The charges and coordinates of two charged particles held fixed in an xy plane are q1 = 2.32 μC, x1 = 3.08 cm, y1 = 0.793 cm and q2 = -3.14 μC, x2 = -1.75 cm, y2 = 1.05 cm. At what (a)x and (b)y coordinates should a third particle of charge q3 = 5.72 μC be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?
To find the x and y coordinates where the net electrostatic force on particle 2 due to particles 1 and 3 is zero, we need to use the concept of Coulomb's law. Coulomb's law states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Let's break down the problem step by step:
Step 1: Calculate the distance between particle 1 and particle 2.
Using the distance formula:
d₁₂ = sqrt((x₂ - x₁)² + (y₂ - y₁)²)
Given:
x₁ = 3.08 cm
y₁ = 0.793 cm
x₂ = -1.75 cm
y₂ = 1.05 cm
Substituting the values:
d₁₂ = sqrt((-1.75 - 3.08)² + (1.05 - 0.793)²)
Evaluating the expression, you should find that d₁₂ ≈ 4.02 cm.
Step 2: Find the force exerted by particle 1 on particle 2.
Using Coulomb's law:
F₁₂ = (k * |q₁ * q₂|) / d₁₂²
Given:
q₁ = 2.32 μC
q₂ = -3.14 μC
k ≈ 8.988 × 10^9 N m²/C²
Substituting the values:
F₁₂ = (8.988 × 10^9 * |2.32 × 10^-6 * -3.14 × 10^-6|) / (4.02 × 10^-2)²
Evaluating the expression, you should find that F₁₂ ≈ 8.362 N (Newton) (Note that the negative sign indicates the force is attractive).
Step 3: Find the x and y components of force F₁₂.
The x component of F₁₂ can be calculated as:
F₁₂,x = F₁₂ * (x₂ - x₁) / d₁₂
The y component of F₁₂ can be calculated as:
F₁₂,y = F₁₂ * (y₂ - y₁) / d₁₂
Substituting the values:
F₁₂,x = 8.362 * (-1.75 - 3.08) / 4.02
F₁₂,y = 8.362 * (1.05 - 0.793) / 4.02
Evaluating the expressions, you should find that F₁₂,x ≈ -2.49 N and F₁₂,y ≈ 0.471 N.
Step 4: Find the x and y coordinates of particle 3.
To make the net force on particle 2 due to particles 1 and 3 zero, the x and y components of force due to particle 3 should balance the x and y components of force due to particle 1.
For the x direction:
F₃,x = -F₁₂,x
Substituting the values:
F₃,x = -(-2.49)
F₃,x ≈ 2.49 N
Using the formula for x component of force:
F₃,x = k * |q₂ * q₃| / d₃₂²
Given:
q₃ = 5.72 μC
Substituting the values:
2.49 = (8.988 × 10^9 * |(-3.14 × 10^-6) * (5.72 × 10^-6)|) / (d₃₂)²
Solving this equation for d₃₂, you can find the distance between particle 2 and particle 3.
For the y direction:
F₃,y = -F₁₂,y
Substituting the values:
0.471 = (8.988 × 10^9 * |(-3.14 × 10^-6) * (5.72 × 10^-6)|) / (d₃₂)²
Solving this equation for d₃₂, you can find the distance between particle 2 and particle 3.
Once you have the distances, you can determine the x and y coordinates of particle 3 by using the given coordinates for particle 2 and subtracting the respective distance values in the x and y directions.