posted by Anonymous .
A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of 8380 N/C. The mass of the water drop is 3.40 10-9 kg.
How many excess electrons or protons reside on the drop?
Enough (n) to make n*e*E equal to the drop's weight, M g
Solve for n.
e is the electron charge, E is the electric field strength, and you know what g is.