Answer:
q = 0.0392 / V, for V= 0.1V q = 0.392 C
Step-by-step explanation:
For this exercise we can assume that the power energy of the drops is transformed into kinetic energy, therefore we use the conservation of energy 
starting point 
 Em₀ = U = q V 
final point 
 Em_f = K = ½ m v² 
 Em₀ = Em_f 
 q V = ½ m v² 
 q = 
 
let's calculate 
 q = ½ 0.40 10⁻³ 14² / V 
 q = 0.0392 / V 
 
The object to be painted is connected to ground therefore its potential is dro, but the gun where it is painted has a given potential, suppose it is
 V = 0.1 V 
 q = 0.0392 / 0.1 
 q = 0.392 C