A semicircular ring of radius 0.5 m is uniformly charged with a total charge of `1.5 xx 10^(-9)` coul. The electric potential at the centre of this ring is :

Options

A. 27 V
B. 13.5 V
C. 54 V
D. 45.5 V

Answer

To find the electric potential at the center of a uniformly charged semicircular ring, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Radius of the semicircular ring, \( r = 0.5 \, \text{m} \)
- Total charge on the ring, \( Q = 1.5 \times 10^{-9} \, \text{C} \)

2. **Understanding Electric Potential**:
- The electric potential \( V \) at a point due to a charge distribution is given by the formula:
\[ V = k \int \frac{dq}{r} \]
- Here, \( k \) is Coulomb's constant, \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \), and \( r \) is the distance from the charge element \( dq \) to the point where the potential is being calculated.

3. **Linear Charge Density**:
- The linear charge density \( \lambda \) for the semicircular ring is given by:
\[ \lambda = \frac{Q}{L} \]
- The length \( L \) of the semicircular ring is \( \pi r \) (since it is half of a full circle).
- Therefore, \( L = \pi \times 0.5 = 1.57 \, \text{m} \).
- Thus, the linear charge density is:
\[ \lambda = \frac{1.5 \times 10^{-9}}{\pi \times 0.5} = \frac{1.5 \times 10^{-9}}{1.57} \approx 9.55 \times 10^{-10} \, \text{C/m} \]

4. **Electric Potential Calculation**:
- The potential \( V \) at the center due to the entire semicircular ring can be calculated by integrating the contributions from each infinitesimal charge element \( dq \):
\[ V = k \int_0^{\pi} \lambda r d\theta \]
- Since \( r \) (the distance from the charge element to the center) is constant and equal to the radius of the ring, we can factor it out:
\[ V = k \lambda r \int_0^{\pi} d\theta = k \lambda r [\theta]_0^{\pi} = k \lambda r \pi \]

5. **Substituting Values**:
- Substitute \( \lambda \) and \( r \) into the equation:
\[ V = k \left(\frac{Q}{\pi r}\right) r \pi = kQ/r \]
- Now substituting the values:
\[ V = \frac{(9 \times 10^9) \times (1.5 \times 10^{-9})}{0.5} \]

6. **Calculating the Final Value**:
- Simplifying:
\[ V = \frac{(9 \times 1.5)}{0.5} = \frac{13.5}{0.5} = 27 \, \text{V} \]

### Final Answer:
The electric potential at the center of the semicircular ring is **27 V**.

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