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[(q^2)/(gamma)]^(1//N)`, where 'K' is a constant. Then N is

Explanation

To solve the problem, we need to analyze the forces acting on the charges and the effect of surface tension on the soap film. Here's a step-by-step breakdown of the solution:

### Step 1: Understanding the Setup
We have four point charges, each of charge \( +q \), fixed at the corners of a square soap film of side length \( a \). The surface tension of the soap film is denoted by \( \gamma \).

### Step 2: Calculate the Force Due to Surface Tension
The surface tension force acts along the edges of the soap film. Since there are two surfaces (top and bottom) and each side has a length \( a \), the total force due to surface tension is:
\[ F_{\text{tension}} = 2 \gamma a \]

### Step 3: Calculate the Electrostatic Forces
Each charge experiences forces due to the other three charges. The force \( F \) between two charges \( q \) separated by a distance \( r \) is given by Coulomb's law:
\[ F = k \frac{q^2}{r^2} \]
where \( k \) is Coulomb's constant.

For a charge at a corner of the square, the distances to the adjacent charges are \( a \) and to the diagonal charge is \( \sqrt{2}a \).

### Step 4: Calculate the Net Electrostatic Force on One Charge
The force due to the two adjacent charges (each at distance \( a \)) is:
\[ F_{\text{adj}} = 2 \left( k \frac{q^2}{a^2} \right) \]
The force due to the diagonal charge (at distance \( \sqrt{2}a \)) is:
\[ F_{\text{diag}} = k \frac{q^2}{(\sqrt{2}a)^2} = k \frac{q^2}{2a^2} \]

### Step 5: Resolve Forces
The forces from the two adjacent charges act at \( 90^\circ \) to each other. The resultant force from these two can be calculated using the Pythagorean theorem:
\[ F_{\text{resultant, adj}} = \sqrt{(F_{\text{adj}})^2 + (F_{\text{adj}})^2} = \sqrt{2 \left( k \frac{q^2}{a^2} \right)^2} = k \frac{q^2}{a^2} \sqrt{2} \]

The total force acting on the charge is then:
\[ F_{\text{total}} = F_{\text{resultant, adj}} + F_{\text{diag}} = k \frac{q^2}{a^2} \sqrt{2} + k \frac{q^2}{2a^2} \]

### Step 6: Equilibrium Condition
For the system to be in equilibrium, the total electrostatic force must balance the surface tension force:
\[ F_{\text{total}} = F_{\text{tension}} \]
Substituting the expressions we derived:
\[ k \frac{q^2}{a^2} \sqrt{2} + k \frac{q^2}{2a^2} = 2 \gamma a \]

### Step 7: Rearranging for \( a \)
We can factor out \( k \frac{q^2}{a^2} \):
\[ k \frac{q^2}{a^2} \left( \sqrt{2} + \frac{1}{2} \right) = 2 \gamma a \]

### Step 8: Solve for \( a \)
Rearranging gives:
\[ a^3 = k \frac{q^2}{\gamma} \left( \sqrt{2} + \frac{1}{2} \right) \]
This implies:
\[ a = k \left( \frac{q^2}{\gamma} \right)^{1/3} \]

### Step 9: Identify \( N \)
From the given equation \( a = k \left( \frac{q^2}{\gamma} \right)^{1/N} \), we see that \( N = 3 \).

### Final Answer
Thus, the value of \( N \) is:
\[ \boxed{3} \]