half-angle $θ$ at $P$ , as shown in the figure. When $θ = 3 0^{\circ}$ , then the electric flux through the curved surface of… Includes diagram2024 · Numerical
An infinitely long thin wire, having a uniform charge density per unit length of $5 nC / m$ , is passing through a spherical shell of radius $1 m$ , as shown in the figure. A $10 nC$ charge is distributed… Includes diagram2024 · Numerical
An electric dipole is formed by two charges $+ q$ and $- q$ located in $x y$ -plane at $(0, 2) mm$ and $(0, - 2) mm$ , respectively, as shown in the figure. The electric potential at point $P (100, 100) mm$ due to the… Includes diagram2023 · MCQ
Six charges are placed around a regular hexagon of side length $a$ as shown in the figure. Five of them have charge $q$ , and the remaining one has charge $x$ . The perpendicular from each charge to the nearest hexagon side passes through… Includes diagram2022 · Multiple correct
A charge $q$ is surrounded by a closed surface consisting of an inverted cone of height $h$ and base radius $R$ , and a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 ϵ _{0}}$ … Includes diagram2022 · Numerical
In the figure, the inner (shaded) region $A$ represents a sphere of radius $r_{A} = 1$ , within which the electrostatic charge density varies with the radial distance $r$ from the center as $ρ_{A} = k r$ , where $k$ is positive. In the… Includes diagram2022 · Multiple correct
A disk of radius $R$ with uniform positive charge density $σ$ is placed on the $x y$ plane with its center at the origin. The Coulomb potential along the $z$ -axis is $V (z) = \frac{σ}{2 ϵ _{0}} (R^{2} + z^{2} - z) .$ …2022 · Multiple correct

View written solutionFree

Correct answer: B

Set up the geometry

Let the fixed bead be at angle $0$ on the circular hoop of radius $a$ , and let the მოძრable bead be at angular position $\theta$ .

The distance between the two beads is the chord length: $r=2a\sin\frac{\theta}{2}$

So the electrostatic potential energy of interaction is

=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a\sin(\theta/2)}$$ 2. **Find the equilibrium position** The movable bead is constrained to move on the hoop, so equilibrium occurs where $U(\theta)$ is minimum. Since $$U(\theta)\propto \frac{1}{\sin(\theta/2)}$$ this is minimum when $\sin(\theta/2)$ is maximum, i.e. $$\theta=\pi$$ So the movable bead oscillates about the diametrically opposite point. 3. **Expand about equilibrium** Let $$\theta=\pi+\phi$$ where $\phi$ is small. Then $$r=2a\sin\left(\frac{\pi+\phi}{2}\right)=2a\sin\left(\frac{\pi}{2}+\frac{\phi}{2}\right)=2a\cos\frac{\phi}{2}$$ Hence $$U(\phi)=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a\cos(\phi/2)}$$ For small $\phi$, $$\cos\frac{\phi}{2}\approx 1-\frac{\phi^2}{8}$$ so $$\frac{1}{\cos(\phi/2)}\approx 1+\frac{\phi^2}{8}$$ Therefore, $$U(\phi)\approx \frac{1}{4\pi\varepsilon_0}\frac{q^2}{2a}\left(1+\frac{\phi^2}{8}\right)$$ So the variable part of potential energy is $$\Delta U=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16a}\phi^2$$ 4. **Compare with SHM form** The kinetic energy of the bead moving on the hoop is $$T=\frac12 m(a\dot\phi)^2=\frac12 ma^2\dot\phi^2$$ For angular coordinate $\phi$, SHM potential is $$\Delta U=\frac12 k_\phi \phi^2$$ Thus $$\frac12 k_\phi=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{16a}$$ which gives $$k_\phi=\frac{1}{4\pi\varepsilon_0}\frac{q^2}{8a}$$ Equation of motion: $$ma^2\ddot\phi+k_\phi\phi=0$$ So $$\omega^2=\frac{k_\phi}{ma^2} =\frac{1}{4\pi\varepsilon_0}\frac{q^2}{8a}\cdot \frac{1}{ma^2}$$ Hence $$\boxed{\omega^2=\frac{q^2}{32\pi\varepsilon_0 a^3 m}}$$ 5. **Match with options** This corresponds to **Option B**.

Electrostatics question

Electrostatics question

More from Electrostatics

Electrostatics question

Electrostatics question

More from Electrostatics