Question

\( -1 \times 10^{-6} C \) charge is on a drop of water having mass \( 10^{-6} kg \). What electric field should be applied on the drop so that it is in the balanced condition with its weight? a) \( 10 V / m \) upward b) \( 10 V / m \) downward c) \( 0.1 V / m \) downward d) \( 0.1 V / m \) upward

Answer 1

By using QE = mg

\(\Rightarrow E=\frac{mg}{Q}=\frac{10^{-6}\times10}{10^{-6}}=10\,V/m\)

upward because charge is positive.