A ring of radius R is charged uniformly with a charge `+Q`. The potential at a point on its axis at a distance 'r' from any point on ring will be-

Options

A. `(kQ)/(sqrt(R^(2)+r^(2)))`
B. `(kQ)/(r)`
C. `(kQ)/((r^(2)+R^(2))^(3//2))`
D. `(kQ)/(r^(2))(r^(2)-R^(2))^(1//2)`

Answer

To find the electric potential at a point on the axis of a uniformly charged ring, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have a ring of radius \( R \) uniformly charged with a total charge \( +Q \).
- We want to find the electric potential at a point \( P \) located on the axis of the ring at a distance \( r \) from the center of the ring.

2. **Identifying the Element of Charge**:
- Consider a small element of charge \( dq \) on the ring. The total charge \( Q \) is uniformly distributed along the ring.

3. **Calculating the Potential due to the Charge Element**:
- The potential \( dV \) at point \( P \) due to the charge element \( dq \) is given by the formula:
\[ dV = \frac{k \, dq}{d} \]
where \( k \) is Coulomb's constant and \( d \) is the distance from the charge element \( dq \) to the point \( P \).

4. **Finding the Distance \( d \)**:
- The distance \( d \) can be calculated using the Pythagorean theorem:
\[ d = \sqrt{R^2 + r^2} \]
where \( R \) is the radius of the ring and \( r \) is the distance from the center of the ring to point \( P \).

5. **Integrating to Find Total Potential**:
- Since the potential due to each infinitesimal charge \( dq \) is the same at point \( P \) (due to symmetry), we can integrate over the entire ring:
\[ V = \int dV = \int \frac{k \, dq}{\sqrt{R^2 + r^2}} \]
- The total charge \( Q \) can be expressed as:
\[ Q = \int dq \]
- Therefore, the total potential \( V \) at point \( P \) becomes:
\[ V = \frac{k \, Q}{\sqrt{R^2 + r^2}} \]

6. **Final Result**:
- The potential at point \( P \) on the axis of the ring is:
\[ V = \frac{kQ}{\sqrt{R^2 + r^2}} \]