Question

Eight charged liquid drops each of radius $1mm$ and charge $0.1nC,$ merge together to form a bigger drop. The electric potential at the surface of single big drop is:

• A. $3.6kV$
• B. $7.2kV$
• C. $1.8kV$
• D. $36kV$

Answer 1

Answer: (A)

$3.6kV$

Let, the density of liquid $=S$

given, radius $=1\times {10}^{-3}m$

charge $=0.1nc$

let mass $=x$

Now, as we know, $S=\frac{mass}{volume}$

The radius of the bigger drop is given by $R$.

and for small drop, $S=\frac{x}{\left(\frac{4}{3}\pi r^3\right)}\Rightarrow r^3=\frac{x}{\frac{4}{3}\pi S}$

For bigger drop, $S=\frac{8x}{\left(\frac{4}{3}\pi R^3\right)}\Rightarrow R^3=\frac{8x}{\frac{4}{3}\pi S}$

So, $\frac{R^3}{r^3}=8\Rightarrow R^3={8x}^3\Rightarrow R=2r$

$\Rightarrow R=2\times {10}^{-3}m$; charge on the bigger drop is given by, $Q=8\times 0.1nc$

$=8\times {10}^{-10}C$

Potential on the surface is determined as,

$=\frac{KQ}{R}=\frac{9\times {10}^9\times 8\times {10}^{-10}}{2\times {10}^{-3}}=3.6kV$

$\therefore$ Option A is the correct answer.