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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?

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    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.

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