Question

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.

Answer 1

$$K=qEd\\K=(1.6\cdot10^{-19})(10^4)0.01=1.6\cdot10^{-17}J\\a=\frac{qE}{m}=\frac{2d}{t^2}\\\\ \frac{(1.6\cdot10^{-19})(10^4)}{(9.1\cdot10^{-31})}=\frac{2(0.01)}{t^2}\\t=3.37\ ns$$