An electron is released from rest at the negative plate of a parallel plate capacitor and...
An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate. The plates are separated by a distance of 1.2 cm and the electric field within the capacitor has a magnitude of 2.1 * 10^6 V/m. What is the speed of the electron just as it reaches the positive plate?
Homework Answers
Voltage across capacitor
V=Ed =2.1*106*0.012=25200 Volts
By Conservation of energy
(1/2)mv2=qV
(1/2)(9.11*10-31)*v2 =(1.6*10-19)*25200
v=9.41*107 m/s
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