A charge Q is spread uniformly in the form of a line charge density `lamda=Q/(3a)` on the sides of an 3a equilateral triangle of perimeter 3a. Calculate the potential at the centroid C of the triangle.

Explanation

To calculate the potential at the centroid \( C \) of an equilateral triangle with a uniform line charge density, we can follow these steps:

### Step 1: Understanding the Geometry
The triangle has a perimeter of \( 3a \), which means each side has a length of \( a \). The centroid of an equilateral triangle is located at a distance of \( \frac{a}{\sqrt{3}} \) from each vertex.

### Step 2: Calculate the Charge Density
Given that the total charge \( Q \) is uniformly distributed along the sides of the triangle, the linear charge density \( \lambda \) is given by:
\[ \lambda = \frac{Q}{3a} \]

### Step 3: Determine the Distance from the Centroid to a Side
To find the potential at the centroid, we need to calculate the distance from the centroid to any side of the triangle. For an equilateral triangle, this distance can be calculated using the formula:
\[ d = \frac{a}{2\sqrt{3}} \]

### Step 4: Calculate the Potential Due to One Side
The potential \( V \) due to a line charge at a distance \( d \) is given by:
\[ V = k \int \frac{dq}{r} \]
where \( k \) is Coulomb's constant, \( dq \) is the charge element, and \( r \) is the distance from the charge element to the point where we want to find the potential.

For a small segment \( dx \) of the line charge, the charge \( dq \) is:
\[ dq = \lambda \, dx = \frac{Q}{3a} \, dx \]
The distance \( r \) from the segment to the centroid is constant and equal to \( d \).

### Step 5: Integrate the Potential
The potential due to one side of the triangle is:
\[ V_{\text{one side}} = k \int \frac{dq}{d} = k \int \frac{\lambda \, dx}{d} \]
Since \( d \) is constant for the side, this simplifies to:
\[ V_{\text{one side}} = \frac{k \lambda}{d} \int dx = \frac{k \lambda}{d} \cdot a \]
Substituting \( \lambda \) and \( d \):
\[ V_{\text{one side}} = \frac{k \cdot \frac{Q}{3a}}{\frac{a}{2\sqrt{3}}} \cdot a = \frac{2kQ\sqrt{3}}{3a^2} \]

### Step 6: Total Potential at the Centroid
Since there are three sides contributing equally to the potential at the centroid, the total potential \( V_C \) at the centroid is:
\[ V_C = 3 \cdot V_{\text{one side}} = 3 \cdot \frac{2kQ\sqrt{3}}{3a^2} = \frac{2kQ\sqrt{3}}{a^2} \]

### Final Answer
The total potential at the centroid \( C \) of the triangle is:
\[ V_C = \frac{2kQ\sqrt{3}}{a^2} \]

---