Question

An electron initially at rest falls a distance of $1.5\, cm$ in a uniform electric field of magnitude $2 \times 10^4\, N/C$. The time taken by the electron to fall this distance is

• A. $1.3 \times 10^2 \,s$
• B. $2.1 \times 10^{-12} \,s$
• C. $1.6 \times 10^{-10} \,s$
• D. $2.9 \times 10^{-9} \,s$

Answer 1

Answer: (D)

Here, the direction of field is upward.
So the negatively charged electron experiences a downward force.

$\therefore $ The acceleration of electron is
$a_{e}=\frac{eE}{m_{e}}\,...\left(i\right)$
The time required by the electron to fall through a distance $h$ is
$t_{e}=\sqrt{\left(\frac{2h}{Aa_{e}}\right)}=\sqrt{\frac{2hm_{e}}{eE}}\,$ (using (i))
$=\left[\frac{2\times1.5\times10^{-2}\times9.11\times10^{-31}}{1.6\times10^{-19}\times2\times10^{4}}\right]^{^{1/2}}$
$=2.9\times10^{-9}\,s$