Question

If 'n' identical water drops (assumed spherical each) each changed to a potential energy U coalesce to form a single drop, the potential energy of the single drop is (Assume that drops are uniformly charged):
A) ${n^{2/3}}U$
B) ${n^{3/2}}U$
C) ${n^{4/3}}U$
D) ${n^{5/3}}U$

Answer 1

Hint
In this question, we know that the potential energy of the one drop is $U = \dfrac{{kqq}}{r}$ here, we use the concept that as the single drop is made by n identical water drops, so the volume of the single drop will be equal to the volume of n drop. As the charge is same so the charge of the single drop which is made by the n drops is nq, after that put the relation which we get by equating the volume of the drops then we will get the desired results.

Complete step by step answer
Here, it is given that 'n' identical water drops (assumed spherical each) each changed to a potential energy U coalesce to form a single drop. We have to find out the value of potential energy of the single drop.
For this let us consider one drop having charge q then potential energy of the drop is $U = \dfrac{{kqq}}{r}$ …………… (1)
Where, q is the charge and r is the distance.
$k = \dfrac{1}{{4\pi { \in _0}}}$ is constant and ∈0 is the permittivity of the free space.
As there are n drops of water, and these drops coalesce to form the single drop so the volume of n drops is equal to the volume of the single drop.
For this let us consider the radius of the small drop is r and radius of the big drop is R
We also know that the water drop is in the form of sphere so the volume of one drop is = $\dfrac{4}{3}\pi {r^3}$
Therefore, volume of the n drops is ${V_1} = n \times \dfrac{4}{3}\pi {r^3}$ ………………… (2)
Similarly, the volume of the big drop is ${V_2} = \dfrac{4}{3}\pi {R^3}$ ………………………… (3)
Equating equations (2) and (3), we get
$\Rightarrow n \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3} $
$\Rightarrow R = {n^{1/3}}r $ …………………………….. (4)
Now, as the potential energy of the small drop is $U = \dfrac{{kqq}}{r} = \dfrac{{k{q^2}}}{r}$ …………………………. (5)
As the single drop is made from n number of small drops and the charge is uniform then the charge of the big drop is equal to the nq, then the potential energy of the big drop is
$ \Rightarrow U' = \dfrac{{k\left( {nq} \right)\left( {nq} \right)}}{R}$
Now, put the value R in above equation from equation (4), we get
$\Rightarrow U' = \dfrac{{k{n^2}{q^2}}}{{{n^{1/3}}r}} $
$\Rightarrow U' = \dfrac{{k{q^2}}}{r}{n^{5/3}} $
Now, using the equation (5), we get
$ \Rightarrow U' = {n^{5/3}}U$
Hence, the potential energy of the big drop is ${n^{5/3}}U$.
Therefore, option (D) is correct.

Note
A charge placed in an electric field possesses potential energy and is measured by the work done in moving the charge from infinity to that point against the electric field. If two charges $q_1$ and $q_2$ are separated by a distance d, the electric potential energy of the system is given as $U = \dfrac{{kqq}}{r}$
The symbol has the same meanings as above.
In this question care must be taken in taking the charge of the big drop as the charge remains constant or uniform throughout the region, therefore we have taken the charge of the big drop as nq. And always remember that always the water droplet is in the form of a sphere, therefore we have to use the volume of the sphere i.e. $\dfrac{4}{3}\pi {r^3}$.