Question

Twenty seven drops of water of the same size are equally and similarly charged. They are then united to from a bigger drop. By what factor will the electrical potential changes ?

• A. $3$ times
• B. $6$ times
• C. $9$ times
• D. $27$ times

Answer 1

Answer: (A)

$3$ times

Net potential due to 27 charged drop is $V=\frac{k\left(27q\right)}{r}$

If $\rho$ be the density of the charges , $q=\frac{4}{3}\pi r^3\rho$

So, $V=27k\frac{\left(4/3\right)\pi r^3\rho }{r}=36k\pi r^2\rho$

Let 27 drops make a big drop of radius R and charge Q.

Thus, $27\times \left(4/3\right)\pi r^3=\left(4/3\right)\pi R^3$

or $R=3r$

Potential due to big drop is $V^{\prime }=\frac{kQ}{R}=k\frac{\left(4/3\right)\pi R^3\rho }{R}=k\left(4/3\right)\pi R^2\rho =k\left(4/3\right)\pi (3r{)}^2\rho =12k\pi r^2\rho$

Thus, $V/V^{\prime }=36/12=3$