Question

$27$ small drops each having charge q and radius $r$ coalesce to from big drop. How many times charge and capacitance will become?

• A. 3, 27
• B. 27, 3
• C. 27, 27
• D. 3, 3

Answer 1

Answer: (B)

Let $R$ and $r$ be the radii of bigger and each smaller drop respectively.
In coalescing into a single drop, charge remains conserved.
Charge on bigger drop
$=27x$ charge on smaller drop
ie, $q^{\prime }=27q$
Now, before and after coalescing, volume remains same.
$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$
$\therefore R=3r$
Hence, capacitance of bigger drop
$C^{\prime }=4\pi {\epsilon }_{0}R$
$=4\pi {\epsilon }_0(2r)$
$=3(4\pi {\epsilon }_{0}r)$
$=3C$