Question

a large insulating thick shet of thickness 2d is charged with uniform volume charge density rho. a particle of mass m carrying a charge q having a sign opposite to that of sheet is released frim surface if sheet. find oscillation frequency of particle inside the sheet

Answer 1

$\omega = \sqrt{\frac{|q|\rho}{\epsilon_0 m}}$

Answer 2

1. Electric field inside the sheet:

Consider a uniformly charged slab of thickness $2d$ (extending from $x=-d$ to $x=+d$) with volume charge density $\rho$. By symmetry, the electric field at a point $x$ inside the slab (with $-d \le x \le d$) is along the $x$-axis. Using Gauss’s law with a pillbox symmetric about $x=0$:

$$\text{Flux: } 2A\,E(x) = \frac{\rho \cdot (2Ax)}{\epsilon_0} \quad \Longrightarrow \quad E(x) = \frac{\rho\,x}{\epsilon_0}.$$

2. Force and simple harmonic motion:

A particle of mass $m$ and charge $q$ (with sign opposite to that of the sheet) experiences an electric force:

$$F = qE(x) = q\frac{\rho\,x}{\epsilon_0}.$$

Since $q$ and $\rho$ have opposite signs, writing $|q|$ for the magnitude of the charge we can express the force as:

$$F = -\frac{|q|\rho}{\epsilon_0}\,x,$$

which is of the form $F=-kx$ (restoring force) with effective spring constant:

$$k=\frac{|q|\rho}{\epsilon_0}.$$

The angular frequency for simple harmonic motion is then:

$$\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{|q|\rho}{\epsilon_0 m}}.$$

3. Initial condition:

The particle is released from the surface (at $x=d$) with zero velocity. Although its amplitude is $d$, the frequency of oscillation is determined solely by the properties of the force.