An electron with a speed of 1.1 * 107 m/s moves horizontally into a region where a constant vertical force of 3.7 * 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 22 mm horizontally.

GIven data:

Speed V = 1.1*10⁷ m/s

Force F = 3.7*10⁻¹⁶N

Mass of electron = m = 9.11*10⁻³¹Kg

Horizontal Distance X = 2.2*10⁻²m

Vertical distance Y = ?

Solution:

As we know that,

F = ma

where a is the acceleration of electron.

(3.7*10⁻¹⁶) = (9.11*10⁻³¹) a

a = 4.06*10¹⁴m/s² upward.

∵opposing weight of electron is negligible.

now we find how long it takes the electron to move 22mm horizontally.

X = V*T

T = X/V

T = 2.2*10⁻²/1.1*10⁷

= 2*10⁻⁹s

now we can find out the vertical distance Y.

Y = 0.5aT²

= (0.5) (4.06*10¹⁴) (2*10⁻⁹) ²

= 8.12*10⁻⁴m