In the figure, a proton is projected horizontally midway between two parallel pl

Question

In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.5 cm. The electrical field due to the plates has magnitude 9.9×105 N/C between the plates away from the edges. If the plates are 3.8 cm long, find the minimum speed of the proton if it just misses the lower plate as it emerges from the field. (e = 1.60 × 10-19 C, 0 = 8.85 × 10-12 C2/N · m2, m pr = 1.67 × 10-27 kg)

I treid 5.233*10^6 and 37.25*10^5 and they weren't correct!

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Explanation / Answer

accroding to the given conditions

Horizontal distance travelled is x = 3.8 cm = 0.038 m

vertical distance travelled is y = 0.5 cm = 0.5*10^-2 m

then Net force acting on the proton is Fe+Fg = (qE+mg) = (1.6*10^-19*9.9*10^5)+(1.67*10^-27*9.81) = 1.58*10^-13 N

accelaration is ay = Fnet /m = 1.58*10^-13/(1.67*10^-27) = 9.46*10^13 m/s^2

time taken to travel is t

then x = (V*t)

0.038 = (V*t)

y = 0.5*ay*t^2

0.5*10^-2 = 0.5*9.46*10^13*t^2

t = 1.02*10^-8 sec

then

x = V*t

0.038 = v*1.02*10^-8

v = 3.72*10^6 m/sec

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