Three rigit rods are joined to form an equilaterla triangle ABC of side 1 m. Three particles carrying charges `20 muC` each are attached to the vertices of the triangle. The whole system is at rest in an inertial frame. The resultant force on the charged particle. A has the magnitude.

Options

A. zero
B. 3.6N
C. `3.6sqrt3N`
D. 7.2N

Answer

To solve the problem, we need to analyze the forces acting on the charged particle at vertex A of the equilateral triangle formed by the three charges.

### Step-by-Step Solution:

1. **Identify the Charges and Configuration**:
- We have three charges, each of \( Q = 20 \, \mu C = 20 \times 10^{-6} \, C \), located at the vertices of an equilateral triangle ABC with side length \( a = 1 \, m \).

2. **Determine the Forces Acting on Charge A**:
- The charge at vertex A experiences electrostatic forces due to the charges at vertices B and C.
- The force between two point charges is given by Coulomb's Law:
\[ F = k \frac{|Q_1 Q_2|}{r^2} \]
where \( k \) is Coulomb's constant (\( k \approx 8.99 \times 10^9 \, N m^2/C^2 \)), \( Q_1 \) and \( Q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

3. **Calculate the Forces**:
- The distance between any two charges (e.g., A and B, A and C) is \( r = 1 \, m \).
- The force between charges A and B:
\[ F_{AB} = k \frac{Q^2}{a^2} = 8.99 \times 10^9 \frac{(20 \times 10^{-6})^2}{1^2} = 8.99 \times 10^9 \times 400 \times 10^{-12} = 3.596 \, N \]
- The force between charges A and C is the same:
\[ F_{AC} = 3.596 \, N \]

4. **Determine the Direction of Forces**:
- The force \( F_{AB} \) acts along the line connecting A and B, pointing away from A towards B.
- The force \( F_{AC} \) acts along the line connecting A and C, pointing away from A towards C.
- Since the triangle is equilateral, the angle between the forces \( F_{AB} \) and \( F_{AC} \) is \( 60^\circ \).

5. **Calculate the Resultant Force**:
- To find the resultant force \( F_R \) on charge A, we can use the law of cosines or vector addition. The magnitude of the resultant force can be calculated as:
\[ F_R = \sqrt{F_{AB}^2 + F_{AC}^2 + 2 F_{AB} F_{AC} \cos(60^\circ)} \]
- Substituting the values:
\[ F_R = \sqrt{(3.596)^2 + (3.596)^2 + 2 \times 3.596 \times 3.596 \times \frac{1}{2}} \]
\[ = \sqrt{2 \times (3.596)^2 + (3.596)^2} = \sqrt{3 \times (3.596)^2} = 3.596 \sqrt{3} \]
\[ \approx 3.596 \times 1.732 = 6.227 \, N \]

6. **Conclusion**:
- The magnitude of the resultant force on the charged particle A is approximately \( 6.227 \, N \).