Answer
$1.3\times10^{-12}V$
Work Step by Step
The electric force has the same magnitude as the electron’s weight.
$$F_{electric}=F_{gravity}$$
The magnitude of the electric force is the magnitude of the electronic charge, multiplied by the electric field strength.
The electric field is the potential difference divided by the distance.
$$q\frac{V}{d}=mg$$
$$V=\frac{mgd}{q}=\frac{(9.11\times10^{-31}kg)(9.80m/s^2)(0.024m)}{1.60\times10^{-19}C}$$
$$=1.3\times10^{-12}V$$
This voltage is minuscule. Electrons can easily move upward against the force of gravity.