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A proton with an initial speed of [tex]8.00 \times 10^5 \, \text{m/s}[/tex] is brought to rest by an electric field. What is the magnitude of the electric field?

1) [tex]2.00 \times 10^5 \, \text{N/C}[/tex]
2) [tex]4.00 \times 10^5 \, \text{N/C}[/tex]
3) [tex]8.00 \times 10^5 \, \text{N/C}[/tex]
4) [tex]16.00 \times 10^5 \, \text{N/C}[/tex]

Answer :

Final answer:

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.

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.

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