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 1.4 cm, and the electric field within the capacitor has a magnitude of 3.5 106 V/m. What is the kinetic energy of the electron just as it reaches the positive plate?

Answer 1

The Voltage change between plates is
(Field)*(Plate separation)
= 3.5*10^6*0.014 = 49*10^3 Volts

Multiply that by the electron charge to get the final kinetic energy.

Answer 2

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