An electron with a speed of 1.2 × 10^7 m/s moves horizontally into a region where a constant vertical force of 5.7 × 10^-16 N acts on it. The mass of the electron is 9.11 × 10^-31 kg. Determine the vertical distance the electron is deflected during the time it has moved 27 mm horizontally.
as Electric force F = Eq
E = F/q = 5.7 e -16/(16 e -19)
E = 356.25 N/C
formula for vertical distance Y = Ee
x^2/(2mv^2)
Y = (356.25 * 1.6 e -19 *0.027 *0.027)/(2 * 9.11 e -31 * 1.2 e7 *1.2 e7)
Y = 0.158 mm -----<<<<<<<<<<<<<<Answer
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