The electric field at the centroid of an equilateral triangle carrying an equal charge q at each of the vertices is :

Options

A. Zero
B. `(sqrt(2) kg)/(x^(2))`
C. `(kq)/(sqrt(2) x^(2))`
D. `(3kq)/(x^(2))`

Explanation

To find the electric field at the centroid of an equilateral triangle with equal charges \( q \) at each vertex, we can follow these steps:

### Step 1: Understand the Configuration
We have an equilateral triangle with vertices A, B, and C, each carrying a charge \( q \). The centroid (point C) is the point where the three medians of the triangle intersect.

### Step 2: Determine the Electric Field Contribution from Each Charge
The electric field \( \vec{E} \) due to a point charge \( q \) at a distance \( r \) is given by:
\[ \vec{E} = \frac{k \cdot q}{r^2} \hat{r} \]
where \( k \) is Coulomb's constant, and \( \hat{r} \) is the unit vector pointing away from the charge.

### Step 3: Calculate the Magnitude of Electric Fields
Let \( R \) be the distance from each vertex to the centroid. The electric field due to each charge at the centroid will have the same magnitude:
\[ E_1 = E_2 = E_3 = \frac{k \cdot q}{R^2} \]

### Step 4: Analyze the Direction of Electric Fields
- The electric field due to charge \( q \) at vertex A will point away from A towards the centroid.
- The electric field due to charge \( q \) at vertex B will point away from B towards the centroid.
- The electric field due to charge \( q \) at vertex C will point away from C towards the centroid.

### Step 5: Resolve the Electric Fields into Components
Since the triangle is equilateral, the angles between the electric field vectors will be \( 120^\circ \). We can resolve these vectors into horizontal and vertical components.

- The horizontal components of \( E_1 \) and \( E_2 \) will cancel each other out since they are equal in magnitude and opposite in direction.
- The vertical components will add up.

### Step 6: Calculate the Resultant Electric Field
Let’s denote the angle between the electric field vector and the vertical as \( 30^\circ \) (since the angles in an equilateral triangle are \( 60^\circ \)).

The vertical component of each electric field is:
\[ E_{\text{vertical}} = E \sin(30^\circ) = E \cdot \frac{1}{2} \]

Since there are two vertical components contributing to the total vertical electric field:
\[ E_{\text{net vertical}} = 2 \cdot E \cdot \frac{1}{2} = E \]

### Step 7: Conclusion
However, since the horizontal components cancel each other out, the net electric field at the centroid is:
\[ \vec{E}_{\text{net}} = 0 \]

Thus, the electric field at the centroid of the triangle is zero.

### Final Answer
The electric field at the centroid of an equilateral triangle carrying an equal charge \( q \) at each of the vertices is:
\[ \vec{E} = 0 \, \text{N/C} \]