Protons are projected with an intial speed v-9.99 km/s from a feid free region t

Question

Protons are projected with an intial speed v-9.99 km/s from a feid free region through a piane and into a region where a uniform electric field E-720j N/C is present above the plane as shown in in the figure below. The intial velooity vector of the protons makes an angle 8 with the plane. The protons are to hit a target that les at a horizontal distance of R 1.19 mm from the point where the protons cross the plane and enter the electric field. We wish to find the angle 8 at which the protons must pass through the plane to strike the target. Proton E- bekow the plane (a) What analysis model describes the horizontal motion of the protons above the plane? o particle under constant velooity e particle under constant acceleration artide in unform aroulr mtion opartice in simgle harmonic motion (b) What analysis model describes the vertical motion of the protons above the plane? particle under constant velocity o particle under constant acceleration o particle in uniform circular motion o particle in simple harmonic motion (c) Arque that R- 2would be aplicable to the protons in this stuation. (Do this on paper, Your instructor may ask you to turn in this work.) (d) Use R 2 to we an expression for R n terms of v,the charge and mass of the proton, and the angle .(Use the following as necessary vi, e.E,, and mo for the mass of proton)

Explanation / Answer

(A) force on proton will be downwards and it is constant.

and there is no force in horizontal direction.

so it will move with constant velocity in horizontal.

ans:: particle under constant velocity.

(B) Ans: particle under constant acceleration.

(C) in vertical, total vertical displacement is zero.

y = v0y t + ay t^2 / 2

0 = v sin(theta) t - a t^2 /2

t = 2 v sin(theta) / a

in horizontal,

R = (v cos(theta)) t

R = ( v cos(theta)) (2 v sin(theta)) / a

R = v^2 (2 sin(theta)cos(theta)) / a   

R = v^2 sin(2 theta) / a

(d) a = q E / mp

R = v^2 mp sin(2 theta) / q E

(e) 1.19 x 10^-3 = (9.99 x 10^3)^2 x 1.67 x 10^-27 x sin(2 theta) / (1.6 x 10^-19 x 720)

sin(2 theta) = 0.8223

2 theta = 55 and 125 deg

theta = 27.5 deg Or 62.5 deg

(f) t = 2 v sin(theta) / a

a = q E / m = 6.90 x 10^10 m/s^2

t = 133.7 x 10^-9 s and 257 x 10^-9 s

t = 133.7 ns Or 257 ns

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