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 question regarding the gravitational field and potential at the center of a uniform ring of mass \( M \) and radius \( R \), we will analyze the gravitational field and gravitational potential step by step.

### Step 1: Understanding Gravitational Field
The gravitational field \( \vec{E} \) at a point is defined as the force experienced by a unit mass placed at that point. For a uniform ring, we need to consider the symmetry of the ring.

### Step 2: Gravitational Field at the Center
At the center of the ring, due to symmetry, the gravitational forces exerted by the mass elements of the ring cancel each other out. Therefore, the net gravitational field \( \vec{E} \) at the center of the ring is:

\[ \vec{E} = 0 \]

### Step 3: Understanding Gravitational Potential
The gravitational potential \( V \) at a point is defined as the work done in bringing a unit mass from infinity to that point against the gravitational field.

### Step 4: Gravitational Potential at the Center
The gravitational potential \( V \) due to a mass \( M \) at a distance \( r \) is given by the formula:

\[ V = -\frac{GM}{r} \]

For a point at the center of the ring, the distance from any mass element of the ring to the center is equal to the radius \( R \) of the ring. Therefore, the gravitational potential at the center of the ring is:

\[ V = -\frac{GM}{R} \]

### Conclusion
Thus, at the center of a uniform ring of mass \( M \) and radius \( R \):
- The gravitational field \( E \) is \( 0 \).
- The gravitational potential \( V \) is \( -\frac{GM}{R} \).

### Final Answer
- Gravitational Field \( E = 0 \)
- Gravitational Potential \( V = -\frac{GM}{R} \)

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