In a regular polygon of `n` sides, each corner is at a distance `r` from the centre. Identical charges are placed at `(n-1)` corners. At the centre, the intensity is `E` and the potential is `V`. The ratio `V//E` has magnitude

Options

A. rn
B. `r(n-1)`
C. `(n-1)//r`
D. `r(n-1)//n`

Explanation

To solve the problem, we need to find the ratio \( \frac{V}{E} \) where \( V \) is the electric potential at the center of a regular polygon with \( n \) sides and \( E \) is the electric field intensity at the center due to \( n-1 \) identical charges placed at the corners.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a regular polygon with \( n \) corners.
- Identical charges \( q \) are placed at \( n-1 \) corners, leaving one corner empty.
- The distance from the center of the polygon to each corner is \( r \).

2. **Electric Field Calculation**:
- The electric field \( E \) at the center due to a single charge \( q \) placed at a distance \( r \) is given by:
\[ E = \frac{kq}{r^2} \]
- Since there are \( n-1 \) charges, we need to consider the vector sum of the electric fields due to these charges. However, due to symmetry, the contributions from the charges will cancel out in certain directions.
- The net electric field \( E \) at the center due to \( n-1 \) charges can be derived from the symmetry of the polygon. The resultant electric field will be directed away from the empty corner.

3. **Potential Calculation**:
- The electric potential \( V \) at the center due to a single charge \( q \) is given by:
\[ V = \frac{kq}{r} \]
- Since there are \( n-1 \) charges, the total potential at the center is:
\[ V = (n-1) \cdot \frac{kq}{r} = \frac{(n-1)kq}{r} \]

4. **Finding the Ratio \( \frac{V}{E} \)**:
- Now, we need to find the ratio \( \frac{V}{E} \):
\[ \frac{V}{E} = \frac{\frac{(n-1)kq}{r}}{\frac{kq}{r^2}} = \frac{(n-1)kq}{r} \cdot \frac{r^2}{kq} \]
- Simplifying this gives:
\[ \frac{V}{E} = \frac{(n-1)r^2}{r} = (n-1)r \]

5. **Final Result**:
- Therefore, the ratio \( \frac{V}{E} \) has a magnitude of:
\[ \frac{V}{E} = r(n-1) \]

### Conclusion:
The correct answer is \( r(n-1) \).