An equilaterial triangle ABC has a side a. Two infinitely long thin straight wires having uniform linear charge densities ' lamda ' and '-lamda' are arranged at A and B perpendicular to the plante of the triangle. The magnitude of electric intensity at the third vertex c will be

Options

A. `(lamda)/(2 pi epsilon_(0)a)`
B. `(lamda)/(pi epsilon_(0)a)`
C. zero
D. `(sqrt(3)lamda)/(2pi epsilon_(0)a)`

Explanation

To find the magnitude of the electric intensity at vertex C of the equilateral triangle ABC, we will follow these steps:

### Step 1: Understand the Setup
We have an equilateral triangle ABC with side length 'a'. There are two infinitely long straight wires with uniform linear charge densities \( \lambda \) at point A and \( -\lambda \) at point B, both perpendicular to the plane of the triangle.

### Step 2: Electric Field Due to Each Wire
The electric field \( E \) due to an infinitely long wire with linear charge density \( \lambda \) at a distance \( r \) is given by the formula:
\[ E = \frac{\lambda}{2\pi \epsilon_0 r} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Calculate the Distance from A and B to C
In our equilateral triangle:
- The distance from A to C and from B to C is equal to the side length \( a \).

### Step 4: Calculate Electric Fields at C
1. **Electric Field due to wire at A (positive charge)**:
- The electric field \( E_A \) at point C due to the wire at A is directed radially outward from A.
- The magnitude is:
\[ E_A = \frac{\lambda}{2\pi \epsilon_0 a} \]

2. **Electric Field due to wire at B (negative charge)**:
- The electric field \( E_B \) at point C due to the wire at B is directed radially inward towards B.
- The magnitude is:
\[ E_B = \frac{\lambda}{2\pi \epsilon_0 a} \]

### Step 5: Analyze the Directions
Since the triangle is equilateral, the angles at vertices A and B are \( 60^\circ \). The electric fields \( E_A \) and \( E_B \) will have components in the horizontal direction (along line AC and BC) and vertical direction (perpendicular to line AB).

### Step 6: Resolve Electric Fields into Components
1. **Horizontal Components**:
- The horizontal component of \( E_A \) at C:
\[ E_{A,x} = E_A \cos(60^\circ) = \frac{\lambda}{2\pi \epsilon_0 a} \cdot \frac{1}{2} = \frac{\lambda}{4\pi \epsilon_0 a} \]
- The horizontal component of \( E_B \) at C:
\[ E_{B,x} = E_B \cos(60^\circ) = \frac{\lambda}{2\pi \epsilon_0 a} \cdot \frac{1}{2} = \frac{\lambda}{4\pi \epsilon_0 a} \]

2. **Vertical Components**:
- The vertical components will cancel each other out since \( E_A \) is directed outward and \( E_B \) is directed inward.

### Step 7: Calculate the Net Electric Field at C
The net electric field \( E_{net} \) at point C is the sum of the horizontal components:
\[ E_{net} = E_{A,x} + E_{B,x} = \frac{\lambda}{4\pi \epsilon_0 a} + \frac{\lambda}{4\pi \epsilon_0 a} = \frac{\lambda}{2\pi \epsilon_0 a} \]

### Final Answer
Thus, the magnitude of the electric intensity at vertex C is:
\[ E_{net} = \frac{\lambda}{2\pi \epsilon_0 a} \]

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