college physics

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An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.3×10^7 m/s. What was the electron's speed as it left the negative plate?

the voltage is 650V
and the charge of q1 is 1.7*10^-9 and the charge of q2 is -1.7*10^-9

  • college physics -

    The kinetic energy gained going from the negative plate to the postivie plate is
    E = e*V,
    where e is the electron charge.

    (1/2)*m*(V2^2 - V1^2) = E

    Solve for the final velocity, V2
    V1 = 2.3*10^7 m/s

    m is the electron mass and e is its charge.

    You don't need to know the charges on the two plates.

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