A charged particles has a mass of 8 mg and a charge of + 5 MC. The particle is s

Question

A charged particles has a mass of 8 mg and a charge of + 5 MC. The particle is shot between the charged plates shown with an initial speed of 300 m/s. The electric field E between the plates has strength of 20000 N/C. The plates are 3 meters long and are separated by 20 cm. The particle enter midway between the two plates. Determine if the particle passes all the way through the region without striking the plates. If the particle does strike one of the plates determine the location and angle of impact. You may ignore the affects of gravity during the particle motion

Explanation / Answer

Electric Force, F = q*E

Vertical Distance = 20/2 = 10 cm = 0.1m
Horizontal distance = 300 m/s
Horizontal acceleration = 0
Vertical acceleration,a = ?
m*a = q*E
a = q*E/m
a = (5*10^6 * 20000)/(8*10^-3) m/s^2
a = 1.25*10^13 m/s^2

Now time taken to travel 0.1 m

s = u*t + 1/2at^2
0.1 = 0 + 1/2 * 1.25*10^13 * t^2
t = 1.26*10^-7 s

Horizontal distance covered,
Sx = u*t
Sx = 300 * 1.26*10^-7
Sx = 0.0000378 m
Location of the strike,  0.0000378 m

So it will strike one of the plates.
Vy = u +a*t
Vy = 0 +  1.25*10^13 * 1.26*10^-7
Vy = 1.575*10^6 m/s

Angle, = tan^-1(1.575*10^6/300)
Angle of impact, = 89.99o

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