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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?

  • Physics -

    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.

  • Physics -


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