An electron traveling at 2 × 106 m/s enters a 0.09 m region with a uniform elect

Question

An electron traveling at 2 × 106 m/s enters a 0.09 m region with a uniform electric field of 164 N/C , as in the figure.
The mass of an electron is 9.10939 × 10-31 kg and the charge on an electron is 1.60218 × 10-19 C .

a.) Find the time it takes the electron to travel through the region of the electric field, assum- ing it doesn’t hit the side walls.
Answer in units of s.

b.) What is the magnitude of the vertical dis- placement ?y of the electron while it is in the electric field?
Answer in units of m.

Explanation / Answer

velocity of th electron u = 2 * 10^ 6 m / s distanc eof the region S = 0.09m electric field E = 164 N /C mass of electron m = 9. 9.10939 × 10-31 kg the charge on an electron q = 1.60218 × 10-19 C . the magnitude of the acceleration of the electron a = Eq / m a = 2.88*10^13 m / s^ 2 (a).required time t =? we know velocity u = distance S / time t from this time t = S / u = 4.5*10^-8 s (b).In vertical direction : ---------------------------- Initial velocity U = 0 accleration a = 2.88*10^13 m / s^ 2 time t= 4.5 * 10^-8 s vertical displacement y = ut + ( 1/ 2) at^ 2 = 0.02916 m

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