Question

Paint droplets initially at rest are dispensed from a paint gun. These droplets are then repelled by the paint sprayer toward the object to be painted. This object is grounded and therefore the electric potential is taken to be zero everywhere on the object. What charge must a 0.400 mg drop of paint have to arrive at the grounded object with a speed of 14.0 m/s

Answer 1

Answer:

q = 0.0392 / V, for V= 0.1V   q = 0.392 C

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 = $\frac{1}{2} \ \frac{m v^2}{V}$

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