The electric field required to just balance a liquid drop of $2 imes 10^{-9} \mathrm{~kg}$ with a

Question

The electric field required to just balance a liquid drop of $2 	imes 10^{-9} \mathrm{~kg}$ with a charge $9.8 	imes 10^{-8} \mu \mathrm{C}$ of is\

Professor PC Thomas Classes · Answer

$\mathrm{qE}=\mathrm{mg}$\ $\mathrm{E}=\frac{\mathrm{mg}}{\mathrm{q}}$\ \ $=\frac{2 	imes 10^{-9} 	imes 9.8}{9.8 	imes 10^{-8} 	imes 10^{-6}}$\