Answer
For proton $E=1.045\times10^{-7}N/C$ direction is upward (away from earth)
For electron $E= 5.6875\times10^{-11}N/C$ direction is downwards (towards the earth )
Work Step by Step
Mass of proton $m_{p}=1.673\times10^{-27}kg$
Acceleration due to gravity near the surface of the earth $g\approx10m/s^2$
force due to gravity on proton (weight of proton)$F_{p}=m_{p}\times g$
$F_{p}=1.673\times10^{-27}kg\times 10m/s^2$
$F_{p}=1.673\times10^{-26}N$
to support downward weight of proton an upward force is needed. So if we choose to give this force through electric field $E$ it should be upward away from the earth.
from the definition of E
$E= \frac{F}{q}$
putting $F=F_{p}=1.673\times10^{-27}N$
charge on proton $q=1.6\times10^{-19}C$
$E= \frac{F}{q}=\frac{1.673\times10^{-26}N}{1.6\times10^{-19}C}=1.045\times10^{-7}N/C$
direction is upward (away from earth)
Mass of electron $m_{e}=9.1\times10^{-31}kg$
Acceleration due to gravity near the surface of the earth $g\approx10m/s^2$
force due to gravity on electron (weight of electron)$F_{e}=m_{e}\times g$
$F_{p}=9.1\times10^{-31}kg\times 10m/s^2$
$F_{p}=9.1\times10^{-30}N$
to support downward weight of electron an upward force is needed. So if we choose to give this force through electric field $E$ , as electron have negative charge it should be a negative electric field whose direction will be towards the earth.
from the definition of E
$E= \frac{F}{q}$
putting $F=F_{e}=9.1\times10^{-30}N$
charge on electron $e=1.6\times10^{-19}C$
$E= \frac{F}{q}=\frac{9.1\times10^{-30}N}{1.6\times10^{-19}C}=5.6875\times10^{-11}N/C$
direction is downwards (towards the earth)
