Protons are projected with an initial speed vi = 9.28 km/s from a field-free reg

Question

Protons are projected with an initial speed vi = 9.28 km/s from a field-free region through a plane and into a region where a uniform electric field = ?720 N/C is present above the plane as shown in in the figure below. The initial velocity vector of the protons makes an angle ? with the plane. The protons are to hit a target that lies at a horizontal distance of R = 1.24 mm from the point where the protons cross the plane and enter the electric field. We wish to find the angle ? at which the protons must pass through the plane to strike the target.

vi2sin 2?i g

Explanation / Answer

Given

Initial velocity vi = 9.28 km/s = 9.28 x 103 m/s

Electric field E = -720 N/C

Range R = 1.24 mm = 1.24 x 10-3 m

Known

Mass of proton mp = 1.67 x 10-27 Kg

Charge of proton e = 1.6 x 10-19 C

Solution

(a)

The horizontal component of velocity

Vix = vicos

The vertical component of velocity

Viy = visin

The net force in x direction

Fx = 0

Hence there is no acceleration in x direction and it travels with constant velocity with no change ( neither in magnitude nor in direction)

The analysis model describes the horizontal motion of the protons above the plane is particle under constant velocity

(b)

The net force in y direction

Fy = Ee - W

Fy = Ee - mpg

Net acceleration

ay = Fy / mp

ay = (Ee/mp) - g

ay = (-720 x 1.6 x 10-19 / 1.67 x 10-27) - 9.8

ay = - 6.9 x 1010 - 9.8

Since g is very much small

ay = a = - 6.9 x 1010 m/s2

‘a’ is the acceleration due to the electric field.

So the proton is under a constant acceleration

The analysis model describes the horizontal motion of the protons above the plane is particle under constant acceleration

(c)

The horizontal distance

R = horizontal velocity x time of flight

R = vicost

The vertical distance travelled

H = viyt + ½at2

Since H = 0

viyt = - ½at2

viy = -at/2

t = -2Visin/a

So

R = vicos( -2Visin/a )

R = -vi2(2sincos)/a

R = -vi2sin2/a

This is similar to equation of free falling objects, which is

R = vi2sin2/g

We just had to consider the effect of electric field on the acceleration

(d)

Since

R = -vi2sin2/a

And a = Ee/mp

R = - vi2sin2mp/eE

(e)

R = - vi2sin2mp/eE

sin2 = -ReE/ vi2mp

Sin2 = - 1.24 x 10-3 x 1.6 x 10-19 x (-720) / (9.28 x 103)2x1.67 x 10-27

Sin2 = 0.993

2 = 83.22o

= 41.61o

Also

Since

Sin2 = sin(180-2)

sin(180-2) = 0.993

2 = 180-83.22

= 96.78/2

= 48.39o

Smallest angle

41.61o

Largest angle

48.39o

(f)

t = -2Visin/a

t = -2Vi mpsin/eE

Smallest period is

tmin = -2 x 9.28 x 103 x 1.67 x 10-27 x sin 41.61 / 1.6 x 10-19 x (-720)

tmin = 0.01787 x 10-5

tmin  = 178.7 ns

Largest period is

tmax = -2 x 9.28 x 103 x 1.67 x 10-27 x sin 48.39 / 1.6 x 10-19 x (-720)

tmax = 0.02012 x 10-5

tmax  =201.2 ns

Smallest duration

178.7 ns

Largest duration

201.2 ns

Smallest angle

41.61o

Largest angle

48.39o

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