Fig shows three charged particles, held in place by forces not shown. What electrostatic
force, due to the other two charges, acts on q1? Take q1 = -1.2μC, q2 = +3.7μC, q3 = -2.3μC,
r12 = 15cm, r13 = 10cm and θ = 32deg.

To determine the electrostatic force acting on charge q1 due to the other two charges, q2 and q3, we can use Coulomb's Law. This law states that the force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Additionally, we need to consider the direction of the forces, as they can either attract or repel depending on the signs of the charges involved.

Calculating the Forces

Let's break this down step by step:

1. Identify the Charges and Distances

• Charge q1 = -1.2 μC
• Charge q2 = +3.7 μC, distance r12 = 15 cm
• Charge q3 = -2.3 μC, distance r13 = 10 cm
• Angle θ = 32 degrees (between the lines connecting q1 to q2 and q1 to q3)

2. Calculate the Force Between q1 and q2

Using Coulomb's Law:

F₁₂ = k * |q1 * q2| / r₁₂²

Where:

• k = 8.99 x 10⁹ N m²/C² (Coulomb's constant)
• r₁₂ = 0.15 m (convert cm to m)

Substituting the values:

F₁₂ = (8.99 x 10⁹) * (1.2 x 10^-6) * (3.7 x 10^-6) / (0.15)²

Calculating this gives:

F₁₂ ≈ 0.224 N

Since q1 is negative and q2 is positive, the force F₁₂ will be attractive, pulling q1 towards q2.

3. Calculate the Force Between q1 and q3

Using the same formula:

F₁₃ = k * |q1 * q3| / r₁₃²

Where:

• r₁₃ = 0.10 m

Substituting the values:

F₁₃ = (8.99 x 10⁹) * (1.2 x 10^-6) * (2.3 x 10^-6) / (0.10)²

Calculating this gives:

F₁₃ ≈ 0.248 N

Since both q1 and q3 are negative, the force F₁₃ will be repulsive, pushing q1 away from q3.

Finding the Resultant Force on q1

Now, we need to consider the direction of these forces. The force F₁₂ acts towards q2, while F₁₃ acts away from q3. To find the net force on q1, we can break down the forces into their components.

4. Resolve Forces into Components

Assuming the x-axis is along the line connecting q1 and q2:

• The x-component of F₁₂ is F₁₂ (since it acts directly along the x-axis).
• The y-component of F₁₂ is 0 (no vertical component).
• The x-component of F₁₃ is F₁₃ * cos(θ).
• The y-component of F₁₃ is F₁₃ * sin(θ).

Calculating the components:

F_13x = 0.248 * cos(32°) ≈ 0.210 N

F_13y = 0.248 * sin(32°) ≈ 0.130 N

5. Combine the Forces

Now, we can find the net force components:

• Net F_x = F₁₂ - F_13x = 0.224 N - 0.210 N = 0.014 N (towards q2)
• Net F_y = 0 - F_13y = -0.130 N (downward)

Resultant Force Magnitude and Direction

To find the magnitude of the resultant force, we use the Pythagorean theorem:

F_net = √(F_x² + F_y²)

Substituting the values:

F_net = √((0.014)² + (-0.130)²) ≈ 0.130 N

The direction can be found using the arctangent function:

θ_net = arctan(F_y / F_x)

θ_net = arctan(-0.130 / 0.014) ≈ -82.5°

This angle indicates that the resultant force is directed downward and slightly towards q2. Thus, the electrostatic force