Question

$n$ charged drops, each of radius $r$ and charge $q$, coalesce to form a big drop of radius $R$ and charge $Q$. If $V$ is the electric potential and $E$ is the electric field at the surface of a drop, then

• A. $E_{big}=n^{1/3}{E}_{small}$
• B. $V_{big}=n^{2/3}{V}_{small}$
• C. $E_{small}=n^{2/3}{E}_{big}$
• D. $V_{big}=n^{1/3}{V}_{small}$

Answer 1

Answer: (A), (B)

The correct options are
A $E_{big}=n^{1/3}{E}_{small}$
B $V_{big}=n^{2/3}{V}_{small}$

From conservation of charge we can say
$nq=Q$----(i)

By conservation of volume
$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$⟹R=n^{1/3}r$ -----(ii)

Let $E_{big}$ be the electric field of bigger drop and $E_{small}$ be the electric field of small droplet.

We can write
$E_{big}=\frac{kQ}{R^2}$ = $\frac{knq}{\left(n^{1/3}\right(r){)}^2}$----(iii)
$E_{small}=\frac{kq}{r^2}$ ----(iv)

On dividing (iii) from (iv) we get
$\frac{E_{big}}{E_{small}}=n^{1/3}$

Let $V_{big}$ and $V_{small}$ be potential of big and small drop respectively.
$V_{big}=\frac{kQ}{R}$ =$\frac{knq}{n^{1/3}r}$----(v)
$V_{small}=\frac{kq}{r}$----(vi)

Dividing (v) from (vi) we get
$\frac{V_{big}}{V_{small}}=n^{2/3}$