Question

Given, ‘n’ charged drops, each of radius r and charge q, coalesce to form a big drop of radius R and charge Q. If V is the electric potential and E is the electric field at the surface of a drop, then:
A. $E_{big}=n^{2/3}{E}_{small}$
B. $V_{big}=n^{1/3}{V}_{small}$
C. $E_{small}=n^{2/3}{E}_{big}$
D. $V_{big}=n^{2/3}{V}_{small}$

Answer 1

Hint: An electric potential is defined as the work needed to move a unit of electric charge from a time varying electric field and vice versa. As ‘n’ drops coalesce to form a big drop, so the volume of the new drop must be equal to the combined volume of each drop. Obtain a relation between the radii of the drops, i.e. initially a single drop and after coalescing into a big one. Later use the relation to check and find the correct option among the given by using the concept of Electric potential and electric field.
Formula Used: $V={\displaystyle \frac{kq}{r}}$

Complete answer:
Let the charge on one drop be ‘q’
Charge on big drop be ‘nq’
According to the question,
$\begin{array}{rl}& n{\displaystyle \frac{4}{3}}\pi r^3={\displaystyle \frac{4}{3}}\pi R^3\\ & \Rightarrow R=n^{\left({\displaystyle \frac{1}{3}}r\right)}\end{array}$
Electric potential is defined as the total potential energy possessed by a unit charge when located at any point in outer space.
Mathematically it is given as, $V={\displaystyle \frac{kq}{r}}$
Where ‘V’ is the electric potential energy, ‘q’ is the point charge, ‘r’ is the distance between some point around the charge and the point charge and ‘k’ is the Coulomb’s constant.
Let the potential on big drop be, $E_{big}={\displaystyle \frac{kQ}{R^2}}$
Let the potential on small drop be, $E_{small}={\displaystyle \frac{kq}{r^2}}$
$E_{big}={\displaystyle \frac{knq}{n^{2/3}{r}^2}}=\left(n^{1/3}\right){\displaystyle \frac{kq}{r^2}}=n^{1/3}{E}_{small}$
$V_{small}={\displaystyle \frac{kq}{r}}$, $V_{big}={\displaystyle \frac{kQ}{R}}$
$\Rightarrow V_{big}={\displaystyle \frac{knq}{n^{1/3}r}}=\left(n^{2/3}\right){\displaystyle \frac{kq}{r}}=n^{2/3}{V}_{small}$

Hence, option (D) is the correct answer.

Note:
The electrostatic potential between any two arbitrary charges ‘$q_1$’ and ‘$q_2$’ separated by a distance ‘r’ is stated by Coulomb’s law. Mathematically it is given as, $U=k\times {\displaystyle \frac{q_1{q}_2}{r^2}}$; where ‘U’ is denoted for the electrostatic potential energy and ‘$q_1$’ and ‘$q_2$’ are two charges. Electric potential difference for an electrical circuit is defined as the potential between two points ‘E’ is given as the amount of work done W’ by an external agent in moving a ‘Q’ unit charge from one point to another. Mathematically, it is given as $E={\displaystyle \frac{W}{Q}}$; where ‘E’ is the electrical potential difference between two points, ‘W’ is the work done in moving a point charge from one point to another and ‘Q’ is the quantity of charge in coulombs.