A particle of mass m and charge -q moves diametrically through a uniformily charged sphere of radius R with total charge Q. The angular frequency of the particle's simple harminic motion, if its amplitude lt R, is given by

Options

A. `sqrt((1)/(4 pi epsilon_(0))(qQ)/(mR))`
B. `sqrt((1)/(4 pi epsilon_(0))(qQ)/(mR^(2)))`
C. `sqrt((1)/(4 pi epsilon_(0))(qQ)/(mR^(3)))`
D. `sqrt((1)/(4 pi epsilon_(0))(m)/(qQ))`

Explanation

To solve the problem of finding the angular frequency of a particle of mass \( m \) and charge \( -q \) moving diametrically through a uniformly charged sphere of radius \( R \) with total charge \( Q \), we can follow these steps:

### Step 1: Understand the System
The particle with charge \( -q \) is moving through a uniformly charged sphere with charge \( Q \). The electric field inside a uniformly charged sphere is given by:
\[ E = \frac{Q r}{4 \pi \epsilon_0 R^3} \]
where \( r \) is the distance from the center of the sphere to the particle.

### Step 2: Determine 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 \( E \):
\[ F = -q \left( \frac{Q r}{4 \pi \epsilon_0 R^3} \right) \]
This force acts towards the center of the sphere (since the charge of the particle is negative).

### Step 3: Relate Force to Simple Harmonic Motion
The force can be expressed in a form similar to Hooke's law, which describes simple harmonic motion (SHM):
\[ F = -kx \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. In our case, \( x = r \) (the distance from the center), and we can identify:
\[ k = \frac{qQ}{4 \pi \epsilon_0 R^3} \]

### Step 4: Set Up the Equation for Angular Frequency
For SHM, the angular frequency \( \omega \) is related to the mass \( m \) and the spring constant \( k \) by the equation:
\[ k = m \omega^2 \]
Substituting our expression for \( k \):
\[ \frac{qQ}{4 \pi \epsilon_0 R^3} = m \omega^2 \]

### Step 5: Solve for Angular Frequency \( \omega \)
Rearranging the equation gives:
\[ \omega^2 = \frac{qQ}{4 \pi \epsilon_0 m R^3} \]
Taking the square root of both sides, we find:
\[ \omega = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}} \]

### Final Answer
Thus, the angular frequency of the particle's simple harmonic motion is:
\[ \omega = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}} \]

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