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    We know that
    eVs (stopping potential in electron volts) = KE of electron

    Does this apply to ANY potential difference, thereby saying that
    eV = KE of electron
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    (Original post by jsmith6131)
    We know that
    eVs (stopping potential in electron volts) = KE of electron.

    Does this apply to ANY potential difference, thereby saying that
    eV = KE of electron
    Not precisely.

    If V is potential difference, then the work done on a particle by electric field is W=qV. This results in change of kinetic energy of a particle.

    Stopping potential difference for an electron is such a potential difference that will make the electron loose all its kinetic energy. So in this special case when the final kinetic energy is 0, you can say that eV=(\text{kinetic energy}).

    However, depending on the values (and signs) of potential difference and the charge of a particle, electric field may cause it to loose or gain kinetic energy.
    In a general case though the movement in an electric field results in a change of kinetic energy.
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    If you want to express it by an equation:

    in any conservative field (such as electric or gravitational field) the change in potential energy \Delta U and change in kinetic energy \Delta T give a sum of 0: \Delta U+\Delta T=qV+\tfrac{1}{2}m\Delta (v^2)=0 (conservation of energy).
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    ok, so
    qV = -KE
    where q = charge on an electron

    So basically if an electron was in a potential different of say 10 kV then the velocity of the electron would be
    about 5.93 * 10^7 m/s
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    (Original post by jsmith6131)
    ok, so
    qV = -KE
    where q = charge on an electron

    So basically if an electron was in a potential different of say 10 kV then the velocity of the electron would be
    about 5.93 * 10^7 m/s
    If the electron has moved through a potential difference of +10kV and its initial velocity was 0, then -eV=-\tfrac{1}{2}m(v-0)^2, and yes, it would gain velocity around 5.9\times 10^{7} \text{ m/s}.
 
 
 
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