Point charge 2.5 is located at x = 0, y = 0.30 , point charge -2.5 is located at x = 0 and y = -0.30 . What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge = 5.0 at x = 0.40 , y = 0?
To calculate the magnitude and direction of the total electric force exerted on the third point charge, we need to consider the forces due to each individual charge and then add them together using vector addition.
Let's start with the force exerted by the first point charge (2.5) on the third point charge (5.0):
(a) Magnitude of the force due to the first charge (q₁):
Using Coulomb's Law, the magnitude of the force (F₁) is given by:
F₁ = k * |q₁ * q₃| / r₁²,
where k is the Coulomb's constant (k ≈ 9.0 × 10^9 N·m²/C²), q₁ is the charge of the first point charge (2.5), q₃ is the charge of the third point charge (5.0), and r₁ is the distance between them (0.4 - 0).
Plug in the values:
F₁ = (9.0 × 10^9 N·m²/C²) * |(2.5 C) * (5.0 C)| / (0.4 m)²
F₁ = [(9.0 × 10^9) * (2.5) * (5.0)] / (0.4)²
(b) Direction of the force due to the first charge (q₁):
Since both charges are positive, the force is repulsive. Therefore, the direction is away from the first charge (along the positive x-axis).
Now let's calculate the force exerted by the second point charge (-2.5) on the third point charge (5.0):
(a) Magnitude of the force due to the second charge (q₂):
Using Coulomb's Law, the magnitude of the force (F₂) is given by:
F₂ = k * |q₂ * q₃| / r₂²,
where q₂ is the charge of the second point charge (-2.5), and r₂ is the distance between them (0.4 - 0).
Plug in the values:
F₂ = (9.0 × 10^9 N·m²/C²) * |(-2.5 C) * (5.0 C)| / (0.4 m)²
F₂ = [(9.0 × 10^9) * (2.5) * (5.0)] / (0.4)²
(b) Direction of the force due to the second charge (q₂):
Since one charge is positive and the other is negative, the force is attractive. Therefore, the direction is towards the second charge (along the positive y-axis).
To find the total force, we need to combine the forces due to each charge. Since these forces act along different directions, we can decompose them into x and y components:
(a) Magnitude of the total force (F):
F = sqrt((F₁x + F₂x)² + (F₁y + F₂y)²)
To find the x and y components of each force, we can use trigonometry:
F₁x = F₁ * cos(θ₁)
F₁y = F₁ * sin(θ₁)
F₂x = F₂ * cos(θ₂)
F₂y = F₂ * sin(θ₂)
(b) Direction of the total force:
The direction of the total force can be given as an angle θ with respect to the positive x-axis. This angle can be calculated using:
θ = atan((F₁y + F₂y) / (F₁x + F₂x))
Substituting the values into the equations above, you can calculate the magnitude and direction of the total electric force.
To find the magnitude and direction of the total electric force on the third point charge, we can use Coulomb's Law.
Coulomb's Law states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The force is also directed along the line connecting the charges.
Let's break down the problem into steps:
Step 1: Calculate the force between the first charge (2.5) and the third charge (5.0).
The distance between them is given by:
r₁₃ = √((x₃ - x₁)² + (y₃ - y₁)²)
r₁₃ = √((0.40 - 0)² + (0 - 0.30)²)
r₁₃ = √((0.40)² + (-0.30)²)
r₁₃ = √(0.16 + 0.09)
r₁₃ = √0.25
r₁₃ = 0.5
Now, we can calculate the magnitude of the electric force using Coulomb's Law:
F₁₃ = (k * |q₁| * |q₃|) / r₁₃²
Where k is the Coulomb's constant (approximately 9.0 x 10^9 N·m²/C²).
F₁₃ = (9.0 x 10^9 * |2.5| * |5.0|) / (0.5)²
F₁₃ = (9.0 x 10^9 * 2.5 * 5.0) / 0.25
F₁₃ = (9.0 x 2.5 x 5.0) x 10^9 / 0.25
F₁₃ = 112.5 x 10^9 / 0.25
F₁₃ = 450 x 10^9 = 4.50 x 10^11 N
Step 2: Calculate the force between the second charge (-2.5) and the third charge (5.0).
The distance between them is the same as in Step 1:
r₂₃ = 0.5
Using Coulomb's Law:
F₂₃ = (k * |q₂| * |q₃|) / r₂₃²
F₂₃ = (9.0 x 10^9 * |-2.5| * |5.0|) / (0.5)²
F₂₃ = (9.0 x 10^9 * 2.5 * 5.0) / 0.25
F₂₃ = (9.0 x 2.5 x 5.0) x 10^9 / 0.25
F₂₃ = 112.5 x 10^9 / 0.25
F₂₃ = 450 x 10^9 = 4.50 x 10^11 N
Step 3: Calculate the net force.
Since the two forces are acting along the same line, they add up.
F_net = F₁₃ + F₂₃
F_net = 4.50 x 10^11 + 4.50 x 10^11
F_net = 9.00 x 10^11 N
So, the magnitude of the total electric force is 9.00 x 10^11 N.
Step 4: Calculate the direction.
The direction of the net force can be determined by the angle between the line connecting the charges and the x-axis.
tanθ = (y₃ - y₁) / (x₃ - x₁)
tanθ = (0 - 0.30) / (0.40 - 0)
tanθ = (-0.30) / 0.40
tanθ = -0.75
θ = atan(-0.75)
Using a calculator, we find that θ ≈ -36.87 degrees.
However, to find the direction, we need to consider the signs of the charges.
Since the magnitude of the first charge is positive and the magnitude of the second charge is negative, the force between them acts from the first charge towards the second charge.
Therefore, the direction of the net force is towards the negative x-axis (opposite direction of the positive x-axis).
So, the magnitude of the total electric force is 9.00 x 10^11 N, and the direction is towards the negative x-axis.