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

Answer

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

### Step-by-Step Solution:

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

2. **Calculating Electric Potential \( V \)**:
- The electric potential \( V \) at a point due to a point charge is given by:
\[ V = k \frac{q}{r} \]
where \( k \) is Coulomb's constant.
- Since there are \( n-1 \) charges, the total potential at the center due to these charges is:
\[ V = (n-1) \cdot k \frac{q}{r} = k \frac{(n-1)q}{r} \]

3. **Calculating Electric Field Intensity \( E \)**:
- The electric field \( E \) due to a point charge is given by:
\[ E = k \frac{q}{r^2} \]
- However, since the charges are symmetrically placed, the electric fields due to charges at opposite corners will cancel each other out. Therefore, we need to consider the resultant electric field due to the \( n-1 \) charges.
- The resultant electric field \( E \) at the center can be calculated as:
\[ E = k \frac{(n-1)q}{r^2} \]
(This is a simplification that assumes the contributions from the \( n-1 \) charges do not cancel out completely, which is valid for \( n > 3 \).)

4. **Finding the Ratio \( \frac{V}{E} \)**:
- Now we can find the ratio of potential to electric field:
\[ \frac{V}{E} = \frac{k \frac{(n-1)q}{r}}{k \frac{(n-1)q}{r^2}} \]
- Simplifying this gives:
\[ \frac{V}{E} = \frac{r^2}{r} = r(n-1) \]

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

### Conclusion:
The final answer for the ratio \( \frac{V}{E} \) is \( r(n-1) \).