An oil drop of 12 excess electrons is held stationary under a constant electric field of ` 2.55 xx 10 ^(4) NC^(-1) ` in Millikan's oil drop experiment. The density of the oils is 1.26 `g cm ^(-3) ` Estimate the radius of the drop ` (g= 9.81 m s ^(-2) and e = 1.60 xx 10 ^(-19) C ) `

Explanation

Here ` E= 2.55 xx 10 ^(4) NC^(-1) `, density of oil `rho =1.26 g cm ^(-3) = 1.26 xx 10^(3) kg m ^(-3) `
As the oil drop has 12 excess electrons hence , charges on oil drop `|q| =12e =12 xx 1.60 xx 10 ^(-19) C.` As the oil drop remain stationary in an electric field, it is possible only if weight of oil drop acting vertically downward is just balanced by force due to electric field which must act in vertically upward direction i.e.
` mg= ( 4)/(3) r^(3) rho g= q_E `
` rArr " " r=[( 3qE)/( 4pi rho g) ] ^(1//3) =[( 3xx12xx1.60 xx 10 ^(-19) xx 2.55 xx 10 ^(4) )/( 4xx 3.14 xx 1.26 xx 10 ^(3) xx 9.81)]^(1//3) `
` =9.81 xx 10 ^(-7) m or 9.81 xx 10 ^(-4) mm. `