Question

a water particle of mass 10 miligramsand having a charge 1.50* 10^-6 stays suspended in a room. what is the magnitude of electric field in the room?what is its direction

Answer 1

$$\begin{gathered} Weightofthechargedwaterparticleisbalancedbythetheelectricfieldtogetitsuspendedi.e., \\ q\times E=mg. \\ Givenq=1.5\times {10}^{-6}C,m=10\times {10}^{-6}Kg,g=9.8m{s}^{-2} \\ Therefore1.5\times {10}^{-6}E=10\times {10}^{-6}\times 9.8 \\ orE=\frac{10\times {10}^{-6}\times 9.8}{1.5\times {10}^{-6}}orE=\frac{9.8\times {10}^{-5}}{1.5\times {10}^{-6}}orE=65.33N{C}^{-1} \end{gathered}$$