Question -
Answer -
Excess electrons on an oildrop,┬аn┬а= 12
Electric field intensity,┬аE┬а= 2.55 ├Ч 104┬аNCтИТ1
Density of oil,┬а╧Б┬а= 1.26 gm/cm3┬а=1.26 ├Ч 103┬аkg/m3
Acceleration due togravity, g = 9.81 m sтИТ2
Charge on an electron,┬аe┬а= 1.6 ├Ч 10тИТ19┬аC
Radius of the oil drop =┬аr
Force (F) due to electric field┬аE┬аis equal to the weight of the oildrop (W)
F = W
Eq┬а=┬аmg
Ene┬а
Where,
q┬а= Net charge on the oil drop =┬аne
m┬а= Mass of the oil drop
= Volume of the oil drop ├ЧDensity of oil
= 9.82 ├Ч 10тИТ4┬аmm
Therefore,the radius of the oil drop is 9.82 ├Ч 10тИТ4┬аmm.