Animated Solution for Physics - Electrostatics: Four point charges, each of $+q$, are rigidly fixed at the four corners of a square planar soap film of side $a$. The surface tension of the soap film is $\gamma$. The system of charges and planar film are in equilibrium, and $a=k{\left[\frac{q^2}{\gamma }\right]}^{1/N}$, where $k$ is a constant. Then $N$ is

The problem of the square soap film with charges at its corners is a beautiful intersection of electrostatics and fluid mechanics. It challenges us to think beyond simple force balancing and step into the elegant world of energy minimization. Let's embark on this thrilling journey!

Imagine a square soap film of side $a$. At each of its four corners, a positive charge $+q$ is rigidly fixed. This entire system is in perfect equilibrium. But what forces are at play here?

On one hand, the four positive charges are fiercely repelling each other, trying to tear the square apart and expand it to infinity. On the other hand, the soap film acts like a stretched rubber sheet. Its surface tension $\gamma$ is desperately trying to pull the corners inward to minimize its surface area.

For the system to be in equilibrium, these two opposing tendencies must perfectly balance out. While we could try to calculate the exact force vectors on a single corner charge, dealing with the continuous pull of the soap film along the boundary is a mathematical nightmare. Instead, we will use a much more powerful tool: The Principle of Virtual Work.

In physics, any system in stable equilibrium will settle into a state where its total potential energy $U$ is at a minimum. The total energy here consists of two components: the electrostatic potential energy $U_e$ and the surface energy $U_s$.

Let's calculate the electrostatic energy first. We have four charges, which means there are $\frac{4\times 3}{2}=6$ unique pairs of interactions. Four of these pairs are along the sides of the square, separated by a distance $a$. The remaining two pairs are across the diagonals, separated by a distance $\sqrt{2}a$.

Summing these up, the total electrostatic energy is:

Next, we calculate the surface energy. Here lies a classic trap! A soap film is not a single surface; it is a thin layer of liquid sandwiched between two layers of soap molecules. Therefore, it has two free surfaces (top and bottom). The surface energy is the surface tension multiplied by the total area:

Now we have our master equation for the total energy:

To find the equilibrium state, we must minimize this energy. We do this by taking the derivative of $U$ with respect to the side length $a$ and setting it to zero:

Differentiating the expression, we get:

Rearranging the terms to solve for $a^3$:

Taking the cube root on both sides reveals the dependency of $a$ on the given parameters:

The problem states that $a=k{\left[\frac{q^2}{\gamma }\right]}^{1/N}$. By directly comparing our result with this expression, it is crystal clear that the exponent is $1/3$.

While the energy method is mathematically rigorous and deeply satisfying, JEE is also about speed. Could we have solved this in 10 seconds? Absolutely!

Enter Dimensional Analysis.

We know that the electrostatic force $F_e$ must balance the surface tension force $F_s$. The electrostatic force scales as:

The surface tension force scales as:

Equating their dimensions:

Instantly, without writing a single complex equation, we see that the power must be $1/3$, giving us $N=3$. This is the hallmark of a true physics ninja!