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. kinetic energy of the electron after traveling 1 cm to the lower plate

Answer 1

We can find the kinetic energy of the electron after traveling 1 cm to the lower plate from the law of conservation of energy:

$$PE_{upper}+KE_{upper}=PE_{lower}+KE_{lower},$$ $$KE_{lower}=PE_{upper}-PE_{lower}+KE_{upper},$$ $$KE_{lower}=qE(y_{upper}-y_{lower})+\dfrac{1}{2}mv_i^2,$$ $$KE_{lower}=1.6\cdot10^{-19}\ C\cdot10^4\ \dfrac{N}{C}\cdot(0\ m-0.01\ m)+0,$$ $$KE_{lower}=-1.6\cdot10^{-17}\ J.$$