Question

The distance travelled by an electron in time of \(3\,{ns}\) when accelerated from rest in an electric field of \(0.9 \times 10^4\, {N/C}\) is: \[ {(mass of electron } = 9 \times 10^{-31} { kg, charge } = 1.6 \times 10^{-19} { C)} \]

• A. 7.2 m
• B. 7.2 cm
• C. 72 cm
• D. 0.72 cm

Hint:

Use \( s = \frac{1}{2} a t^2 \) for motion from rest and calculate acceleration using \( a = \frac{F}{m} \).

Answer 1

The Correct Option is D

Force \( F = eE = 1.6 \times 10^{-19} \cdot 0.9 \times 10^4 = 1.44 \times 10^{-15} \, {N} \) 
Acceleration: \[ a = \frac{F}{m} = \frac{1.44 \times 10^{-15}}{9 \times 10^{-31}} = 1.6 \times 10^{15} \, {m/s}^2 \] Time: \( t = 3\,{ns} = 3 \times 10^{-9} \, {s} \) Distance from rest: \[ s = \frac{1}{2} a t^2 = \frac{1}{2} \cdot 1.6 \times 10^{15} \cdot (3 \times 10^{-9})^2 = 0.72 \times 10^{-2} = 0.72\,{cm} \]