Three equal charges, each `+q ` are placed on the corners of an equilatreal triangel . The electric field intensity at the centroid of the triangle is

Options

A. `kq//r^(2)`
B. `3kq//r^(2)`
C. `sqrt(3)kq//r^(2)`
D. zero

Explanation

To find the electric field intensity at the centroid of an equilateral triangle with three equal charges placed at its corners, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
We have three equal charges, each of magnitude \( +q \), placed at the vertices of an equilateral triangle. Let's denote the vertices as A, B, and C.

2. **Identifying the Centroid**:
The centroid (G) of an equilateral triangle is the point where the three medians intersect. It is also the center of symmetry for the triangle.

3. **Calculating the Electric Field due to One Charge**:
The electric field \( \vec{E} \) due to a point charge \( +q \) at a distance \( r \) is given by the formula:
\[ \vec{E} = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant.

4. **Distance from Vertices to Centroid**:
In an equilateral triangle, the distance from each vertex to the centroid is \( \frac{2}{3} \) of the height of the triangle. If the side length of the triangle is \( a \), the height \( h \) can be calculated as:
\[ h = \frac{\sqrt{3}}{2} a \]
Therefore, the distance \( r \) from each vertex to the centroid is:
\[ r = \frac{2}{3} h = \frac{2}{3} \cdot \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{3} a \]

5. **Direction of Electric Fields**:
The electric field vectors due to each charge will point away from the charges since they are positive. The angles between the electric fields at the centroid due to each charge will be \( 120^\circ \).

6. **Vector Addition of Electric Fields**:
When we add the electric fields vectorially, we can use the symmetry of the triangle. The electric fields due to the three charges will have equal magnitudes and will be separated by \( 120^\circ \).

7. **Net Electric Field Calculation**:
The symmetry of the arrangement means that the components of the electric fields in the horizontal direction will cancel out. Similarly, the vertical components will also cancel out. Therefore, the net electric field \( \vec{E}_{\text{net}} \) at the centroid is:
\[ \vec{E}_{\text{net}} = \vec{E}_A + \vec{E}_B + \vec{E}_C = 0 \]

### Conclusion:
The net electric field intensity at the centroid of the triangle is zero.