Answer:
Vfx = 0.42 √x and Vfy = - 0.729 √x
Explanation:
To solve this problem let's use the coulomb law applied to each load
F = k q1 q2 / r²
Let's look for the distance from a corner to the center of the square
r² = (a / 2) ² + (a / 2)²
r = √ (a / 2)² 2
r = a / 2 √2
Let's write the forces, in the attached there is a diagram of the charges
Charge q in the corner
F1 = k q q / r²
F1 = k q² / r²
F1 = k q² 2 / a²
F1 = 8.99 10⁹ (2.8 10⁻⁶)² 2 / 1.5²
F1 = 62.65 10⁻³ N = 6.265 10⁻² N
charge 2q in another corner
F2 = k 2q q / r²
F2 = 2 k q² / r²
Notice that this force is twice the force F1, since the distances are equal
F2 = 2 F1
3q charge
F3 = k 3q q / r²
F3 = 3 F1
6q charge
F4 = k 6q q / r²
F4 = 6 F1
Let's calculate the total force, for this it is useful to break down each force into its components and then add them, let's use trigonometry
cos θ = Fx / F
Fx = F cos θ
sin θ = Fy / F
Fy = F sin θ
Let's add each force
X axis
Fx = F1x - F2x - F3x + F4x
Fx = F1 cos 45 - 2F1 cos 45 - 3F1 cos 45 + 6 F1 cos 45
Fx = 2 F1 cos 45
Fx = 2 (6,265 10-2) cos 45
Fx = 8.86 10-2 N
Axis y
Fy = F1y + F2y -F3y -F4y
Fy = F1 sin 45 + 2 F1 sin 45 - 3 F1 sin 45 - 6 F1 sin 45
Fy = - 6Fi sin 45
Fy = -6 (6,265 10⁻²) sin45
Fy = - 26.54 10⁻² N
As we have the accelerations, we can find the speed on each axis
X axis
Vf² = Vo² + 2aₓ x
Vfx² = 2aₓ x
Vfx² = 2 8.86 10⁻² x
Vfx = 0.42 √x
Axis y
Vfy² = vfy² + 2ay Y
Vfy² = 2 ay Y
Vfy² = -2 26.54 10-2 x
Vfy = - 0.729 √x
