Determine the final velocity of an alpha particle (3.2x10^-19) if it is placed in a 7.2x10^6 n/c uniform field. The mass of the alpha particle is 6.64x10^-27
For what distance is is accelerated?
It does not say in the question
Huh ? Are you doing relativity?
The force on it is constant, q*E
so the acceleration is constant, forever if it were not limited by approaching the speed of light.
I don't think so, we just started on parallel plates in school and he assigned this question for homework and i have no clue how I would get the final velocity from what was given.
To determine the final velocity of an alpha particle in a uniform electric field, you can use the equation:
F = q * E
where F is the force experienced by the particle, q is the charge of the particle, and E is the electric field strength.
In this case, the force experienced by the alpha particle is given by:
F = q * E = (3.2x10^-19 C) * (7.2x10^6 N/C)
Next, you can use Newton's second law of motion to relate force and acceleration:
F = m * a
where m is the mass of the alpha particle and a is its acceleration. Rearranging the equation, we have:
a = F / m = (3.2x10^-19 C) * (7.2x10^6 N/C) / (6.64x10^-27 kg)
Now that you have the acceleration, you can calculate the final velocity using the equation of motion:
v^2 = u^2 + 2 * a * d
where v is the final velocity, u is the initial velocity (which is assumed to be zero in this case), a is the acceleration, and d is the distance traveled.
Since the initial velocity is zero, the equation simplifies to:
v^2 = 2 * a * d
Solving for v gives:
v = sqrt(2 * a * d)
Note that we need to know the distance traveled by the alpha particle in order to find the final velocity. If the distance is not provided in the question, this calculation cannot be completed.
So, to find the final velocity, you need to know the distance traveled by the alpha particle.