If the voltage between the two plates of an electric air cleaner is 500 V, how fast would a 10-12 kg soot particle with -1 x 10-11 C of charge on it be moving if it went from the negative plate to the positive?

Answer 1

The electrical potential energy loss,
q*V = 5*10^-9 J ,
would equal the kinetic energy gain when it hits the opposite plate,
(1/2) M V^2

V = sqrt(2 q V/M) = 100 m/s