Four charges are at the corners of a square centered at the origin as follows q at a a 2q at a a 3q at a a and 6q at a a A fifth

Question

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Four charges are at the corners of a square centered at the origin as follows q at a a 2q at a a 3q at a a and 6q at a a A fifth

charge q is placed at the origin and released from rest Find its speed when it is a great distance from the origin if a 1.5 m q 2.8 mu or micro CC and its mass is 0.4 kg

Physics

1 answer:

Wewaii [24] · Answer 1 · 2022-08-25T08:09:40+03:00

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