A proton beam enters a region permeated by a uniform elec- tric field of magnitu

Question

A proton beam enters a region permeated by a uniform elec- tric field of magnitude E oriented along the -y direction Beneath the y 0 plane the neld vanishes: the protons pro- ceed undeflected at a speed inclined at an angle with respect to the y = 0 plane. Upon crossing the plane, they are accelerated downward and strike a target on the plane a distance R from their point of entry. (a) Provide a succinct beam argument as to why the four usual equations for kinematics apply here. (b) Starting with these, calculate the distance R. (c) Would a more massive particle travel further than the proton in the same field? Compare with gravity (i.e., how does mass affect projectile motion due to gravity on Earth?). (d) Determine the possible angles between the beam and the y-0 plane if the impact point is 1.27mm away fronm the point of entry, given that the field's magnitude is 720 N/C and the protons move at 9.55 km/s E =-7201 N/C ×Target ProtonI E 0 below the plane

Explanation / Answer

the net force acting on the particle is electric force Fe = E*q

as the proton moves the force remains same

from newtons second law Fnet = m*a

acceleration a = Eq/m

as E , q , mare constant

acceleration is constant

the usual kinematic equations can be applied for constant accelrations

so here the kinematics equations can be applied

==============

(b)

acceleration ay = Eq/m = -(Ey*q/m)*j

after the proton reaches the target

the total displacement   y = 0

y = voy*t + (1/2)*ay*t^2

0 = v0*sintheta*t - (1/2)*(Ey*q/m)*t^2

t = 2*v0*sintheta*m/(Ey*q)

along horizontal

acceleration ax = 0

displacement x = R

x = vox*t + (1/2)*ax*t^2

R = vo*costheta*2*v0*sintheta*m/(Ey*q)

R = vo^2*2*sintheta*costheta*m/(Ey*q)

R = vo^2*sin(2theta)*m/(Ey*q)

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YES as m increases R increases

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(d)

1.27*10^-3 = (9.55*1000)^2*sin(2theta)*1.67*10^-27/(720*1.6*10^-19)

sin(2 theta ) = 0.961

2 theta = sin^-1(0.961) = 74

theta = 37 degrees <<<<----ANSWER

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