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29 April, 22:55

Beams of high-speed protons can be produced in "guns" using electric fields to accelerate the protons. A certain gun has an electric field of 1.76 x 10 4 N/C. What speed would the proton obtain if the field accelerated the proton through a distance of 1.00 cm?

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  1. 30 April, 00:07
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    Answer: v = 2.75 * 10^7 m/s

    Explanation: Since the electric field is a uniform one and the distance is small, the motion of the electron is of a constant acceleration, hence newton's laws of motion is applicable.

    From the question

    E=strength of electric field = 214N/c

    q=magnitude of an electronic charge = 1.609 * 10^-16c

    m = mass of an electronic charge = 9.11*10^-31kg

    v = velocity of electron.

    S = distance covered = 1cm = 0.01m

    a = acceleration of electron.

    F = ma but F=Eq

    Eq = ma.

    a = Eq/m

    a = 214 * 1.609*10^-16 / 9.11 * 10^-31

    a = 344.32 * 10^-16 / 9.11 * 10 ^-31

    a = 3. 779 * 10^16 m/s²

    Before the electron is accelerated, they are always not moving, hence initial velocity (u) = 0.

    By using the equation of motion, we have

    v² = u² + 2aS

    But u = 0

    v² = 2aS

    v² = 2 * 3.779*10^16 * 0.01

    v² = 7.558 * 10^14

    v = √7.558 * 10^14

    v = 2.75 * 10^7 m/s.
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