A particle with mass m = 1.20g and charge q = 345ìC is traveling in the x-direction with an initial velocity of v0 = 68.0m/s. The particle passes through a constant electric field in the y-direction with magnitude |E| = 263N/C over a distance of d = 12.0m. (a) What is the magnitude of the velocity after exiting the constant electric field? (b) What angle does the trajectory of the particle make with the horizontal after the particle exits the constant electric field?
To solve this problem, we need to use the principles of kinematics and the equation for the force experienced by a charged particle in an electric field.
(a) To find the magnitude of the velocity after exiting the constant electric field, we need to calculate the acceleration experienced by the particle in the y-direction. The force experienced by a charged particle in an electric field is given by:
F = qE
where F is the force, q is the charge, and E is the electric field.
In this case, the force experienced by the particle is the only force acting on it in the y-direction, so we can equate the force to the mass of the particle times its acceleration:
ma = qE
Rearranging the equation, we can solve for the acceleration:
a = (qE) / m
Now we can use the kinematic equation to find the final velocity in the y-direction. Since the particle was initially traveling only in the x-direction, the velocity in the y-direction before entering the electric field is 0 m/s.
The kinematic equation for velocity with constant acceleration is:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time.
In this case, vi = 0 m/s, a is the acceleration we found earlier, and t is the time it takes for the particle to travel the distance d in the electric field. We can calculate the time using the equation:
d = vit + (1/2)at^2
Since the initial velocity in the y-direction is 0 m/s, the equation simplifies to:
d = (1/2)at^2
Solving for t:
t^2 = (2d) / a
t = sqrt[(2d) / a]
Substituting the acceleration value we found earlier:
t = sqrt[(2d) / ((qE) / m)]
Now we can use this time value to calculate the final velocity in the y-direction:
vf = 0 + at
Substituting the known values, we can solve for vf.
(b) To find the angle that the trajectory of the particle makes with the horizontal after exiting the constant electric field, we can use trigonometry.
The horizontal component of the final velocity is the same as the initial velocity in the x-direction (v0). The vertical component of the final velocity is the final velocity in the y-direction that we found earlier (vf).
Using these two components, we can calculate the angle using the equation:
θ = tan^(-1)(vf / v0)
Substituting the known values, we can solve for θ.