Question

There are 27 drops of a conducting fluid. Each drop has a radius $r$, and each of them is charged with the same potential ${V_1}$. They are all combined to form a bigger drop. The potential of the bigger drop is ${V_2}$. Find the ratio ${V_2}/{V_1}$. Ignore the change in density of the fluid on combining the drops.

Answer 1

First, find the radius of the bigger drop which is formed after combining the smaller drops. We can use the fact that the volume of the drops remains constant before and after merging. Then, use the concept of potential of a spherical charged body. The potential of a spherically charged body is given by the formula, $V = k\dfrac{q}{r}$

Complete answer:
First, we have to find the radius of the bigger drop to find the potential of the drop.
We will use the fact that the volume of the drops is constant before and after merging all the drops.
Combined volume of smaller drops $= 27 \times \dfrac{4}{3}\pi {r^3}$
Let the radius of the bigger drop be $R$
Volume of the bigger drop after combining$= \dfrac{4}{3}\pi {R^3}$
Equating both the volumes,
$\Rightarrow 27 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$
$\Rightarrow R = 3r$
Potential of one small drop is given as ${V_1}$.
We know that,
Potential of a charged sphere is given by the formula,
$V = k\dfrac{q}{r}$
Where,
$q$ is the charge of the sphere
$r$ is the radius of the sphere
$\Rightarrow {V_1} = k\dfrac{q}{r}$
Potential of the bigger drop is given as ${V_2}$.
$\Rightarrow {V_2} = k\dfrac{{27q}}{{3r}}$
Dividing ${V_2}$by ${V_1}$
$\Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{k\dfrac{{27q}}{{3r}}}}{{k\dfrac{q}{r}}}$

$\therefore$ The required values is $\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{9}{1}$

Note:
It is given in the question that the density of the fluid is constant while mixing. This is the reason we have conserved the volume of the drops otherwise we cannot do that. We need not to be confused while doing calculations and in between the terms $R$ and $r$.