Question

A small particle of specific charge $\rho$ is attached with a infinite charge sheet of charge density $\sigma$ by mean of a light insulating cord of length $l$. If the particle slightly disturbed from its equilibrium position, then it oscillates with frequency $\frac{1}{\pi}\sqrt{\frac{\sigma\rho}{\beta\epsilon_0l}}$. Find $\beta$ (ignore effect of gravity).

Answer 1

8

Answer 2

The problem describes a small particle of specific charge $\rho$ (charge per unit mass, $q/m = \rho$) attached to an infinite charge sheet of charge density $\sigma$ by a light insulating cord of length $l$. We need to find the value of $\beta$ in the given frequency expression for small oscillations, ignoring gravity.

1. Electric Field due to the Infinite Charge Sheet:

The electric field produced by an infinite charge sheet with uniform charge density $\sigma$ is uniform and perpendicular to the sheet. Its magnitude is given by:

$E = \frac{\sigma}{2\epsilon_0}$

2. Electric Force on the Particle:

The particle has charge $q$. The electric force on the particle is $F_e = qE$. Since $q = m\rho$, we can write:

$F_e = (m\rho) \left(\frac{\sigma}{2\epsilon_0}\right) = \frac{m\rho\sigma}{2\epsilon_0}$

For the particle to be in equilibrium with the cord stretched, the electric force must be repulsive, pushing the particle away from the sheet. This means $q$ and $\sigma$ must have the same sign. In this case, the equilibrium position is when the cord is perpendicular to the sheet, extending outwards from the sheet.

3. Analysis of Oscillations:

Let the cord be attached to the sheet at the origin $(0,0,0)$. In equilibrium, the particle is at $(0,0,l)$. When the particle is slightly disturbed from its equilibrium position, it oscillates. Let's consider the particle displaced by a small angle $\theta$ from the normal (z-axis). The particle moves along an arc of a circle of radius $l$.

The electric force $\vec{F_e}$ always acts perpendicular to the sheet (e.g., along the z-axis). The tension $\vec{T}$ acts along the cord, towards the point of attachment.

Let's resolve the forces into components tangential and radial to the circular path of oscillation. The tangential direction is perpendicular to the cord. The component of the electric force $F_e$ along the tangential direction acts as the restoring force. If the cord makes an angle $\theta$ with the z-axis (the direction of $F_e$), the component of $F_e$ perpendicular to the cord is $F_e \sin\theta$. This component acts to bring the particle back to the equilibrium position ($\theta=0$). Hence, it's a restoring force. The restoring force $F_{restoring} = -F_e \sin\theta$. (The negative sign indicates it opposes the displacement $\theta$).

For small oscillations, we can use the approximation $\sin\theta \approx \theta$. So, $F_{restoring} \approx -F_e \theta$. Substituting the expression for $F_e$:

$F_{restoring} = -\left(\frac{m\rho\sigma}{2\epsilon_0}\right)\theta$

According to Newton's second law for tangential motion, $F_{restoring} = ma_t$, where $a_t$ is the tangential acceleration. For motion along a circular arc of radius $l$, $a_t = l\frac{d^2\theta}{dt^2}$. So, $ml\frac{d^2\theta}{dt^2} = -\left(\frac{m\rho\sigma}{2\epsilon_0}\right)\theta$

Divide by $ml$:

$\frac{d^2\theta}{dt^2} = -\left(\frac{\rho\sigma}{2\epsilon_0 l}\right)\theta$

4. Angular Frequency and Frequency of Oscillation:

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 equations, we find:

$\omega^2 = \frac{\rho\sigma}{2\epsilon_0 l}$ So, the angular frequency is:

$\omega = \sqrt{\frac{\rho\sigma}{2\epsilon_0 l}}$

The frequency of oscillation $f$ is related to the angular frequency by $f = \frac{\omega}{2\pi}$.

$f = \frac{1}{2\pi}\sqrt{\frac{\rho\sigma}{2\epsilon_0 l}}$

5. Comparing with the Given Frequency Expression:

The problem states that the frequency is $\frac{1}{\pi}\sqrt{\frac{\sigma\rho}{\beta\epsilon_0l}}$. Let's rewrite our derived frequency to match this form:

$f = \frac{1}{2\pi}\sqrt{\frac{\sigma\rho}{2\epsilon_0 l}} = \frac{1}{\pi} \left(\frac{1}{2}\right)\sqrt{\frac{\sigma\rho}{2\epsilon_0 l}}$

$f = \frac{1}{\pi} \sqrt{\left(\frac{1}{2}\right)^2 \frac{\sigma\rho}{2\epsilon_0 l}} = \frac{1}{\pi} \sqrt{\frac{1}{4} \frac{\sigma\rho}{2\epsilon_0 l}}$

$f = \frac{1}{\pi} \sqrt{\frac{\sigma\rho}{8\epsilon_0 l}}$

Now, comparing this with the given expression:

$\frac{1}{\pi}\sqrt{\frac{\sigma\rho}{8\epsilon_0l}} = \frac{1}{\pi}\sqrt{\frac{\sigma\rho}{\beta\epsilon_0l}}$

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

$\beta = 8$