Question

A proton with an initial speed of 8.00×10⁵ m/s is brought to rest by an electric field. What is the magnitude of the electric field?

1) 2.00×10⁵ N/C
2) 4.00×10⁵ N/C
3) 8.00×10⁵ N/C
4) 16.00×10⁵ N/C

Answer 1

The magnitude of the electric field that can bring a proton to rest can be calculated using the equation final velocity = initial velocity + acceleration * time. By rearranging the equation to solve for force, we can find that the magnitude of the electric field is approximately 1.34×10^-21 N/C.

Step-by-step explanation:

To calculate the magnitude of the electric field that can bring a proton to rest, we can use the equation:

final velocity = initial velocity + acceleration * time

Since the proton is brought to rest, the final velocity is 0 m/s. The initial velocity is 8.00×10^5 m/s. The acceleration can be calculated using the equation:

acceleration = force / mass

Since the mass of a proton is approximately 1.67×10^-27 kg and the force is unknown, we can rearrange the equation to solve for the force:

force = acceleration * mass

Plugging in the known values, we get:

force = (8.00×10^5 m/s) * (1.67×10^-27 kg)

force ≈ 1.34×10^-21 N

Therefore, the magnitude of the electric field is approximately 1.34×10^-21 N/C.