Particle 1 carrying -4.0 μC of charge is fixed at the origin of an xy coordinate system, particle 2 carrying +8.0 μC of charge is located on the x axis at x = 2.0 m , and particle 3, identical to particle 2, is located on the x axis at x = -2.0 m .
Part A
What is the vector sum of the electric forces exerted on particle 3? Determine the x and y components of the vector sum.
All of your particles lie on the x axis. The force is therefore also along the x axis !
F = k Qa Qb /d^2, repulsive if same sign of Q
so
Fx at 3 = k (8 μC) (4 μC)/4 pulling right - k (8 μC) (8 μC)/16 pushing left
= k (+8-4) = +4 k μ^2
q=2.0*10^-6 C
Well, since particle 1 is fixed at the origin, it doesn't really contribute to the electric forces on particle 3. So, let's focus on particle 2.
The electric force between particle 2 and particle 3 can be calculated using Coulomb's law:
F = k * (|q1| * |q2|) / r^2
Where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between the charges.
In this case, q1 = +8.0 μC, q2 = +8.0 μC, and r = 4.0 m (since particle 2 is at x = 2.0 m and particle 3 is at x = -2.0 m).
Now, let's break down the force into its x and y components. Since particle 2 and particle 3 are on the x-axis, the only force component is in the x-direction.
F_x = F * cos(θ)
F_y = F * sin(θ)
Since the x-axis is horizontal, the angle θ is 0 degrees. Therefore, cos(0) = 1 and sin(0) = 0.
So, the x-component of the force on particle 3 is F_x = F * 1 = F, and the y-component is F_y = F * 0 = 0.
In conclusion, the vector sum of the electric forces exerted on particle 3 has an x-component equal to the electric force F, and a y-component of 0.
To determine the vector sum of the electric forces exerted on particle 3, we need to calculate the electric force between particle 3 and particles 1 and 2 separately. The electric force between two charged particles can be calculated using Coulomb's law:
F = k * |q1 * q2| / r^2 ,
where F is the electric force, k is Coulomb's constant (k ≈ 9 × 10^9 N·m^2/C^2), q1 and q2 are the charges of the particles, and r is the distance between them.
Let's first calculate the electric force exerted on particle 3 by particle 1 at the origin.
Given:
Charge of particle 1, q1 = -4.0 μC
Charge of particle 3, q3 = +8.0 μC
Distance between particle 1 and particle 3, r1 = -2.0 m
Using Coulomb's law, we can calculate the electric force:
F1 = k * |q1 * q3| / r1^2.
Now let's calculate the electric force exerted on particle 3 by particle 2.
Given:
Charge of particle 2, q2 = +8.0 μC
Distance between particle 2 and particle 3, r2 = 4.0 m
Using Coulomb's law, we can calculate the electric force:
F2 = k * |q2 * q3| / r2^2.
To find the vector sum of the forces, we need to add the x and y components of the forces separately.
The x component of the force exerted by particle 1 on particle 3 is F1x = F1 * cos(θ1), where θ1 is the angle between the x-axis and the force. Since particle 3 is on the x-axis, θ1 = 0°, and cos(0°) = 1. Thus, F1x = F1.
The y component of the force exerted by particle 1 on particle 3 is F1y = 0, as particle 3 is located on the x-axis.
The x component of the force exerted by particle 2 on particle 3 is F2x = -F2, as particle 2 is located to the right of particle 3.
The y component of the force exerted by particle 2 on particle 3 is F2y = 0, as particle 3 is located on the x-axis.
To find the vector sum of the forces, we add the x components and y components separately:
Fx = F1x + F2x
Fy = F1y + F2y
Finally, the vector sum of the electric forces exerted on particle 3 is the resulting force vector with components Fx and Fy.