Answer
a)Field point downwards
b)$\sigma = 52.7C/m^2$
Work Step by Step
a) The electrical force on the drop must be upward to counter the weight of the droplet.
The field should point downward since the drop is negative to create an upward force on the droplet.
b) $F_B = qE = mg$
$\frac{\sigma}{\epsilon}(5 \times 1.6\times10^{-19} ) = 486 \times10^{-9} \times 9.8 $
Solving the equation for $\sigma,$
$\sigma = 52.7C/m^2$
