Question

Eight Charged Drops of water each of radius 1 mm and having a charge of 10^-10 coulomb combine to form a bigger drop. Determime the potential of the bigger drop

Answer 1

Let radius of smaller drop is r and bigger drop is R.

8 × volume of smaller drop = volume of bigger drop

8 × 4/3 πr³ = 4/3 πR³

or, 8r³ = R³ => R = 2r

capacitance of bigger drop, C = $4\pi\epsilon_0R$

= 4 × 3.14 × 8.85 × 10^-12 × 2 × 10^-3

[ as r = 1mm = 10^-3 m, $\epsilon$ = 8.85 × 10^-12 ]

= 222.312 × 10^-15

= 2.223 × 10^-13 F

charge on bigger drop , Q = 8 × charge on smaller drop

= 8 × 10^-10 C

potential on bigger drop , V = Q/C

= 8 × 10^-10/(2.223 × 10^-13) volts

= 3.598 × 10³

= 3598 volts