Animated Solution for Physics - Electrostatics: 512 identical drops of mercury are charged to a potential of 2 V each. The drops are joined to form a single drop. The potential of this drop is ............ V.

Imagine a fascinating scenario: you have $512$ tiny, identical drops of liquid mercury. Each of these drops carries a small electrostatic charge and is maintained at a potential of $2\text{ V}$. Suddenly, they all coalesce—merging together to form one single, massive drop. The question that arises is: what happens to the electric potential of this new, giant drop?

This classic problem is a beautiful intersection of basic geometry, fluid mechanics, and electrostatics. To solve it, we must rely on two fundamental conservation laws: the conservation of volume and the conservation of charge.

Conservation of Volume

When liquid drops merge, assuming no liquid is lost or evaporated in the process, the total volume of the liquid remains strictly conserved. The volume of the single large drop must be exactly equal to the sum of the volumes of all the individual small drops.

Let the radius of a single small drop be $r$, and the radius of the newly formed large drop be $R$. Since there are $n=512$ small drops, we can write the volume conservation equation as:

By canceling out the common geometric factors $\frac{4}{3}\pi$ from both sides, we are left with a simple cubic relationship:

To find the relationship between the radii, we take the cube root of both sides. Recognizing that $512$ is a perfect cube ($8^3=512$), we get:

This tells us that the radius of the giant drop is exactly $8$ times the radius of a single small drop. Notice how the radius does not scale linearly with the number of drops; it scales with the cube root of the number of drops ($n^{1/3}$).

Conservation of Charge

Just as matter cannot be created or destroyed out of nowhere, electric charge is also a conserved quantity. When the $512$ drops merge, all their individual charges pool together onto the surface of the new, larger drop.

Let the charge on a single small drop be $q$, and the total charge on the large drop be $Q$. The conservation of charge dictates that:

Unlike the radius, the total charge scales linearly with the number of drops. The big drop holds $512$ times more charge than a single small drop.

Now, let's bring in the electrostatics. The electric potential $V$ at the surface of a charged spherical conductor is given by the formula:

where $k$ is Coulomb's constant ($k=\frac{1}{4\pi {ϵ}_0}$), $Q$ is the total charge on the sphere, and $R$ is its radius.

For a single small drop, we are given that its potential is $2\text{ V}$. Therefore, we can write:

This equation serves as our foundational reference point. We need to express the potential of the big drop in terms of this known quantity.

Calculating the New Potential

Let's set up the potential equation for the newly formed giant drop, which we will call $V_{\text{big}}$. Using the formula for potential, we substitute the new total charge $Q$ and the new radius $R$:

Now, we substitute the relationships we derived from our conservation laws ($Q=512q$ and $R=8r$):

We can separate the numerical constants from the physical variables to see the structure more clearly:

Dividing $512$ by $8$ gives us $64$. And if you look closely at the second term in the parentheses, $\frac{kq}{r}$, you will recognize that it is exactly the expression for the potential of the small drop, $V_{\text{small}}$!

This is a profound result. It shows that when $n$ drops merge, the potential scales by a factor of $n^{2/3}$ (since $512^{2/3}=(8^3{)}^{2/3}=8^2=64$).

Finally, we substitute the given numerical value for the small drop's potential ($V_{\text{small}}=2\text{ V}$):

A Thought Experiment

This problem is a gateway to deeper physical insights. While we calculated the potential, what happens to the electrostatic potential energy during this merging process?

The energy of a charged sphere is given by $U=\frac{1}{2}QV$. If you calculate the total initial energy of the $512$ drops and compare it to the final energy of the single big drop, you will find that the energy has actually increased.

Where does this extra energy come from? It comes from the mechanical work done by the surface tension of the mercury. As the drops merge, their total surface area decreases, releasing surface energy, which in turn does work against the electrostatic repulsion of the charges, packing them into a tighter configuration relative to their total amount. Physics is beautifully interconnected!