A charge of 5 nC is uniformly distributed on a ring of radius 4 cm. Find the potential at the axis at a distance of 3 cm from the center.

Options

A. `0.09 V`
B. 600 V
C. 900 V
D. 300 V

Answer

To find the electric potential \( V \) at a point along the axis of a uniformly charged ring, we can use the formula:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{\sqrt{R^2 + x^2}} \]

where:
- \( Q \) is the total charge on the ring,
- \( R \) is the radius of the ring,
- \( x \) is the distance from the center of the ring along the axis,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.854 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step-by-Step Solution:

1. **Convert the given values to SI units:**
- Charge \( Q = 5 \, \text{nC} = 5 \times 10^{-9} \, \text{C} \)
- Radius of the ring \( R = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Distance from the center along the axis \( x = 3 \, \text{cm} = 0.03 \, \text{m} \)

2. **Calculate \( R^2 + x^2 \):**
\[ R^2 = (0.04)^2 = 0.0016 \, \text{m}^2 \]
\[ x^2 = (0.03)^2 = 0.0009 \, \text{m}^2 \]
\[ R^2 + x^2 = 0.0016 + 0.0009 = 0.0025 \, \text{m}^2 \]

3. **Calculate \( \sqrt{R^2 + x^2} \):**
\[ \sqrt{R^2 + x^2} = \sqrt{0.0025} = 0.05 \, \text{m} \]

4. **Substitute the values into the potential formula:**
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{\sqrt{R^2 + x^2}} \]
\[ V = \frac{1}{4\pi (8.854 \times 10^{-12})} \cdot \frac{5 \times 10^{-9}}{0.05} \]

5. **Calculate the constant term:**
\[ \frac{1}{4\pi (8.854 \times 10^{-12})} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

6. **Calculate the potential:**
\[ V = (9 \times 10^9) \cdot \frac{5 \times 10^{-9}}{0.05} \]
\[ V = (9 \times 10^9) \cdot (1 \times 10^{-7}) = 0.9 \, \text{V} \]

### Final Answer:
The potential at the axis at a distance of 3 cm from the center of the ring is \( 0.9 \, \text{V} \).