Animated Solution for Physics - Electrostatics: 27 similar drops of mercury are maintained at 10 V each. All these spherical drops combine into a single big drop. The potential energy of the bigger drop is ............ times that of a smaller drop.

Imagine you are in a lab, watching tiny, shimmering drops of liquid mercury. You have exactly $27$ of these identical drops. Because mercury is a metal, each drop acts as a tiny conducting sphere. Now, imagine pushing them all together until they coalesce into one single, massive drop.

This isn't just a cool visual; it's a classic physics scenario that tests your understanding of conservation laws and electrostatic properties. Let's break down exactly what happens to the potential energy during this transformation.

The Conservation Laws

When multiple drops combine, two fundamental quantities remain absolutely conserved: Volume and Charge.

First, let's look at the volume. The total volume of the $27$ small drops must equal the volume of the new, big drop. If a small drop has a radius $r$ and the big drop has a radius $R$, we can write:

By canceling out the common terms and taking the cube root of both sides, we find a beautiful, simple relationship between the radii:

So, the big drop is exactly three times wider than a small drop.

Next, we apply the conservation of charge. Charge is an additive property. If each small drop carries a charge $q$, the total charge $Q$ of the big drop is simply the sum of all the individual charges:

Electrostatic Potential Energy

Now, we need to find the potential energy. Here is a crucial pro-tip: Mercury is a conducting metal. This means any excess charge will immediately repel itself and spread out evenly over the surface of the drop. Electrostatically, a solid conducting sphere behaves exactly like a thin spherical shell.

The potential energy $U$ of a conducting sphere with charge $Q$ and radius $R$ is given by:

Let's write down the energy of a single small drop ($U_s$) for reference:

The Energy Ratio

We are ready to find the potential energy of the big drop ($U_B$). We just need to substitute our conserved quantities $Q=27q$ and $R=3r$ into the energy formula:

Let's carefully expand the numerator. The square of $27$ is $729$:

Notice how we isolated the expression for the energy of the small drop! Now, we just perform the final division:

The potential energy of the bigger drop is exactly $243$ times that of a single smaller drop.

Why stop at energy? This setup is a goldmine for variations. What if the question asked for the new potential $V$? Since $V=\frac{kQ}{R}$, the new potential would be $\frac{27}{3}=9$ times the original potential. What about capacitance? Since $C=4\pi {ϵ}_{0}R$, the new capacitance is simply $3$ times the original. Mastering this single framework unlocks the answers to a whole family of JEE problems!