Question

Two horizontal metal plates, each 10.0 cm square, are aligned 1.00 cm apart with one above the other. They are given equal-magnitude charges of opposite sign so that a uniform downward electric field of 2.02*10^3 N/C exists in the region between them. A particle of mass 2.00*10^16kg and with a positive charge of 1.03*10^6 C leaves the center of the bottom negative plate with an initial speed of 1.02*10^5 m/s at an angle of 37.0° above the horizontal.
(a) Describe the trajectory of the particle.
a parabolaa circle a helixa straight lineYour answer is correct.
(b) Which plate does it strike?
negative platepositive plate It does not strike either plate.Your answer is correct.
(c) Where does it strike, relative to its starting point? (Enter the horizontal distance from the initial position.)
= m.

Answer 1

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