Question

Twenty seven charged water droplets each with a diameter of 2mm and a charge of 10-12 c coalesce to form a single drop . Calculate the potential of bigger drop

Answer 1

Hope it will help u

Answer 2

Answer:

The potential of the bigger drop is $V=81V$

Explanation:

Given that :

diameter$=2mm$

Then radius will be $\frac{2}{2}=1mm$

Charge is given to be $Q=10^{-12}C$

Here we have to first find the volume of $27$ charged water droplets.

This will be the volume of the bigger drop. From that, we can figure out the radius of the bigger drop.

$n*\frac{4}{3} *$ π $*r^3=\frac{4}{3}*$ π $*R^3$

$27*\frac{4}{3} *\frac{22}{7} *(1)^3=\frac{4}{3} *\frac{22}{7} *(R)^3$

$27=R^3$

$R=\sqrt[3]{27} \\R=3mm\\R=3*10^{-3}m$

After finding out the radius, next we find the total charge of the bigger drop.

Total charge of the bigger drop $=27*10^{-12}C$

We know that the formula for potential $V=\frac{1}{4*\frac{22}{7} *E_{0} } \frac{Q}{R}$

$=\frac{9*10^9*27*10^{-12}}{3*10^{-3}} \\=\frac{81*10^{-3}}{10^{-3}} \\\\V=81V$

Hence, the total potential is $81V$