a) You need to move a test particle across a floor. Given source that has a charge of q = 3e, What is the displacement between the source and the particle if you need the particle to travel 2.5m in t = 2.5s? Note: the source emits this charge only the INSTANT you pull the trigger (its off for rest of experiment), and the particle has q = e and m = 1.67 * 10-27 Kg
b) What is the initial velocity the particle experiences and what is its final velocity before reaching desitination?
c) What is the electric field magnitude and direction of the source?
To answer these questions, we need to use the equations related to the motion of charged particles in electric fields. Here are the step-by-step explanations on how to get the answers:
a) To calculate the displacement between the source and the test particle, we can use the equation:
Δx = (1/2) * at^2
where Δx is the displacement, a is the acceleration, and t is the time. In this case, we need to find the acceleration.
The acceleration of a charged particle in an electric field can be calculated using the equation:
a = (q * E) / m
where q is the charge of the particle, E is the electric field, and m is the mass of the particle.
In this scenario, the charge of the particle, q, is e (equal to the elementary charge). The mass, m, is given as 1.67 * 10^-27 kg. However, we need to find the electric field, E.
To calculate the electric field, we can use the equation:
E = F / q
where F is the force experienced by the particle. The force acting on a charged particle in an electric field is given by:
F = q * E
This means that the force acting on the particle is equal to the charge multiplied by the electric field. In this case, the force is provided by the source that emits a charge of 3e.
Substituting the values into the equation, we have:
F = (3e) * E
Now, we can solve for E by rearranging the equation:
E = F / (3e) = (3e * q) / (3e) = q
Since the electric field is equal to the charge of the source, we have found the value of E.
Now, we can calculate the acceleration by substituting the values of q and m into the equation for acceleration:
a = (q * E) / m
b) To find the initial velocity, we can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Since the particle starts from rest, its initial velocity, u, will be 0.
To find the final velocity, v, we can rearrange the equation:
v = u + at = 0 + a * t
c) To find the electric field magnitude and direction of the source, we already determined that the electric field is given by the charge of the source. In this case, the source has a charge of 3e. The magnitude of the electric field is equal to the charge of the source. The direction of the electric field will be directed away from the source since it is emitting a positive charge.
Remember to substitute the given values into the equations and perform the necessary calculations to get the final answers.