Answer:
a) motion is PARABOLIC, b) positive particle is accelerated towards the negative plate, c) x = 6.19 10⁹ m
Explanation:
This exercise looks at the motion of a positively charged particle in an electric field.
a) Since the field is vertical the acceleration in this direction is
F = m a
the electric force is
F = q E
we substitute
q E = m a
a = qE / m
the mass of the particle is m = 2.00 10-16 kg
a = 1.6 10⁻¹⁹ 2.02 10³ / 2.00 10⁻¹⁶ kg
a = 1,616 m / s²
on the x-axis there are no relationships because there are no forces.
Since the particle has velocities in both axes, its motion is PARABOLIC,
b) the positive particle is accelerated towards the negative plate,
The field is descending, for which the event is down
c) where hit the particle on the x-axis
they indicate that the particle leaves the center of the negative plate, for which we will fix our reference system at this point.
Let's find the components of the initial velocity.
sin θ = v_{oy} / v
cos θ = v₀ₓ / v
v_{oy} = v₀ sin θ
v₀ₓ = v₀ cos θ
v_{oy) = 1.02 10⁵ sin 37 = 0.6139 10⁵ m / s
v₀ₓ = 1.02 10⁵ cos 37 = 0.8146 10⁵ m / s
Let's find the time it takes to hit the negative plate
y = y₀I + v_{oy} t + ½ a and t2
in this case the positions are y = y₀ = 0 and the accelerations
a = - 1,616m/s2,
we substitute
0 = 0 + v_{oy} t - ½ a_y t²
v_{oy}= ½ a_y t
t = 2v_{oy} / a_y
let's calculate
t = 2 0.6139 10⁵ / 1.616
t = 7.597 10⁴s
in this time the particle travels a horizontal distance
x = v₀ₓ t
x = 0.8145 10⁵ 7.597 10⁴
x = 6.19 10⁹ m
the particle falls off the plate
