Question

If 'n' identical water drops assumed spherical each charged to a potential energy U coalesce to a single drop, the potential energy of the single drop is: (Assume that drops are uniformly charged).

• A. $n^{2/3}U$
• B. $n^{3/2}U$
• C. $n^{4/3}U$
• D. $n^{5/3}U$

Answer 1

The correct option is A $n^{2/3}U$

Potential drop in 1 drop be= $k\frac{q}{r}$= U (Let)

According to question, volume of n drops is equal to volume of the bigger drop.

So, $n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$-----(1)

where (r= radius of smaller drop and R= radius of bigger drop)

From (1) R= r$n^{\frac{1}{3}}$

The total charge remains constant.

Charge on 1 drop= q

Charge on the bigger drop= nq

Potential on the bigger drop, U'= k$\frac{nq}{r{n}^{\frac{1}{3}}}$

= $n^{\frac{2}{3}}$U