Question

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

Answer 1

To find the speed of the fifth charge at a great distance, we use conservation of energy by equating the total initial electric potential energy with the final kinetic energy, then solve for velocity.

Step-by-step explanation:

Finding the final speed of a charge

To find the speed of the fifth charge when it is a great distance from the origin, we will apply conservation of energy principles. Initially, the charge has only electric potential energy due to the positions of the other charges. When the charge is released and moves a great distance away, all of this electric potential energy will have been converted into kinetic energy (assuming negligible external forces like resistance or gravity). The total initial electric potential energy can be calculated by summing the individual potential energies due to each charge at the corners of the square. Once we have the total potential energy, we set it equal to the final kinetic energy to solve for the final speed.

The initial potential energy (U) due to a pair of charges is given by the equation U = k * q1 * q2 / r, where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them.

The kinetic energy (KE) of the charge when it has moved away to a distance much greater than the dimensions of the square is given by KE = 0.5 * m * v^2, where m is the mass of the charge and v is its final speed.

By equating the total initial potential energy with the final kinetic energy and solving for v, we can find the speed of the charge at a great distance.

Answer 2

Vfx = 0.42 √x and Vfy = - 0.729 √x

Step-by-step 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