Answer:
W = 2.3 10² [tex]F_{e}[/tex]
Explanation:
The force of the weight is
W = m g
let's use the concept of density
ρ= m / v
the volume of a sphere is
V = [tex]\frac{4}{3}[/tex] π r³
V = [tex]\frac{4}{3}[/tex] π (1.0 10⁻³)³
V = 4.1887 10⁻⁹ m³
the density of water ρ = 1000 kg / m³
m = ρ V
m = 1000 4.1887 10⁻⁹
m = 4.1887 10⁻⁶ kg
therefore the out of gravity is
W = 4.1887 10⁻⁶ 9.8
W = 41.05 10⁻⁶ N
now let's look for the electric force
F_e = q E
F_e = 12 10⁻¹² 15000
F_e = 1.8 10⁻⁷ N
the relationship between these two quantities is
[tex]\frac{W}{F_e}[/tex] = 41.05 10⁻⁶ / 1.8 10⁻⁷
\frac{W}{F_e} = 2,281 10²
W = 2.3 10² [tex]F_{e}[/tex]
therefore the weight of the drop is much greater than the electric force
