Equilibrium of Soap Film with Four Point Charges

To find the value of $N$, we analyze the total potential energy of the system at equilibrium.

1. Electrostatic Potential Energy

The system consists of four point charges, each of $+q$, placed at the four corners of a square of side $a$. The total electrostatic potential energy $U_e$ of the four charges is given by the sum of interactions between all pairs of charges:
$U_e = \frac{1}{4\pi\varepsilon_0} \left[ 4 \left( \frac{q^2}{a} \right) + 2 \left( \frac{q^2}{\sqrt{2}a} \right) \right] = \frac{q^2}{4\pi\varepsilon_0 a} \left( 4 + \sqrt{2} \right)$

2. Surface Energy of the Soap Film

A planar soap film has two exposed surfaces. The total surface area of both sides of the square film of side $a$ is $2a^2$. The surface potential energy $U_s$ associated with the surface tension $\gamma$ is:
$U_s = 2 \gamma a^2$

3. Equilibrium Condition

The total potential energy of the system is:
$U = U_e + U_s = \frac{q^2 (4 + \sqrt{2})}{4\pi\varepsilon_0 a} + 2\gamma a^2$

For the system to be in equilibrium, the total potential energy must be minimized with respect to the side length $a$:
$\frac{dU}{da} = 0$

Differentiating $U$ with respect to $a$:
$-\frac{q^2 (4 + \sqrt{2})}{4\pi\varepsilon_0 a^2} + 4\gamma a = 0$

Rearranging the terms:
$4\gamma a = \frac{q^2 (4 + \sqrt{2})}{4\pi\varepsilon_0 a^2}$

$a^3 = \left[ \frac{4 + \sqrt{2}}{16\pi\varepsilon_0} \right] \frac{q^2}{\gamma}$

Taking the cube root on both sides:
$a = k \left[ \frac{q^2}{\gamma} \right]^{1/3}$

where $k = \left[ \frac{4 + \sqrt{2}}{16\pi\varepsilon_0} \right]^{1/3}$ is a constant.

4. Conclusion

Comparing this result with the given equation:
$a = k \left[ \frac{q^2}{\gamma} \right]^{1/N}$

We find:
$N = 3$