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17 January, 04:50

A uniform electric field exists in the region between two oppositely charged parallel plates 1.70 cm apart. A proton is released from rest at the surface of the positively charged plate and strikes the surface of the opposite plate in a time interval of 1.54*10-6 s.

a. Find the magnitude of the electric field.

b. Find the speed of the proton at the moment it strikes the negatively charged plate.

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  1. 17 January, 05:19
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    Given Information:

    distance between parallel plates = r = 1.70 cm = 0.0170

    time = t = 1.54*10⁻⁶ s

    Required Information:

    a) Electric field = E = ?

    b) Speed = v = ?

    Answer:

    a) Electric field = 149.44 N/C

    b) speed = 22076.67 m/s

    Explanation:

    Part (a)

    The relation between electric field, time and distance is given by

    E = 2mr/qt ²

    E = 2*1.67x10⁻²⁷*0.0170*/1.602x10⁻¹⁹ * (1.54*10⁻⁶) ²

    E = 149.44 N/C

    Part (b)

    From the kinematics,

    v = v₀ + at

    From Newton second law

    F = ma

    also we have,

    F = qE

    qE = ma

    a = qE/m

    v = v₀ + (qE/m) t

    Where v₀ is zero

    v = (1.602x10⁻¹⁹*149.44/1.67x10⁻²⁷) * 1.54*10⁻⁶

    v = 22076.67 m/s
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