For a uniform ring of mass M and radius R at its centre

Options

A. field and potential both are zero
B. field is zero but potential is `(GM)/(R)`
C. field is zero but potential is `-GM//R`
D. magnitude of field is `(GM)/(R^(2))` and potential `-(GM)/(R)`

Explanation

To solve the problem regarding the gravitational field and gravitational potential at the center of a uniform ring of mass \( M \) and radius \( R \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a uniform ring with mass \( M \) and radius \( R \). We need to determine the gravitational field and gravitational potential at the center of this ring.

2. **Gravitational Field Calculation**:
- Consider an infinitesimal mass element \( dm \) on the ring. The gravitational field \( d\vec{E} \) due to this mass element at the center of the ring points towards the mass element.
- Now, consider another mass element \( dm \) located directly opposite to the first one on the ring. The gravitational field \( d\vec{E'} \) due to this mass element will also point towards it but in the opposite direction.
- Since both mass elements are equidistant from the center and have the same mass \( dm \), the magnitudes of the gravitational fields \( d\vec{E} \) and \( d\vec{E'} \) will be equal, but their directions will be opposite.

3. **Cancellation of Gravitational Fields**:
- When we sum up the contributions of all such pairs of mass elements around the ring, we find that they will cancel each other out. Therefore, the net gravitational field \( \vec{E} \) at the center of the ring is:
\[ \vec{E} = 0 \]

4. **Gravitational Potential Calculation**:
- The gravitational potential \( V \) due to a mass element \( dm \) at a distance \( R \) from the center is given by:
\[ dV = -\frac{G \, dm}{R} \]
- Since gravitational potential is a scalar quantity, we can sum the contributions from all mass elements without worrying about their directions. Thus, we integrate over the entire ring:
\[ V = \int dV = \int -\frac{G \, dm}{R} \]
- The total mass of the ring is \( M \), so:
\[ V = -\frac{G}{R} \int dm = -\frac{G M}{R} \]

5. **Final Results**:
- The gravitational field at the center of the ring is:
\[ \vec{E} = 0 \]
- The gravitational potential at the center of the ring is:
\[ V = -\frac{G M}{R} \]

### Conclusion:
The gravitational field at the center of a uniform ring is zero, and the gravitational potential at the center is \(-\frac{G M}{R}\).