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 \( +q \) placed at its corners, we can follow these steps:

### Step 1: Understand the Configuration
We have three equal charges \( +q \) located at the vertices of an equilateral triangle. Let's denote the vertices as A, B, and C. The centroid (point C) of the triangle is the point where we need to calculate the electric field intensity.

### Step 2: Calculate the Distance from the Charges to the Centroid
In an equilateral triangle, the distance from each vertex to the centroid can be calculated using the formula:
\[ R = \frac{a}{\sqrt{3}} \]
where \( a \) is the length of a side of the triangle.

### Step 3: Calculate the Electric Field Due to One Charge
The electric field \( E \) due to a point charge \( +q \) at a distance \( R \) is given by:
\[ E = \frac{kq}{R^2} \]
Substituting \( R = \frac{a}{\sqrt{3}} \):
\[ E = \frac{kq}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{kq \cdot 3}{a^2} = \frac{3kq}{a^2} \]

### Step 4: Determine the Direction of the Electric Fields
The electric field due to each charge will point away from the charge since they are all positive. We can denote the electric fields due to charges at A, B, and C as \( E_A \), \( E_B \), and \( E_C \), respectively.

### Step 5: Resolve the Electric Fields into Components
Due to symmetry, the horizontal components of the electric fields from charges A and B will cancel each other out. The vertical components will add up.

- The angle between the line connecting the charge to the centroid and the vertical line is \( 30^\circ \) (since the angles in an equilateral triangle are \( 60^\circ \)).
- The vertical component of each electric field is:
\[ E_{vertical} = E \cdot \cos(30^\circ) = \frac{3kq}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}kq}{2a^2} \]

### Step 6: Calculate the Net Electric Field
Since there are three charges, the total vertical component of the electric field at the centroid will be:
\[ E_{net} = E_A + E_B + E_C = 3 \cdot E_{vertical} = 3 \cdot \frac{3\sqrt{3}kq}{2a^2} = \frac{9\sqrt{3}kq}{2a^2} \]
However, since the horizontal components cancel out, the net electric field in the horizontal direction is zero.

### Step 7: Conclusion
The net electric field at the centroid of the triangle is zero because the vertical components from each charge cancel each other out due to symmetry.

Thus, the electric field intensity at the centroid of the triangle is:
\[ \text{Electric Field Intensity} = 0 \]