Question

A tiny ball weighs 0.0055 kg and carries a charge of of 3.25 x10-6 C. What electric field (magnitude and direction) is needed for the ball to remain suspended in air?

 

Answer 1

At the equilibrium (the ball remains suspended in air)

$$mg=qE$$

Hence, the electric field

$$E=\frac{mg}{q}=\frac{0.0055*9.8}{3.25*10^{-6}}=1.66*10^4\:\rm V/m$$

The field must be directed upward.