Question

Two identical mercury drops, each of radius r are charged to the same potential V.If the mercury drops coalesce to form a big drop, each of radius R then the potential of the combined drop will be

• A. $(3{)}^{1/2}V$
• B. $(3{)}^{2/3}V$
• C. $(2{)}^{2/3}V$
• D. $(2{)}^{3/2}V$

Answer 1

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

Given

number drops =2 with radius of small drop=r and potential =V.

Radius of bigger drop=R

Now potential each small drop $V=\frac{kq}{r}$

Volume of 2 Hg drop=Volume of one large drop

$2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

$R={\left(2\right)}^{\frac{1}{3}}r$ -------------(A) {here charge remains conserved}

Now charge on small drop $=q$ and charge on large drop containing 2 drop$=2q$

Thus potential of 1 large drop $V_l=\frac{kq}{R}=\frac{k2q}{2^{1/3}r}=2^{2/3}\frac{kq}{r}$

Therefore $V_l=2^{2/3}V$