An electron is released from rest in a uniform electric field E~ = 4.0 N C1yˆ. a

Question

An electron is released from rest in a uniform electric field E~ = 4.0 N C1yˆ.

a.) Calculation What is the magnitude of the acceleration of the electron at the moment it is released?

b.) Conceptual What direction is the electron accelerated at the moment it is released? (Please explain the physics to support your answer.)

c.) Calculation Consider the position of the electron at the moment it is released and at a time t = 2 µs after it is released. What is the electric potential difference between these two points?

Explanation / Answer

a)

F=qE=1.6*10^-19*4.0=6.4*10^-19N

a=F/m=6.4/9.11*10^12=7.03*10^11m/s^2

B)

the direction of force is I +y ,

thus its acceleration is in +y direction

C)

y=1/2at^2=0.5*7.03*4*0.1=1.41m

v=-4.0*1.41=-5.64V

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