Given the arrangement of charged particles in the figure below, find the net electrostatic force on the q1 = 5.15-µC charged particle. (Assume q2 = 12.33 µC
and q3 = −25.12 µC. Express your answer in vector form.)
Located on the coordinate: q1 at (-2.00cm,0), q2 at (1.00cm,1.00cm) and q3 at (0,-1.00cm)
To find the net electrostatic force on the q1 particle, we need to find the individual forces between q1 and q2, and q1 and q3, and then add them together.
Let's first calculate the force between q1 and q2.
Using the formula for electric force:
F₁₂ = k * |q1 * q2| / r₁₂²
where k is the electrostatic constant (k = 8.99 × 10^9 Nm²/C²), q1 and q2 are the charges, and r₁₂ is the distance between the charges, we can calculate the magnitude and direction of the force.
The distance between q1 and q2 can be found using the distance formula:
r₁₂ = √((x₂ - x₁)² + (y₂ - y₁)²)
where (x₁, y₁) are the coordinates of q1 and (x₂, y₂) are the coordinates of q2.
Substituting the values, we have:
r₁₂ = √((1.00cm - (-2.00cm))² + (1.00cm - 0)²)
= √(3.00cm² + 1.00cm²)
= √(10.00cm²)
= 3.162cm
Substituting the values into the electrostatic force formula:
F₁₂ = (8.99 × 10^9 Nm²/C²) * |5.15 × 10^-6C * 12.33 × 10^-6C| / (3.162cm)²
F₁₂ = (8.99 × 10^9 Nm²/C²) * (6.34 × 10^-11C²) / (3.162cm)²
F₁₂ ≈ 1.81 × 10^-4N
The direction of the force can be determined by considering the signs of the charges. Since q1 and q2 have opposite signs, the force is attractive and acts from q1 towards q2.
Now, let's calculate the force between q1 and q3.
Using the same formula as before with the corresponding values:
r₁₃ = √((0 - (-2.00cm))² + (-1.00cm - 0)²)
= √(2.00cm² + 1.00cm²)
= √(5.00cm²)
= 2.236cm
F₁₃ = (8.99 × 10^9 Nm²/C²) * |-25.12 × 10^-6C * 5.15 × 10^-6C| / (2.236cm)²
F₁₃ = (8.99 × 10^9 Nm²/C²) * (1.29 × 10^-10C²) / (2.236cm)²
F₁₃ ≈ 2.94 × 10^-4N
The direction of the force can be determined by considering the signs of the charges. Since q1 and q3 have the same sign (both negative), the force is repulsive and acts from q1 away from q3.
To find the net force, we add the forces together using vector addition:
F_net = F₁₂ + F₁₃
F_net = (1.81 × 10^-4N) + (2.94 × 10^-4N)
F_net = 4.75 × 10^-4N
So, the net electrostatic force on the q1 = 5.15 µC charged particle is approximately 4.75 × 10^-4N in the direction shown by the vector addition of the individual forces.