An electron traveling at 5 × 106 m/s enters a 0.07 m region with a uniform elect

Question

An electron traveling at 5 × 106 m/s enters a 0.07 m region with a uniform electric field of 271 N/C.

Find the magnitude of the acceleration of the electron while in the electric field. The mass of an electron is 9.109 × 1031 kg and the fundamental charge is 1.602 × 1019 C .

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.

What is the magnitude of the vertical dis- placement y of the electron while it is in the electric field?

Explanation / Answer

accelration a = q*E/m
= (1.6*10^-19)* (271) / (9.109*10^-31)
= 4.76*10^13 m/s^2
This acceleration will be perpendicular to initial velocity. Let us assume that initial velocity was in horizontal direction. SO this a will be in vertical acceleration
So horizontal velocity will not be affected by this a.
time = horizontal distance / horizontal veocity
=0.007 / (5*10^6)
= 1.4*10^-9 s

vertical displacement = 0.5*a*t^2
= 0.5 * 4.76*10^13 *(1.4*10^-9 )^2
=4.66*10^-5 m

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