Consider three charges `q, - q,` (in `Sl` units) at the vetrices of an equilateral triangle with side length b. The magnitude of electric field at the centroid of the triangle is :

Options

A. `(3q)/(2pi epsilon_(0)b^(2))`
B. `(3q)/(4pi epsilon_(90)b^(2))`
C. `(q)/(2pi epsilon_(0)b^(2))`
D. 0

Explanation

To find the magnitude of the electric field at the centroid of an equilateral triangle with charges at its vertices, we can follow these steps:

### Step 1: Understand the Configuration
We have three charges located at the vertices of an equilateral triangle:
- Charge \( q \) at vertex A
- Charge \( -q \) at vertex B
- Charge \( q \) at vertex C

The side length of the triangle is \( b \).

### Step 2: Locate the Centroid
The centroid of an equilateral triangle is located at a distance of \( \frac{b}{\sqrt{3}} \) from each vertex.

### Step 3: Calculate the Electric Field Due to Each Charge
The electric field \( E \) due to a point charge is given by the formula:
\[ E = \frac{k |q|}{r^2} \]
where \( k \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge to the point where the field is being calculated.

For our configuration:
- Distance \( r \) from each charge to the centroid is \( \frac{b}{\sqrt{3}} \).

Thus, the electric field due to each charge is:
- For charge \( q \) at vertex A:
\[ E_A = \frac{k q}{\left(\frac{b}{\sqrt{3}}\right)^2} = \frac{3k q}{b^2} \]
- For charge \( -q \) at vertex B:
\[ E_B = \frac{k (-q)}{\left(\frac{b}{\sqrt{3}}\right)^2} = -\frac{3k q}{b^2} \]
- For charge \( q \) at vertex C:
\[ E_C = \frac{k q}{\left(\frac{b}{\sqrt{3}}\right)^2} = \frac{3k q}{b^2} \]

### Step 4: Determine the Direction of Each Electric Field
- The electric field \( E_A \) due to charge \( q \) at A points away from A.
- The electric field \( E_B \) due to charge \( -q \) at B points towards B.
- The electric field \( E_C \) due to charge \( q \) at C points away from C.

### Step 5: Resolve the Electric Fields into Components
Since the triangle is equilateral, we can resolve the electric fields into horizontal and vertical components.

- The horizontal components of \( E_A \) and \( E_C \) will cancel each other out because they are equal in magnitude and opposite in direction.
- The vertical components of all three electric fields will add up.

### Step 6: Calculate the Vertical Components
The vertical component of each electric field is given by:
\[ E_{vertical} = E \cdot \sin(60^\circ) = E \cdot \frac{\sqrt{3}}{2} \]
Thus, the total vertical component \( E_{total} \) is:
\[ E_{total} = E_A \cdot \frac{\sqrt{3}}{2} + E_B \cdot \frac{\sqrt{3}}{2} + E_C \cdot \frac{\sqrt{3}}{2} \]
Substituting the values:
\[ E_{total} = \left( \frac{3k q}{b^2} \cdot \frac{\sqrt{3}}{2} \right) + \left( -\frac{3k q}{b^2} \cdot \frac{\sqrt{3}}{2} \right) + \left( \frac{3k q}{b^2} \cdot \frac{\sqrt{3}}{2} \right) \]
The contributions from \( E_A \) and \( E_B \) cancel out, leaving:
\[ E_{total} = \frac{3k q \sqrt{3}}{2b^2} \]

### Step 7: Final Expression
The final expression for the magnitude of the electric field at the centroid is:
\[ E_{centroid} = \frac{3k q}{2b^2} \]