JEE Advanced 2024 Paper 2 Q05 Physics Electrostatics & Circuits Electric Field Hard

JEE Advanced 2024 Paper 2 · Q05 · Electric Field

A small electric dipole $\vec{p}_0$, 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 $\sigma$ is uniformly distributed on the spherical shell. The dipole is initially oriented at a small angle $\theta$ as shown in the figure. While staying at a distance $r$, the dipole is free to rotate about its center.

If released from rest, then which of the following statement(s) is(are) correct?

[$\varepsilon_0$ is the permittivity of free space.]

  1. A. The dipole will undergo small oscillations at any finite value of $r$.
  2. B. The dipole will undergo small oscillations at any finite value of $r > R$.
  3. C. The dipole will undergo small oscillations with an angular frequency of $\sqrt{\dfrac{2\sigma p_0}{\varepsilon_0 I}}$ at $r = 2R$.
  4. D. The dipole will undergo small oscillations with an angular frequency of $\sqrt{\dfrac{\sigma p_0}{100\varepsilon_0 I}}$ at $r = 10R$.
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:B, D

Solution

For a uniformly charged spherical shell, the field inside ($rR$) the shell behaves like a point charge $Q = 4\pi R^2\sigma$ at the centre. The radial electric field at distance $r$ is

$E(r) = \dfrac{Q}{4\pi\varepsilon_0 r^2} = \dfrac{\sigma R^2}{\varepsilon_0 r^2}$.

For a small angular displacement $\theta$ of the dipole from radial alignment, restoring torque $\tau = -p_0 E\sin\theta \approx -p_0 E\,\theta$, giving $I\ddot\theta = -p_0 E\,\theta$ ⇒ small oscillations at any $r > R$. So (B) is correct.

Angular frequency: $\omega = \sqrt{\dfrac{p_0 E}{I}} = \sqrt{\dfrac{p_0 \sigma R^2}{\varepsilon_0 I r^2}}$.

At $r = 2R$: $\omega = \sqrt{\dfrac{p_0\sigma R^2}{\varepsilon_0 I (2R)^2}} = \sqrt{\dfrac{p_0\sigma}{4\varepsilon_0 I}}$. Option (C) gives $\sqrt{2\sigma p_0/(\varepsilon_0 I)}$ — wrong by factor of $\sqrt{8}$. So (C) is incorrect.

At $r = 10R$: $\omega = \sqrt{\dfrac{p_0\sigma R^2}{\varepsilon_0 I (10R)^2}} = \sqrt{\dfrac{p_0\sigma}{100\varepsilon_0 I}}$. Matches (D).

Units check: $[\sigma p_0/(\varepsilon_0 I)] = $ (C/m$^2$)(C$\cdot$m)/((C$^2$/N$\cdot$m$^2$)(kg$\cdot$m$^2$)) $…[truncated]

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