A charge q is unifomly distrybuted over a nonconducting ring of radius R. The ring is rotated about an axis passing through its centre and perpendicular to the plane of the ring with frequency f. The ratio of electric potential to the magnetic field at the centre of the ring depends on.

Options

A. `q,f, and R`
B. `q and f only `
C. `f and R `
D. `f`

Answer

To solve the problem, we need to find the ratio of electric potential (V) to the magnetic field (B) at the center of a non-conducting ring with a uniformly distributed charge \( q \) that is rotating with a frequency \( f \).

### Step-by-Step Solution:

1. **Determine the Electric Potential (V) at the Center of the Ring:**
The electric potential \( V \) at the center of a ring with charge \( q \) uniformly distributed over it and radius \( R \) is given by the formula:
\[ V = \frac{kq}{R} \]
where \( k \) is Coulomb's constant.

2. **Determine the Magnetic Field (B) at the Center of the Ring:**
When the ring rotates with frequency \( f \), it creates a current \( I \). The current can be expressed as:
\[ I = \frac{q}{T} = qf \]
where \( T \) is the period of rotation, and \( f \) is the frequency. The magnetic field \( B \) at the center of the ring is given by:
\[ B = \frac{\mu_0 I}{2R} \]
Substituting \( I \) into the equation:
\[ B = \frac{\mu_0 (qf)}{2R} \]

3. **Find the Ratio of Electric Potential to Magnetic Field:**
Now, we need to find the ratio \( \frac{V}{B} \):
\[ \frac{V}{B} = \frac{\frac{kq}{R}}{\frac{\mu_0 (qf)}{2R}} \]
Simplifying this expression:
\[ \frac{V}{B} = \frac{kq}{R} \cdot \frac{2R}{\mu_0 (qf)} = \frac{2k}{\mu_0 f} \]
Here, the \( R \) and \( q \) terms cancel out.

4. **Conclusion:**
The ratio \( \frac{V}{B} \) depends on the constants \( k \) and \( \mu_0 \), and the frequency \( f \). It does not depend on the radius \( R \) or the charge \( q \).

### Final Answer:
The ratio of electric potential to the magnetic field at the center of the ring depends on \( f \).