A ring of radius 3 m has charge density `lambda`. Electric potential at 4 m from its centre on its axis is ____ `[k=(1)/(4pi epsilon_(0))]`

Options

A. `(2)/(5) pi k lambda`
B. `(1)/(5) pi k lambda`
C. `(6)/(5) pi k lambda`
D. `(3)/(5) pi k lambda`

Answer

To find the electric potential at a distance of 4 m from the center of a ring of radius 3 m with a linear charge density \( \lambda \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the parameters:**
- Radius of the ring, \( r = 3 \, \text{m} \)
- Distance from the center along the axis, \( z = 4 \, \text{m} \)
- Linear charge density, \( \lambda \)

2. **Formula for electric potential due to a ring:**
The electric potential \( V \) at a distance \( z \) from the center of a ring of radius \( r \) with a linear charge density \( \lambda \) is given by:
\[ V = k \lambda \frac{2 \pi r}{\sqrt{r^2 + z^2}} \]
where \( k = \frac{1}{4 \pi \epsilon_0} \).

3. **Substituting the values into the formula:**
We substitute \( r = 3 \, \text{m} \) and \( z = 4 \, \text{m} \) into the formula:
\[ V = k \lambda \frac{2 \pi (3)}{\sqrt{(3)^2 + (4)^2}} \]

4. **Calculating the denominator:**
Calculate \( \sqrt{(3)^2 + (4)^2} \):
\[ \sqrt{9 + 16} = \sqrt{25} = 5 \]

5. **Final expression for potential:**
Substitute the value of the denominator back into the equation:
\[ V = k \lambda \frac{2 \pi (3)}{5} \]
Simplifying this gives:
\[ V = \frac{6 \pi k \lambda}{5} \]

### Final Answer:
The electric potential at a distance of 4 m from the center of the ring on its axis is:
\[ V = \frac{6 \pi k \lambda}{5} \]