Answer to Question #216670 in Physics for Gee

Question #216670
Two large parallel plates are placed horizontally with respect to each other and are then connected to a 100-V battery. If the distance between the plates is 1 cm, the uniform electric field between the plates is 104 N/C in the upward direction as shown by the vectors in the figure. If the electron is released from rest at the upper plate, calculate: i. ii. kinetic energy of the electron after traveling 1 cm to the lower plate, and the time required for it to travel this distance.
1
Expert's answer
2021-07-14T10:15:41-0400
K=qEdK=(1.61019)(104)0.01=1.61017Ja=qEm=2dt2(1.61019)(104)(9.11031)=2(0.01)t2t=3.37 nsK=qEd\\K=(1.6\cdot10^{-19})(10^4)0.01=1.6\cdot10^{-17}J\\a=\frac{qE}{m}=\frac{2d}{t^2}\\\\ \frac{(1.6\cdot10^{-19})(10^4)}{(9.1\cdot10^{-31})}=\frac{2(0.01)}{t^2}\\t=3.37\ ns


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