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(\left[\right. \frac{q^{2}}{\gamma} \left]\right.\right)^{1 / N}, where 'k' is a constant. Then N is

Engineering

Physics

Equilibrium

Surface Tension and Surface Energy

Charge and Coulombs Law

Question

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 $γ$ . The system of charges and planar film are in equilibrium, and $a = k {[\frac{q^{2}}{γ}]}^{1 / N}$ , where 'k' is a constant. Then N is

$F_{AC} = \frac{q^{2}}{8 {πε}_{0} a^{2}}$

$F_{AD} = F_{AB} = \frac{q^{2}}{4 {πε}_{0} a^{2}}$

$F_{R} = \frac{q^{2}}{4 {πε}_{0} a^{2}} (2 \cos 45 ° + \frac{1}{2})$

$= r (2) (BD) = 2 r (\sqrt{2} a)$

$\Rightarrow a^{3} = \frac{q^{2} (\sqrt{2} + \frac{1}{2})}{8 \sqrt{2} {πε}_{0} r}$

$a = k {(\frac{q^{2}}{r})}^{1 / 3} \Rightarrow N = 3$

Where $k = {(\frac{\sqrt{2} + \frac{1}{2}}{8 \sqrt{2} π})}^{1 / 3}$

This problem involves finding the equilibrium condition for a square soap film with fixed point charges at its corners. The surface tension of the soap film tries to contract the film, while the electrostatic repulsion between the charges tries to expand it. At equilibrium, these forces balance each other.

Step 1: Identify the Forces

The soap film has a surface tension γ, which acts to minimize its surface area. For a square film of side 'a', the contractile force due to surface tension can be thought of as acting along the perimeter. However, a more precise approach is to consider the energy.

The total energy of the system has two parts:

Surface energy due to surface tension: $U s = γ \times A = γ a^{2}$ (since area A = a²).
Electrostatic potential energy due to the four charges.

Step 2: Calculate the Electrostatic Potential Energy

There are four identical charges, each of magnitude +q, fixed at the corners of the square. The electrostatic potential energy (Ue) for a system of point charges is given by the sum of the potential energies for each pair:

U e = \frac{1}{4 π ε 0} \sum_{i < j} \frac{q i q j}{r i j}

For our square:

There are 4 sides: each of length 'a'. Number of pairs at this distance: 4.
There are 2 diagonals: each of length $\sqrt{2} a$ . Number of pairs at this distance: 2.

Therefore, the total potential energy is:

U e = \frac{1}{4 π ε 0} [\frac{4 q^{2}}{a} + \frac{2 q^{2}}{\sqrt{2} a}] = \frac{1}{4 π ε 0} \frac{q^{2}}{a} [4 + \sqrt{2}]

Let's define a constant $C = \frac{1}{4 π ε 0} [4 + \sqrt{2}]$ for simplicity. So, $U e = C \frac{q^{2}}{a}$ .

Step 3: Total Energy and Equilibrium Condition

The total energy U of the system is the sum of surface energy and electrostatic energy:

U = U s + U e = γ a^{2} + C \frac{q^{2}}{a}

For the system to be in equilibrium, this total energy must be at a minimum. Therefore, we differentiate U with respect to 'a' and set the derivative to zero:

\frac{d U}{d a} = 2 γ a - C \frac{q^{2}}{a^{2}} = 0

Solving for equilibrium:

2 γ a = C \frac{q^{2}}{a^{2}}

2 γ a^{3} = C q^{2}

a^{3} = \frac{C}{2 γ} q^{2}

Now, take the cube root on both sides to solve for 'a':

a = {(\frac{C}{2 γ})}^{1 / 3} q^{2 / 3}

We can combine the constant terms into a single constant k':

a = k^{'} {(\frac{q^{2}}{γ})}^{1 / 3}

Comparing this with the given form $a = k {(\frac{q^{2}}{γ})}^{1 / N}$ , we find that the exponent N is 3.

Final Answer: N = 3

Related Topics and Formulae:

Surface Tension: The property of a liquid surface that allows it to resist an external force, due to the cohesive nature of its molecules. The surface energy is given by U = γ × A, where γ is the surface tension and A is the area.

Electrostatic Potential Energy: For a system of point charges, the total potential energy is the work done to assemble the charges from infinity. It is calculated by summing the potential energy for each unique pair: $U = \frac{1}{4 π ε 0} \sum_{i < j} \frac{q i q j}{r i j}$ .

Equilibrium Condition: A system is in stable equilibrium when its total potential energy is at a local minimum. This is found by setting the first derivative of the potential energy function with respect to the coordinate to zero.

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