An arc of radius r carries charge. The linear density of charge is `lamda` and the arc subtends an angle `(pi)/(3)` at the centre.

Options

A. `(lamda)/(4 epsilon_(0))`
B. `(lamda)/(8epsilon_(0))`
C. `(lamda)/(12 epsilon_(0))`
D. `(lamda)/(16 epsilon_(0))`

Answer

To solve the problem, we need to calculate the electric potential \( V \) at the center of an arc of radius \( r \) that carries a linear charge density \( \lambda \) and subtends an angle of \( \frac{\pi}{3} \) at the center.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the arc, \( r \)
- Linear charge density, \( \lambda \)
- Angle subtended at the center, \( \theta = \frac{\pi}{3} \)

2. **Calculate the Length of the Arc:**
The length \( L \) of the arc can be calculated using the formula:
\[ L = r \theta \]
Substituting the value of \( \theta \):
\[ L = r \cdot \frac{\pi}{3} = \frac{r\pi}{3} \]

3. **Calculate the Total Charge \( Q \) on the Arc:**
The total charge \( Q \) on the arc can be found using the linear charge density:
\[ Q = \lambda \cdot L \]
Substituting the expression for \( L \):
\[ Q = \lambda \cdot \frac{r\pi}{3} = \frac{\lambda r \pi}{3} \]

4. **Use the Formula for Electric Potential \( V \):**
The electric potential \( V \) at a distance \( r \) from a point charge is given by:
\[ V = k \frac{Q}{r} \]
where \( k = \frac{1}{4\pi \epsilon_0} \).

5. **Substitute \( Q \) into the Potential Formula:**
Substituting the expression for \( Q \):
\[ V = k \frac{\frac{\lambda r \pi}{3}}{r} \]
The \( r \) in the numerator and denominator cancels out:
\[ V = k \frac{\lambda \pi}{3} \]

6. **Substitute the Value of \( k \):**
Now substitute \( k = \frac{1}{4\pi \epsilon_0} \):
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{\lambda \pi}{3} \]

7. **Simplify the Expression:**
Simplifying the expression gives:
\[ V = \frac{\lambda}{12 \epsilon_0} \]

8. **Final Result:**
Therefore, the electric potential \( V \) at the center of the arc is:
\[ V = \frac{\lambda}{12 \epsilon_0} \]

### Conclusion:
The correct option is \( \frac{\lambda}{12 \epsilon_0} \).