ABC is an equillatreal triangle of each side 5cm. Two charges of `+- (50)/(3) xx 10^(-3) mu C` are placed at A and B respectively. Calculate magnitude and direction of resultant intensity at C.

Explanation

To solve the problem, we need to calculate the electric field intensity at point C due to the charges placed at points A and B in the equilateral triangle ABC.

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions:**
- Charge at A (Q1) = \( +\frac{50}{3} \times 10^{-3} \, \mu C = +\frac{50}{3} \times 10^{-9} \, C \)
- Charge at B (Q2) = \( -\frac{50}{3} \times 10^{-3} \, \mu C = -\frac{50}{3} \times 10^{-9} \, C \)
- Side length of the triangle (L) = 5 cm = \( 5 \times 10^{-2} \, m \)

2. **Calculate the Electric Field Due to Each Charge at Point C:**
- The formula for electric field \( E \) due to a point charge is given by:
\[ E = \frac{k \cdot |Q|}{r^2} \]
where \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) and \( r \) is the distance from the charge to the point where the field is being calculated.

3. **Calculate Electric Field at C due to Charge at A (E_A):**
- Distance from A to C = L = \( 5 \times 10^{-2} \, m \)
- Using the formula:
\[ E_A = \frac{9 \times 10^9 \cdot \left(\frac{50}{3} \times 10^{-9}\right)}{(5 \times 10^{-2})^2} \]
- Simplifying:
\[ E_A = \frac{9 \times 10^9 \cdot \frac{50}{3} \times 10^{-9}}{25 \times 10^{-4}} = \frac{9 \times 50}{3 \times 25} \times 10^4 = 6 \times 10^4 \, N/C \]
- Direction: Away from charge A (since it is positive).

4. **Calculate Electric Field at C due to Charge at B (E_B):**
- Distance from B to C = L = \( 5 \times 10^{-2} \, m \)
- Using the formula:
\[ E_B = \frac{9 \times 10^9 \cdot \left(\frac{50}{3} \times 10^{-9}\right)}{(5 \times 10^{-2})^2} \]
- Since the charge at B is negative, the direction of the electric field will be towards charge B:
\[ E_B = 6 \times 10^4 \, N/C \]

5. **Determine the Resultant Electric Field at C:**
- The electric fields \( E_A \) and \( E_B \) are at an angle of 60 degrees to each other (since ABC is an equilateral triangle).
- The resultant electric field \( E_{net} \) can be calculated using the vector addition:
\[ E_{net} = \sqrt{E_A^2 + E_B^2 + 2E_AE_B \cos(60^\circ)} \]
- Substituting the values:
\[ E_{net} = \sqrt{(6 \times 10^4)^2 + (6 \times 10^4)^2 + 2(6 \times 10^4)(6 \times 10^4) \cdot \frac{1}{2}} \]
- Simplifying:
\[ E_{net} = \sqrt{(6 \times 10^4)^2 + (6 \times 10^4)^2 + (6 \times 10^4)^2} = \sqrt{3(6 \times 10^4)^2} = 6 \times 10^4 \sqrt{3} \, N/C \]

6. **Direction of the Resultant Electric Field:**
- The direction of the resultant electric field will be along the bisector of the angle between the two electric fields, pointing towards the direction of the negative charge (B).

### Final Result:
- Magnitude of the resultant electric field at C:
\[ E_{net} = 6 \sqrt{3} \times 10^4 \, N/C \approx 1.04 \times 10^5 \, N/C \]
- Direction: Towards point B.