Explanation

(d) [image:1_image343.emf]
As the field is upward, so the negatively charged electron experiences a downward force of the magnitude of $eE$ where $E$ is the magnitude of the electric field.
The acceleration of the electron is $a_{e} = \frac{eE}{m_{e}}$
Where, $m_{e}$ is the mass of the electron
Starting from rest, the time taken by the electron to fall through a distance $h$ is given by
$t_{e} = \sqrt{\frac{2h}{a_{e}}} = \sqrt{\frac{2hm_{e}}{eE}} = \sqrt{\frac{2 \times 1.5 \times 10^{- 2} \times 9.1 \times 10^{- 31}}{1.6 \times 10^{- 19} \times 2 \times 10^{4}}}$
$= 2.9 \times 10^{- 9}s$