Question

Eight charged water drops each of radius 1mm & charge 10^-10 C form a single drop the potential of big drop is?

Answer 1

solution:

[tex]volume of the resultant water drop=volume of all the drops together\\ 8\times\frac{4}{3}\pi r^3=\frac{4}{3}\times\pi\times r^3\\ r=2r potential = \frac{Kq}{R}\\ k\times8\times(\frac{10^-9}{2\times10^-2})\\ =3600[/tex]

Answer 2

Answer:

The potential of the big drop is 360 V.

Explanation:

Given that,

Radius r = 1mm

Charge $q=10^{-10}\ C$

Volume of the resultant of small water drop = Volume of the big drop of water

$8\times\dfrac{4}{3}\pi\times r^3=\dfrac{4}{3}\pi\times R^3$

$(2r)^3=R^3$

$R= 2r$

The potential of big drop is

$V= \dfrac{kQ}{R}$

$V= \dfrac{9\times10^{9}\times8\times10^{-10}}{2\times10^{-2}}$

$V=360 volt$

Hence, The potential of the big drop is 360 V.