Question

Solve it for me.But don't waste answers if you don't know.

Answer 1

Number of excess electron n=12

Electric field E=2.55×1064N/c

Density ρ=1.26g/cm3

=1.26×103kg/m3

Electrostatic force on drop =qE

q=ne

qE=neE

gravitational force on the drop =mg

mass = volume × density

Volume v′=43πr3

gravitation force =43πr3×ρ×g

the oil drop is held stationary.

Hence the net force on the drop is zero.

∴ Electrostatic force = gravitational force

neE=43πr3eg

r3=3neE4πeg

Substituting the values

r3=3×12×1.6×106−19×2.55×1044×3.14×1.26×103×9.8

r3=0.94×10−18

r=(0.94×10−18)1/3

=9.81×10−7m

Hence the radius of the drop is 9.81×10−7m

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