A large insulating thick sheet of thickness `2`d is charged with a uniform volume charge density p. A particle of mass m, carrying a charge q having a sign opposite to that of the sheet, is released from the surface of the sheet. The sheet does not offer any mechanical resistance to the motion of the particle. Find the oscillation frequency v of the particle inside the sheet

Options

A. `v = (1)/(2pi) sqrt((qp)/(m epsi_(0))`
B. `v = (1)/(2pi)sqrt((2qp)/(m epsi_(0)))`
C. `v = (1)/(2)sqrt((qp)/(m epsi_(0)))`
D. `v = (1)/(4pi) sqrt((2qp)/(m epsi_(0)))`

Explanation

To solve the problem step by step, we will analyze the situation involving the charged particle and the insulating sheet.

### Step 1: Understand the System
We have a thick insulating sheet of thickness \(2d\) with a uniform volume charge density \(\rho\). A particle of mass \(m\) and charge \(q\) (opposite in sign to the sheet) is released from the surface of the sheet.

### Step 2: Identify the Electric Field Inside the Sheet
The electric field \(E\) inside a uniformly charged thick sheet can be derived from the volume charge density \(\rho\). The electric field due to an infinite sheet with surface charge density \(\sigma\) is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \(\epsilon_0\) is the permittivity of free space.

Since we have a volume charge density \(\rho\), we can relate it to the surface charge density \(\sigma\):
\[ \sigma = \rho \cdot d \]
Thus, the electric field at a distance \(r\) from the center of the sheet (where \(r\) varies from \(-d\) to \(d\)) can be expressed as:
\[ E = \frac{\rho \cdot r}{2\epsilon_0} \]

### Step 3: Calculate the Force on the Particle
The force \(F\) acting on the particle due to the electric field is given by:
\[ F = qE \]
Substituting the expression for \(E\):
\[ F = q \cdot \frac{\rho \cdot r}{2\epsilon_0} \]
This force acts in the direction opposite to the displacement \(r\) because the charge \(q\) is opposite in sign to the charge of the sheet.

### Step 4: Set Up the Equation of Motion
According to Newton's second law, the net force is equal to mass times acceleration:
\[ F = ma \]
Substituting the expression for force:
\[ ma = -\frac{q \rho r}{2\epsilon_0} \]
This can be rearranged to give the acceleration \(a\):
\[ a = -\frac{q \rho}{2m\epsilon_0} r \]

### Step 5: Identify the Oscillation Frequency
The equation \(a = -\omega^2 r\) resembles the equation of simple harmonic motion, where \(\omega\) is the angular frequency. By comparing:
\[ \omega^2 = \frac{q \rho}{2m\epsilon_0} \]
Thus, the angular frequency \(\omega\) is:
\[ \omega = \sqrt{\frac{q \rho}{2m\epsilon_0}} \]

### Step 6: Calculate the Frequency
The frequency \(v\) is related to the angular frequency by:
\[ v = \frac{\omega}{2\pi} \]
Substituting the expression for \(\omega\):
\[ v = \frac{1}{2\pi} \sqrt{\frac{q \rho}{2m\epsilon_0}} \]

### Final Result
Thus, the oscillation frequency \(v\) of the particle inside the sheet is:
\[ v = \frac{1}{2\pi} \sqrt{\frac{q \rho}{2m\epsilon_0}} \]

---