Three charge `-q,+q` and `-q` are placed at the cornors of an equilateral triangle of side `a` The resultant electric froce an a charge `+q` placed at a the centroid `O` of the triangle is

Options

A. `(3q^(2))/(4piepsilon_(0)a^(2))`
B. `(q^(2))/(4piepsilon_(0)a^(2))`
C. `(q^(2))/(2piepsilon_(0)a^(2))`
D. `(3q^(2))/(2piepsilon_(0)a^(2))`

Explanation

To solve the problem of finding the resultant electric force on a charge \( +q \) placed at the centroid \( O \) of an equilateral triangle with charges \( -q, +q, -q \) at its corners, we can follow these steps:

### Step 1: Identify the Geometry
We have an equilateral triangle with three charges:
- Charge \( -q \) at point A
- Charge \( +q \) at point B
- Charge \( -q \) at point C

The centroid \( O \) of the triangle is the point where we will place the charge \( +q \).

### Step 2: Calculate the Distance from the Centroid to the Corners
The distance from the centroid \( O \) to any vertex (corner) of the triangle can be calculated using the formula:
\[ r = \frac{a}{\sqrt{3}} \]
where \( a \) is the side length of the triangle.

### Step 3: Calculate the Electric Forces Due to Each Charge
The electric force \( F \) between two point charges is given by Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
where \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

For the charge \( +q \) at the centroid \( O \):
1. **Force due to charge at A (\(-q\))**:
\[ F_A = k \frac{|-q \cdot +q|}{r^2} = k \frac{q^2}{\left(\frac{a}{\sqrt{3}}\right)^2} = k \frac{3q^2}{a^2} \]
This force is attractive and directed towards charge \( -q \) at A.

2. **Force due to charge at B (\(+q\))**:
\[ F_B = k \frac{|+q \cdot +q|}{r^2} = k \frac{q^2}{\left(\frac{a}{\sqrt{3}}\right)^2} = k \frac{3q^2}{a^2} \]
This force is repulsive and directed away from charge \( +q \) at B.

3. **Force due to charge at C (\(-q\))**:
\[ F_C = k \frac{|-q \cdot +q|}{r^2} = k \frac{q^2}{\left(\frac{a}{\sqrt{3}}\right)^2} = k \frac{3q^2}{a^2} \]
This force is attractive and directed towards charge \( -q \) at C.

### Step 4: Resolve Forces into Components
Since the triangle is equilateral, we can resolve the forces into components. The angle between the line connecting the centroid to the vertex and the horizontal axis is \( 30^\circ \) for charges at A and C, and \( 60^\circ \) for charge B.

- **Force components**:
- For \( F_A \) and \( F_C \):
- Horizontal components: \( F_A \cos(30^\circ) \) and \( F_C \cos(30^\circ) \)
- Vertical components: \( F_A \sin(30^\circ) \) and \( F_C \sin(30^\circ) \)

- For \( F_B \):
- Horizontal component: \( F_B \)
- Vertical component: \( 0 \)

### Step 5: Calculate the Resultant Force
The net force in the horizontal direction will cancel out due to symmetry, while the vertical components will add up.

The total vertical force \( F_{net} \) will be:
\[ F_{net} = F_A \sin(30^\circ) + F_C \sin(30^\circ) - F_B \]
Substituting the values:
\[ F_{net} = 2 \cdot k \frac{3q^2}{a^2} \cdot \frac{1}{2} - k \frac{3q^2}{a^2} \]
This simplifies to:
\[ F_{net} = 0 \]

### Final Result
The resultant electric force on the charge \( +q \) placed at the centroid \( O \) is:
\[ \text{Resultant Electric Force} = 0 \]