The first electron has the initial velocity v0, which makes an angle =45° with t

Question

The first electron has the initial velocity v0, which makes an angle =45° with the lower plate and has a magnitude of 8.78×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

Another electron has an initial velocity which has the angle =45° with the lower plate and has a magnitude of 5.86×106 m/s. Will this electron strike one of the plates? If so, what is the horizontal distance from the left edge? If not enter the vertical position at which the particle leaves the space between the plates.

PLEASE USE MY NUMBERS

Explanation / Answer

consider the motion in horizontal direction

X = horizontal distance travelled = L = 0.04 m

Vox = velocity in horizontal direction = (8.78 x 106) Cos45

t = time taken

X = Vox t

0.04 = ((8.78 x 106) Cos45) t

t = 6.44 x 10-9 sec

consider the motion along the Y-direction

Voy = (8.78 x 106) Sin45

a = acceleration = - qE/m = - (1.6 x 10-19) (4500)/(9.1 x 10-31) = - 7.91 x 1014

t = 6.44 x 10-9 sec

Y = ?

using the equation

Y = Voy t + (0.5) a t2

Y = (8.78 x 106) Sin45 (6.44 x 10-9) + (0.5) (- 7.91 x 1014 ) (6.44 x 10-9)2

Y = 0.0236 m

since Y > d

hence the electron hits the upper plate

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