Question

A small electric dipole is placed at the centre of uniformly charged hemispherical shell of charge density $\sigma$ and radius $R$. The dipole is fixed at the centre and free to rotate. The dipole has dipole moment $p$ and moment of inertia $I$. Now dipole is slightly displaced and released from its equilibrium position. The period of oscillation of dipole is $2\pi\sqrt{\frac{\alpha\epsilon_0 I}{p\sigma}}$. Find $\alpha$

Answer 1

4

Answer 2

The problem asks us to find the value of $\alpha$ in the expression for the period of oscillation of a small electric dipole placed at the center of a uniformly charged hemispherical shell.

1. Calculate the Electric Field (E) at the center of the hemispherical shell:

Consider a uniformly charged hemispherical shell of radius $R$ and surface charge density $\sigma$. We want to find the electric field at its center.

Let's consider a differential ring element on the hemisphere. Let this ring be at an angle $\theta$ from the axis of symmetry (the axis passing through the center and perpendicular to the base).

The radius of this ring is $r = R\sin\theta$.

The width of the ring is $Rd\theta$.

The area of this differential ring is $dA = (2\pi r)(Rd\theta) = (2\pi R\sin\theta)(Rd\theta) = 2\pi R^2\sin\theta d\theta$.

The charge on this ring is $dQ = \sigma dA = 2\pi R^2\sigma\sin\theta d\theta$.

The electric field produced by this ring at the center of the hemisphere (which is also the center of the ring's base) will have components that cancel out perpendicular to the axis of symmetry. Only the component along the axis of symmetry will contribute to the net electric field. The distance from any point on the ring to the center is $R$.

The electric field $dE$ due to a point charge $dQ$ at a distance $R$ is $dE' = \frac{1}{4\pi\epsilon_0}\frac{dQ}{R^2}$.

The component of this field along the axis of symmetry is $dE_z = dE' \cos\theta = \frac{1}{4\pi\epsilon_0}\frac{dQ}{R^2}\cos\theta$.

Substitute $dQ = 2\pi R^2\sigma\sin\theta d\theta$:

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{(2\pi R^2\sigma\sin\theta d\theta)}{R^2}\cos\theta$

$dE_z = \frac{\sigma}{2\epsilon_0}\sin\theta\cos\theta d\theta$.

To find the total electric field $E$ at the center, we integrate $dE_z$ over the entire hemisphere. The angle $\theta$ ranges from $0$ to $\pi/2$.

$E = \int_0^{\pi/2} \frac{\sigma}{2\epsilon_0}\sin\theta\cos\theta d\theta$

We can use the identity $\sin(2\theta) = 2\sin\theta\cos\theta$, so $\sin\theta\cos\theta = \frac{1}{2}\sin(2\theta)$.

$E = \int_0^{\pi/2} \frac{\sigma}{2\epsilon_0}\left(\frac{1}{2}\sin(2\theta)\right) d\theta$

$E = \frac{\sigma}{4\epsilon_0} \int_0^{\pi/2} \sin(2\theta) d\theta$

$E = \frac{\sigma}{4\epsilon_0} \left[-\frac{\cos(2\theta)}{2}\right]_0^{\pi/2}$

$E = \frac{\sigma}{4\epsilon_0} \left(-\frac{\cos(\pi)}{2} - \left(-\frac{\cos(0)}{2}\right)\right)$

$E = \frac{\sigma}{4\epsilon_0} \left(-\frac{-1}{2} + \frac{1}{2}\right)$

$E = \frac{\sigma}{4\epsilon_0} \left(\frac{1}{2} + \frac{1}{2}\right) = \frac{\sigma}{4\epsilon_0}$.

The electric field at the center of the hemispherical shell is $E = \frac{\sigma}{4\epsilon_0}$ and points along the axis of symmetry, away from the shell (assuming $\sigma$ is positive).

2. Period of Oscillation of the Dipole:

An electric dipole with dipole moment $p$ placed in an electric field $E$ experiences a torque $\tau = pE\sin\theta$, where $\theta$ is the angle between the dipole moment vector and the electric field vector.

The equilibrium position for the dipole is when its moment $p$ is aligned with the electric field $E$ ($\theta=0$).

When the dipole is slightly displaced by a small angle $\theta$ from its equilibrium position, the restoring torque is given by $\tau = -pE\sin\theta$.

For small oscillations, $\sin\theta \approx \theta$.

So, the restoring torque is $\tau \approx -pE\theta$.

According to Newton's second law for rotational motion, $\tau = I\frac{d^2\theta}{dt^2}$, where $I$ is the moment of inertia of the dipole.

Therefore, $I\frac{d^2\theta}{dt^2} = -pE\theta$.

This can be rewritten as $\frac{d^2\theta}{dt^2} = -\left(\frac{pE}{I}\right)\theta$.

This is the equation for Simple Harmonic Motion (SHM) of the form $\frac{d^2\theta}{dt^2} = -\omega^2\theta$, where $\omega$ is the angular frequency of oscillation.

Comparing the two equations, we get $\omega^2 = \frac{pE}{I}$.

So, the angular frequency is $\omega = \sqrt{\frac{pE}{I}}$.

The period of oscillation $T$ is given by $T = \frac{2\pi}{\omega}$.

$T = 2\pi\sqrt{\frac{I}{pE}}$.

3. Substitute E and find $\alpha$:

Substitute the calculated value of $E = \frac{\sigma}{4\epsilon_0}$ into the expression for $T$:

$T = 2\pi\sqrt{\frac{I}{p\left(\frac{\sigma}{4\epsilon_0}\right)}}$

$T = 2\pi\sqrt{\frac{4I\epsilon_0}{p\sigma}}$.

The problem states that the period of oscillation is $T = 2\pi\sqrt{\frac{\alpha\epsilon_0 I}{p\sigma}}$.

Comparing our derived expression with the given form:

$2\pi\sqrt{\frac{4I\epsilon_0}{p\sigma}} = 2\pi\sqrt{\frac{\alpha\epsilon_0 I}{p\sigma}}$.

By comparing the terms inside the square root, we can see that:

$\frac{4I\epsilon_0}{p\sigma} = \frac{\alpha\epsilon_0 I}{p\sigma}$

$\alpha = 4$.

The final answer is $\boxed{4}$.

Explanation of the solution:

1. Electric Field Calculation: The electric field at the center of a uniformly charged hemispherical shell of charge density $\sigma$ and radius $R$ is calculated by integrating the contributions from differential rings. The resulting field is $E = \frac{\sigma}{4\epsilon_0}$ along the axis of symmetry.

2. Dipole Oscillation: A dipole with moment $p$ in an electric field $E$ experiences a torque $\tau = pE\sin\theta$. For small displacements $\theta$ from the equilibrium position (where $p$ is aligned with $E$), the restoring torque is $\tau \approx -pE\theta$.

3. SHM Equation: Equating the torque to $I\frac{d^2\theta}{dt^2}$ (where $I$ is the moment of inertia) yields the SHM equation $\frac{d^2\theta}{dt^2} = -\left(\frac{pE}{I}\right)\theta$.

4. Period Derivation: From the SHM equation, the angular frequency is $\omega = \sqrt{\frac{pE}{I}}$, and the period is $T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{I}{pE}}$.

5. Substitution and Comparison: Substituting $E = \frac{\sigma}{4\epsilon_0}$ into the period expression gives $T = 2\pi\sqrt{\frac{4I\epsilon_0}{p\sigma}}$. Comparing this with the given form $T = 2\pi\sqrt{\frac{\alpha\epsilon_0 I}{p\sigma}}$, we find $\alpha = 4$.