Equal charges `q` are placed at the vertices `A` and `B` of an equilatral triangle `ABC` of side `a`. The magnitude of electric field at the point `C` is

Options

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

Explanation

To find the magnitude of the electric field at point C due to equal charges \( q \) placed at vertices A and B of an equilateral triangle ABC with side length \( a \), we can follow these steps:

### Step 1: Understand the Configuration
We have an equilateral triangle ABC with equal charges \( q \) at points A and B. We need to determine the electric field at point C due to these charges.

### Step 2: Calculate the Electric Field Due to Charge at A
The electric field \( E_A \) at point C due to charge \( q \) at point A can be calculated using Coulomb's law:
\[ E_A = \frac{k \cdot q}{r^2} \]
where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)) and \( r \) is the distance from charge A to point C. In this case, \( r = a \).

Thus,
\[ E_A = \frac{k \cdot q}{a^2} \]

### Step 3: Calculate the Electric Field Due to Charge at B
Similarly, the electric field \( E_B \) at point C due to charge \( q \) at point B is also given by:
\[ E_B = \frac{k \cdot q}{a^2} \]

### Step 4: Determine the Direction of the Electric Fields
The electric field due to a positive charge points away from the charge. Therefore:
- The electric field \( E_A \) from charge A at point C points towards the direction from A to C.
- The electric field \( E_B \) from charge B at point C points towards the direction from B to C.

### Step 5: Resolve the Electric Fields into Components
Since the triangle is equilateral, the angle between the line joining A to C and the horizontal line (line joining A to B) is \( 60^\circ \). Therefore, we can resolve \( E_A \) and \( E_B \) into horizontal and vertical components.

The horizontal components \( E_{Ax} \) and \( E_{Bx} \) are:
\[ E_{Ax} = E_A \cos(60^\circ) = \frac{k \cdot q}{a^2} \cdot \frac{1}{2} = \frac{k \cdot q}{2a^2} \]
\[ E_{Bx} = E_B \cos(60^\circ) = \frac{k \cdot q}{a^2} \cdot \frac{1}{2} = \frac{k \cdot q}{2a^2} \]

The vertical components \( E_{Ay} \) and \( E_{By} \) are:
\[ E_{Ay} = E_A \sin(60^\circ) = \frac{k \cdot q}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{k \cdot q \sqrt{3}}{2a^2} \]
\[ E_{By} = E_B \sin(60^\circ) = \frac{k \cdot q}{a^2} \cdot \frac{\sqrt{3}}{2} = \frac{k \cdot q \sqrt{3}}{2a^2} \]

### Step 6: Combine the Electric Fields
Since the horizontal components \( E_{Ax} \) and \( E_{Bx} \) are in opposite directions, they will cancel each other out. The vertical components \( E_{Ay} \) and \( E_{By} \) will add up:
\[ E_{net} = E_{Ay} + E_{By} = \frac{k \cdot q \sqrt{3}}{2a^2} + \frac{k \cdot q \sqrt{3}}{2a^2} = \frac{k \cdot q \sqrt{3}}{a^2} \]

### Final Result
Thus, the magnitude of the electric field at point C is:
\[ E_{net} = \frac{k \cdot q \sqrt{3}}{a^2} \]