Question

what are magnitude and direction of electric field that will just support an electron so that it long in air

Answer 1

There are two forces that act on the electron: the electric force $F_e$ directed upward and the force of gravity $F_g$ directed downward. In order to electron to be long in air, the force of gravity must be balanced by the electric force:

$$F_e=F_g,$$ $$qE=m_eg,$$ $$E=\dfrac{m_eg}{q},$$ $$E=\dfrac{9.1\cdot10^{-31}\ kg\cdot 9.8\ \dfrac{m}{s^2}}{-1.6\cdot10^{-19}\ C}=-5.6\cdot10^{-11}\ \dfrac{N}{C}.$$

The magnitude of the electric field is $E=5.6\cdot10^{-11}\ \dfrac{N}{C}.$ The sign minus indicates that the direction of the electric field is downward.

Answer:

$E=5.6\cdot10^{-11}\ \dfrac{N}{C},$ downward.