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
$ = 27 x$ charge on smaller drop
ie, $q' = 27 q$
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
$C' = 4 \pi \varepsilon_0 R$
$ = 4 \pi \varepsilon_0 (2r)$
$ = 3 (4 \pi \varepsilon_0 r)$
$= 3C$