Question

Eight drops fo water each having a charge of `3xx10^(-9) C` having surface potential `2 V` coalesce to from a single drop. What is the surface potentail (in volt) of new drop ?

Answer 1

Correct Answer - 8
Let `r` be the radius of each small drop `R` be the radius of one big drop.
`:. (4)/(3) pi R^(3) = 8xx (4)/(3) pi r^(3)`
`R^(3) = (2r)^(3)`
or `R = 2r`
Now `q = CV = 4pi in_(0) r V`
`r = (q)/(4pi in_(0) V) = (9xx10^(9)x3xx10^(-9))/(2) = (27)/(2) m`
`R = 2r = 2xx (27)/(2) = 27 m`
Surface potential of new drop
`V = ("total charge")/("capacity") = (8q)/(4pi in_(0) R)`
`V = (8xx3xx10^(-9)xx10^(9))/(27) = 8` volt