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 find the electric potential \( V \) at the center of an arc of radius \( r \) that carries a linear charge density \( \lambda \) and subtends an angle \( \frac{\pi}{3} \) at the center.

### Step-by-Step Solution:

1. **Identify the Length of the Arc**:
The length \( l \) of the arc can be calculated using the formula:
\[ l = r \theta \]
where \( \theta \) is the angle in radians. Here, \( \theta = \frac{\pi}{3} \).
\[ l = r \cdot \frac{\pi}{3} = \frac{\pi r}{3} \]

**Hint**: Remember that the length of an arc is given by the product of the radius and the angle in radians.

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

**Hint**: The total charge is the product of the linear charge density and the length of the arc.

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

**Hint**: Electric potential due to a point charge depends on the distance from the charge and the magnitude of the charge.

4. **Substitute for \( Q \)**:
Now substitute \( Q \) into the potential formula:
\[ V = k \frac{\lambda \cdot \frac{\pi r}{3}}{r} \]

**Hint**: Simplifying the expression will help you to find the final form of the potential.

5. **Simplify the Expression**:
The \( r \) in the numerator and denominator cancels out:
\[ V = k \cdot \frac{\lambda \pi}{3} \]

**Hint**: Always look for opportunities to simplify expressions by canceling common factors.

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

**Hint**: When substituting constants, ensure to keep track of all factors.

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

**Hint**: Double-check your arithmetic to ensure accuracy in the final expression.

### Final Answer:
The electric potential \( V \) at the center of the arc is:
\[ V = \frac{\lambda}{12 \epsilon_0} \]