Answer:
[tex]1.49\cdot 10^{-17}C[/tex]
Explanation:
The oil drop remains stationary when the electric force on it and the gravitational force are balanced, so we have:
[tex]F_E = F_G\\qE = mg[/tex]
where
q is the charge of the oil drop
E is the electric field strength
m is the mass of the drop
g is the acceleration due to gravity
here we have
[tex]E=1\cdot 10^6 N/C[/tex]
[tex]m=1.51837\cdot 10^{-12} kg[/tex]
[tex]g=9.8 m/s^2[/tex]
So the charge of the drop is
[tex]q=\frac{mg}{E}=\frac{(1.51837\cdot 10^{-12} kg)(9.8 m/s^2)}{1\cdot 10^6 N/C}=1.49\cdot 10^{-17}C[/tex]