An alpha particle (charge +2e and mass 6.64 x10-27 kg) is traveling to the right at 1.50 km/s. What uniform electric field (magnitude and direction) is needed to cause it to travel to the left at the same speed after 2.65 s.?

Plz solve the qurstion

To determine the uniform electric field required to cause the alpha particle to change its direction, we can use the principles of electromagnetism.

1. First, let's analyze the situation. The alpha particle has a charge of +2e and a mass of 6.64 x10-27 kg. It is initially traveling to the right at 1.50 km/s.

2. Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, the electric force will provide the acceleration needed to change the direction of the alpha particle.

3. The electric force acting on a charged particle in an electric field is given by the equation F = qE, where F is the force, q is the charge, and E is the electric field.

4. The acceleration of the alpha particle can be determined using the equation F = ma, where a is the acceleration and m is the mass of the particle.

5. Combining equations (3) and (4), we have qE = ma.

6. We know that the particle needs to change direction but maintain the same speed. This means that the acceleration should be in the opposite direction to the initial velocity, and its magnitude should be the same as the initial acceleration.

7. The initial acceleration can be determined using the equation a = Δv / Δt, where Δv is the change in velocity and Δt is the time taken.

8. Δv = -1.50 km/s - 1.50 km/s = -3.00 km/s (negative because it is in the opposite direction).

9. Converting from km/s to m/s, Δv = - 3.00 km/s x 1000 m/km = -3000 m/s.

10. Δt = 2.65 s.

11. Therefore, the initial acceleration is a = (-3000 m/s) / (2.65 s) = -1132.08 m/s^2.

12. Since the acceleration is in the opposite direction, the force acting on the alpha particle will also be in the opposite direction.

13. Substituting the known values into the equation qE = ma, we have (2e)(E) = (6.64 x10-27 kg)(-1132.08 m/s^2).

14. Rearranging the equation, we can solve for the electric field E:

E = (6.64 x10-27 kg)(-1132.08 m/s^2) / (2e).

15. Plugging in the values, we have E = (-7.50 x 10^-23 N) / (2e).

16. The magnitude of the electric field is given by E = |(-7.50 x 10^-23 N) / (2e)|.

17. Finally, substituting the value of e (charge of an electron) as 1.60 x 10^-19 C, we can calculate the magnitude of the electric field.

E = |(-7.50 x 10^-23 N) / (2 x (1.60 x 10^-19 C))|.

18. Simplifying this expression, we find that the magnitude of the electric field required is approximately 2.34 x 10^5 N/C.

19. Since the electric field acts in the opposite direction to the initial velocity, the direction of the electric field is to the left.