Question

Calculate the number of electrons carried by an oil drop of radius 1.2 104 m bela
stationary between two parallel plates maintained at a vertical distance of 10 cm from each
other. Charge of an electron is 1.6 x 10-19 C and density of the oil is 910 kg m The voltage
between the plates is 8233V

Answer 1

Answer

Charge on drop =4×1.6×10−19C

                            =6.4×10−19C

Radius of drop =10−6m

Mass of the drop = volume × density

                             =34π10−18×2000

                             =8.3775×10−15kgs

For equilibrium, weight = electric force

⇒8.3775×10−15×g=5×10−3v×6.4×1019

⇒5×6.48.3775×10×g=v

⇒v=641.04volts