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12 October, 16:20

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x

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  1. 12 October, 17:51
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    E = - 4556.18 N/m

    Explanation:

    Given data

    u = 3.6*10^6 m/sec

    angle = 34°

    distance x = 1.5 cm = 1.5*10^-2 m (This data has been assumed not given in

    Question)

    from the projectile motion the horizontal distance traveled by electron is

    x = u*cosA*t

    ⇒t = x / (u*cos A)

    We also know that force in an electric field is given as

    F = qE

    q = charge, E = strength of electric field

    By newton 2nd law of motion

    ma = qE

    ⇒a = qE/m

    Also, y = u*sinA*t - 0.5*a*t^2

    ⇒y = u*sinA*t - 0.5 * (qE/m) * t^2

    if y = 0 then

    ⇒t = 2mu*sinA / (qE) = x / (u*cosA)

    Also, E = 2mu^2*sinA*cosA / (x*q)

    Now plugging the values we get

    E = 2*9.1*10^{-31}*3.6^2*10^{12} * (sin34°) * (cos34°) / (1.5*10^{-2} * (-1.6) * 10^{-19})

    E = - 4556.18 N/m
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