Answer :
The acceleration experienced by a proton due to a 2.00×10 N/C electric field. the final speed of the proton after accelerating for 10 mm is approximately 1.55×10^10 m/s.
To calculate the acceleration experienced by a proton in an electric field, we can use the equation:
acceleration = electric field / mass
Given that the electric field is 2.00×10^4 N/C, we need to know the mass of the proton. The mass of a proton is approximately 1.67×10^-27 kg.
acceleration = (2.00×10^4 N/C) / (1.67×10^-27 kg)
acceleration ≈ 1.20×10^23 m/s²
Next, we can calculate the final speed of the proton using the equation of motion:
final velocity² = initial velocity² + 2 * acceleration * displacement
The proton starts from rest, so the initial velocity is 0. The displacement is given as 10 mm, which is 0.01 m.
final velocity² = 0 + 2 * (1.20×10^23 m/s²) * (0.01 m)
final velocity² ≈ 2.40×10^21 m²/s²
Taking the square root of both sides:
final velocity ≈ √(2.40×10^21 m²/s²)
final velocity ≈ 1.55×10^10 m/s
Therefore, the final speed of the proton after accelerating for 10 mm is approximately 1.55×10^10 m/s.
To know more about acceleration refer here:
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