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
We have a uniformly charged sphere of radius \( R \) with total charge \( Q \). A particle with mass \( m \) and charge \( -q \) is moving through the sphere along its diameter.

### Step 2: Determine the Electric Field Inside the Sphere
Using Gauss's law, the electric field \( E \) at a distance \( r \) from the center of a uniformly charged sphere (for \( r < R \)) is given by:

\[ E = \frac{Q}{4 \pi \epsilon_0} \cdot \frac{r}{R^3} \]

This shows that the electric field inside the sphere is linearly proportional to the distance \( r \) from the center.

### 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 \( E \):

\[ F = -q \cdot \left(\frac{Q}{4 \pi \epsilon_0} \cdot \frac{r}{R^3}\right) \]

This simplifies to:

\[ F = -\frac{qQ}{4 \pi \epsilon_0 R^3} r \]

### Step 4: Recognize the Form of Simple Harmonic Motion
The force can be rewritten in the form:

\[ F = -k r \]

where \( k = \frac{qQ}{4 \pi \epsilon_0 R^3} \). This indicates that the particle is undergoing simple harmonic motion (SHM) about the center of the sphere.

### Step 5: Relate the Force Constant to Angular Frequency
For SHM, the relationship between the force constant \( k \) and the angular frequency \( \omega \) is given by:

\[ F = -m \omega^2 r \]

Setting the two expressions for \( F \) equal gives:

\[ -\frac{qQ}{4 \pi \epsilon_0 R^3} r = -m \omega^2 r \]

### Step 6: Solve for Angular Frequency
By equating the coefficients of \( r \):

\[ \frac{qQ}{4 \pi \epsilon_0 R^3} = m \omega^2 \]

Rearranging gives:

\[ \omega^2 = \frac{qQ}{4 \pi \epsilon_0 m R^3} \]

Taking the square root yields:

\[ \omega = \sqrt{\frac{qQ}{4 \pi \epsilon_0 m R^3}} \]

### 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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