Consider three charges `q_(1),q_(2)` and `q_(3)` each equal to `q`, at the vertices of an equilateral triangle of side l. What is the force on a charge Q placed at the centroid of the triangle?

Options

A. `(3Qq)/(4piepsilon_(0)l^(2))`
B. `(2Qq)/(4piepsilon_(0)l^(2))`
C. `(Qq)/(2piepsilon_(0)l^(2))`
D. zero

Explanation

To solve the problem of finding the force on a charge \( Q \) placed at the centroid of an equilateral triangle with charges \( q_1, q_2, q_3 \) each equal to \( q \) at the vertices, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Geometry and Symmetry:**
- The charges \( q_1, q_2, q_3 \) are at the vertices of an equilateral triangle with side length \( l \).
- The centroid of an equilateral triangle is equidistant from all three vertices.

2. **Calculate the Distance from Centroid to a Vertex:**
- The distance from the centroid to any vertex of an equilateral triangle is given by:
\[ r = \frac{l}{\sqrt{3}} \]

3. **Determine the Force due to One Charge:**
- The force \( F \) on charge \( Q \) due to one charge \( q \) at a distance \( r \) is given by Coulomb's law:
\[ F = k \frac{qQ}{r^2} \]
- Substituting \( r = \frac{l}{\sqrt{3}} \):
\[ F = k \frac{qQ}{\left(\frac{l}{\sqrt{3}}\right)^2} = k \frac{qQ}{\frac{l^2}{3}} = 3k \frac{qQ}{l^2} \]

4. **Resolve the Forces into Components:**
- Due to symmetry, the forces exerted by the three charges on \( Q \) will have components that cancel out.
- Each force \( F \) can be resolved into horizontal and vertical components.
- Since the triangle is equilateral, the angle between any two forces is \( 120^\circ \).

5. **Calculate the Net Force:**
- The horizontal components of the forces due to \( q_1, q_2, q_3 \) will cancel each other out.
- Similarly, the vertical components will also cancel out due to symmetry.
- Therefore, the net force \( \vec{F}_{\text{net}} \) on \( Q \) is zero.

### Final Result:
\[ \vec{F}_{\text{net}} = 0 \]