Eight drops of water of the same size are equally and similarly charged. They combine together to form a bigger drop. Compare the capacitances of a bigger drop with that of the smaller drop.

Let $r$ be the radius of the smaller drop and $q$ be the charge on it, then charge on bigger drop $Q = 8q$.
Now, $R$ be the radius of a bigger drop, then

$\frac{4}{3}\; \pi \; R^3 = 8 \; \frac{4}{3} \; \pi \; r^3$ 

or, $R^3 = 8\;r^3$ 

⇒ $R = 2\;r$ ......... (i)

Now,

Capacitance of bigger drop $C_b = 4\; \pi\; \epsilon_0\; R$

Capacitance of smaller drop $C_s = 4\; \pi\; \epsilon_0\; r$

To compare the capacitance of bigger and smaller drop is,

$\frac{C_b}{C_s} = \frac{R}{r} =  \frac{2\;r}{r}$ $ = 2$

∴ $\frac{C_b}{C_s} $ $ = 2$

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