Question

a charged oil drop is suspended in a uniform electric field 27.5*10⁴volt m^-1sothatit neither rises nor falls .find the charge on the drop, given mass of the drop is 9*10^-5kg(neglect air resistance).

Answer 1

$$\begin{gathered} \mathrm{Here},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{case}\mathrm{equilibrium}.\mathrm{The}\mathrm{force}\mathrm{of}\mathrm{gravity}\mathrm{is}\mathrm{balanced}\mathrm{by}\mathrm{electric}\mathrm{force}. \\ \mathrm{qE}=\mathrm{mg} \\ \mathrm{or}\mathrm{q}=\frac{\mathrm{mg}}{\mathrm{E}} \\ \mathrm{Putting}\mathrm{the}\mathrm{values}, \\ \mathrm{q}=\frac{9\times {10}^{-5}\times 10}{27.5\times {10}^4}=3.27\times {10}^{-9}\mathrm{C} \end{gathered}$$