An electron traveling at 5 times 106 m/s enters a 0.08 m region with a uniform e

Question

An electron traveling at 5 times 106 m/s enters a 0.08 m region with a uniform electric field of 87 N/C , as in the figure. The mass of an electron is 9.10939 Times 10-31 kg and the charge on an electron is 1.60218 Times 10-19 C. Find the magnitude of the acceleration of the electron while in the electric field. Answer in units of m/s2 Find the time it takes the electron to travel through the region of the electric field, assuming it doesn't hit the side walls. Answer in units of s What is the magnitude of the vertical displacement delta y of the electron while it is in the electric field? Answer in units of m

Explanation / Answer

(a) Force on electron = eE

(e = charge on electron, E = electric field)

Accl = force/mass = eE/m

= (1.6 x 10^-19 x 87)/(9.10939 x 10^-31)

= 1.5281 x 10^13 m/s2

(b) The horizontal velocity of the elctron remains unchanged

Time = horizontal distance/horizontal velocity

= 0.08/(5 x 10^6) = 1.6 x 10^(-8) seconds

(c) Vertical displacement (y) = 0.5at^2 (a = accl)

y = 0.5 x 1.5281 x 10^13 x (1.6 x 10^(-8))^2

= 1.9959 x 10^-3 m

= 1.9959 mm

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