An electron is accelerated from rest across the gap of a capacitor (two parallel plates charged...
An electron is accelerated from rest across the gap of a capacitor (two parallel plates charged -Q and +Q respectively). A hole in the top plate allows the electron to emerge with a constant velocity of v = 71 m/s. If the gap between the plates is d = 0.47 m, what is the magnitude of the electric field between the plates?
Homework Answers
From
1/2 mv^2 =q delta v
delta v =(0.5*9.1*10^-31*71^2)/(1.6*10^-19)
=1.43*10^-8
E=V/d = (1.43*10^-8)/(0.47)
E=3.04*10^-8 V/m
I hope help you !!
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