Question

The plates of a parallel plate capacitor are separated by a distance of 1.2cm, and the electric field within the capacitor has a magnitude of 2.1x 10^6 V/m. An electron starts from rest at the negative plate and accelerates to the positive plate. What is the kinetic energy of the electron just as the electron reaches the positive plate?

Answer 1

The field strength E multiplied by the plate separation equals the potential energy change, per Coulomb. That would be 25,200 J/C.
Set that equal to (1/2)(m/e) V^2, the kinetic energy per Coulomb, and solve for V.
m and e are the electron mass and charge.