Question

$n$ small drops of small size are charged to $V$ volt each. If they collapse to form a single large drop, then its potential will be

• A. $\left(V/n\right)$
• B. $Vn$
• C. $V{n}^{2/3}$
• D. None of these

Answer 1

The correct option is C $V{n}^{2/3}$

Volume of a bigger drop from n drop of same size $=n\times \left(4/3\right)\pi r^3=\left(4/3\right)\pi R^3$

Where, r and R are the radius of small drop and bigger drop respectively.

Hence,

$R=n^{1/3}\times r$

If each small drop have the charge q coulomb then charge on bigger drop will be nq coulomb.

Potential $V_b$ of bigger drop is given by,

$V_b=\frac{nq}{4\pi {\epsilon }_0{n}^{1/3}r}=n^{2/3}\frac{q}{4\pi {\epsilon }_{0}r}=n^{2/3}V$