Answer:
The electric field to balance the weight is approximately equals to 3.49x10^5 Newton/Coulumb
Explanation:
In order to be stationary position, magnitude of the total force due to electric field should be equal to the gravitational force that is, [tex]|F_{electrostatic}| = |F_{gravitational}|[/tex]
where
[tex]F_{electrostatic}=q.E[/tex]
where q is the charge of the droplet and E is the electric field. On the other hand
[tex]F_{gravitational}=m.g=V.d.g=\frac{4}{3}.\pi.r^3.d.g[/tex]
where m,V ,d and r are the mass, volume, density and radius of the oil droplet respectively and g is the gravitational acceleration (g=9,80665 m/sn^2). By using the first equation and solving it for the electric field we can write,
[tex]q.E=\frac{4}{3}.\pi.r^3.d.g\\E=\frac{4}{3}.\pi.\frac{r^3.d.g}{q}\\E=\frac{4}{3}.\pi\frac{(1,6x10^{-4})^3.(0,85x10^{-3}).(9,81)}{1,6x10^{-19}}\\E\approx 3,49x10^{5} (N/C)[/tex]
(Note: d=0.85 x 10^-3 kg/cm^3 and the unit of electric field is Newton per coulumb (N/C))
