Question

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see the drawing). The plates are separated by a distance of 1.7 cm, and the electric field within the capacitor has a magnitude of 2.7 x 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate? The figure shows a vertical plate on the left that is negatively charged and another vertical plate on the right that is positively charged. The electric field lines originate at the positive plate and terminate at the negative plate. An electron is shown between the plates. The force on the electron is vector F pointing horizontally to the right.

+ + + + + + + + + + Electric field Electron ІІІ ТІГІ

Answer 1

electric field = 2.7*10^6 V/m

distance = 0.017 m

acceleration of electron = Force / mass

= qE/m

= 1.6*10^-19*2.7*10^6 / 1.67*10^-27

= 2.58*10^14 m/s^2

v^2 - u^2 = 2aS

v^2 - 0 = 2*2.58*10^14*0.017

v^2 = 8.8*10^12

Kinetic energy = 0.5mv^2

= 0.5*1.67*10^-27*8.8*10^12

= 7.34*10^-15 J