27 drops of water, each of radius 3 mm and having equal charge are combined to form a large drop. Find the ratio of the surface density of charge on the large drop to that on each small drop.

Explanation

Suppose, charge on each small drop of water is q. So the charge in the combined drop will be 27q.
In the first case, surface density of charge,

`sigma_1 = q/(4pir^2) = q/(4pi(0.3)^2) ` here, r =3mm = 0.3 cm
If R be the radius of the large drop, we have,
`4/3piR^3 = 27xx4/3pi(0.3)^3`
or, R=0.9 cm
In the second case,
`sigma_2=(27q)/(4piR^2) = (27q)/(4pi(0.9)^2)`
`therefore sigma_2/sigma_1=(27q)/(4pi(0.9)^2) xx (4pi(0.3)^2)/q = (27xx0.09)/0.81 = 3/1`
`therefore sigma_2:sigma_1=3:1`