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.
Question
Explanation:
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