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/(4piepsi_(0)a^(2))`
B. `(sqrt(3) q)/(4pi epsi_(0)a^(2))`
C. `(sqrt(2) q)/(4pi epsi_(0)a^(2))`
D. `(2q)/(4piepsi_(0)a^(2))`

Answer

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: Identify the Electric Fields Due to Each Charge
The electric field at point C due to the charge at point A (denoted as \( E_2 \)) and the charge at point B (denoted as \( E_1 \)) can be calculated using Coulomb's law. The electric field \( E \) due to a point charge is given by:

\[ E = \frac{k \cdot q}{r^2} \]

where \( k = \frac{1}{4 \pi \epsilon_0} \) and \( r \) is the distance from the charge to the point where the field is being calculated.

Since the distance from A to C and from B to C is equal to \( a \), we have:

\[ E_1 = \frac{k \cdot q}{a^2} \]
\[ E_2 = \frac{k \cdot q}{a^2} \]

### Step 2: Determine the Directions of the Electric Fields
The electric field \( E_1 \) due to charge at B will point from B to C, while the electric field \( E_2 \) due to charge at A will point from A to C. In an equilateral triangle, the angle between the lines from A and B to C is \( 60^\circ \).

### Step 3: Calculate the Resultant Electric Field
To find the net electric field \( E_{\text{net}} \) at point C, we can use the vector addition of the two electric fields. The formula for the resultant of two vectors \( E_1 \) and \( E_2 \) at an angle \( \theta \) is given by:

\[ E_{\text{net}} = \sqrt{E_1^2 + E_2^2 + 2 E_1 E_2 \cos \theta} \]

Substituting \( E_1 = E_2 = \frac{k \cdot q}{a^2} \) and \( \theta = 60^\circ \):

\[ E_{\text{net}} = \sqrt{E_1^2 + E_1^2 + 2 E_1 E_1 \cos 60^\circ} \]

### Step 4: Simplify the Expression
Since \( \cos 60^\circ = \frac{1}{2} \):

\[ E_{\text{net}} = \sqrt{E_1^2 + E_1^2 + 2 E_1^2 \cdot \frac{1}{2}} \]
\[ = \sqrt{2E_1^2 + E_1^2} = \sqrt{3E_1^2} = \sqrt{3} E_1 \]

### Step 5: Substitute \( E_1 \) Back into the Equation
Now substituting \( E_1 = \frac{k \cdot q}{a^2} \):

\[ E_{\text{net}} = \sqrt{3} \cdot \frac{k \cdot q}{a^2} \]

### Final Result
Thus, the magnitude of the electric field at point C is:

\[ E_{\text{net}} = \frac{\sqrt{3} k q}{a^2} \]