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