Question

An electron with a speed of 1.1 ✕ 107 m/s moves horizontally into a region where a constant vertical force of 5.1 ✕ 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 29 mm horizontally.

Answer 1

Sv = 1/2 a t^2 equation describing vertical distance traveled

Sv = 1/2 F / m * t^2 need to find t

Sh = .029m horizontal distance traveled

.029 m = Vh t time to travel 29 mm

t = 29E-3 m / 1.1E7 m/s = 2.64E-9 s

Sv = 1/2 * 5.1E-16 / 9.11E-31 * (2.64E-9)^2 = 5.1 / 2 * 2.64^2 / 9.11 * 10^-3 m

Sv = .00195 m = 1.95 mm