Question

An $e^{-}$ of mass $m$ is released from rest under gravity. An electric field $E=\frac{mg}{e}$ is applied upwards, the acceleration of electron is $N\times g$ downward, then value of $N$ is

Answer 1

$F_{net}=qE+mg=eE+mg$ (since electron has negative charge, therefore accleration will be acting in downward direction)
Substituting E,
$F_{net}=mg+mg=2mg$ (downwards)

$\overrightarrow{a}=\frac{F_{net}}{m}=2g\downarrow$