Question

A charged drop of oil of radius 2.76 micrometer and density 920 kg m^-3 is held stationary in a vertically downward electric field of 1.65*10^6 nC^-1. (1) Find a magnitude and sign of the charge on the drop. (2) If the drop captures two electrons on the same electric field what would be its acceleration, what would be its acceleration? Ignore viscous frag. (Step-by-Step solution please)

Answer 1

Answer :The weight of the oil drop is balanced by the electrostatic force applied on it .Number of excess electron n=12n=12Electric field E=2.55×1064N/cE=2.55×1064N/cDensity ρ=1.26g/cm3ρ=1.26g/cm3=1.26×103kg/m3=1.26×103kg/m3Electrostatic force on drop =qE=qEq=neq=neqE=neEqE=neEgravitational force on the drop =mgmass = volume ×× densityVolume v′=43v′=43πr3πr3gravitation force =43=43πr3×ρ×gπr3×ρ×gthe oil drop is held stationary.Hence the net force on the drop is zero.∴∴ Electrostatic force = gravitational forceneE=43neE=43πr3egπr3egr3=3neE4πegr3=3neE4πegSubstituting the valuesr3=3×12×1.6×106−19×2.55×1044×3.14×1.26×103×9.8r3=3×12×1.6×106−19×2.55×1044×3.14×1.26×103×9.8r3=0.94×10−18r3=0.94×10−18r=(0.94×10−18)1/3r=(0.94×10−18)1/3=9.81×10−7m=9.81×10−7mHence the radius of the drop is 9.81×10−7m

Answer 2

Answer:

Explanation:

Answer :The weight of the oil drop is balanced by the electrostatic force applied on it .Number of excess electron n=12n=12Electric field E=2.55×1064N/cE=2.55×1064N/cDensity ρ=1.26g/cm3ρ=1.26g/cm3=1.26×103kg/m3=1.26×103kg/m3Electrostatic force on drop =qE=qEq=neq=neqE=neEqE=neEgravitational force on the drop =mgmass = volume ×× densityVolume v′=43v′=43πr3πr3gravitation force =43=43πr3×ρ×gπr3×ρ×gthe oil drop is held stationary.Hence the net force on the drop is zero.∴∴ Electrostatic force = gravitational forceneE=43neE=43πr3egπr3egr3=3neE4πegr3=3neE4πegSubstituting the valuesr3=3×12×1.6×106−19×2.55×1044×3.14×1.26×103×9.8r3=3×12×1.6×106−19×2.55×1044×3.14×1.26×103×9.8r3=0.94×10−18r3=0.94×10−18r=(0.94×10−18)1/3r=(0.94×10−18)1/3=9.81×10−7m=9.81×10−7mHence the radius of the drop is 9.81×10−7m

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