Question

There are $27$ drops of a conducting fluid. each has a radius r and they are charged to a potential $V_0$. These are combined to form a bigger drop. Its potential will be:

• A. $V_0$
• B. $3V_0$
• C. $9V_0$
• D. $29V_0$

Answer 1

The correct option is C $9V_0$

If $Q$ be the charge on each drop.

Change on big drop is $Q=27g$

if $R$ be the radius of bigger drop,

$\begin{array}{c}\frac{4}{3}\pi R^3=27\left(\frac{4}{3}\pi R^3\right)\\ \Rightarrow R=3r\end{array}$

Thus,

$V_1=k\frac{q}{r}$

and

$\begin{array}{c}{V}_2=k\frac{Q}{R}=k\frac{27q}{3r}=6k\frac{q}{r}=9V_1\\ \therefore \frac{V_2}{V_1}=9\end{array}$