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 \over \gamma ]^1/N , where k is a constant. Then N is \underline\hspace2cm .

Electrostatics PYQ — JEE Advanced 2011 · Shift 1 · Q67 with Solution

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Correct answer: 3

Forces trying to expand the square

Each corner has a charge $+q$ . Consider one side of the square.

If the side length increases by a small amount $da$ , the electrostatic potential energy changes. At equilibrium, the outward electrostatic tendency is balanced by the inward pull due to surface tension.

Electrostatic potential energy of the four charges

There are:

$4$ pairs along the sides, each separated by $a$
$2$ pairs along the diagonals, each separated by $a\sqrt{2}$

So,

U_e = 4\cdot \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a} + 2\cdot \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a\sqrt{2}}

U_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{a}(4+\sqrt{2})

Let

K = \frac{1}{4\pi\varepsilon_0}(4+\sqrt{2})

Then

U_e = K\frac{q^2}{a}

Surface energy of soap film

A soap film has two surfaces, so the surface energy is

U_s = 2\gamma \times \text{area} = 2\gamma a^2

Equilibrium condition

Total energy:

U(a)=K\frac{q^2}{a}+2\gamma a^2

At equilibrium,

\frac{dU}{da}=0

So,

- K\frac{q^2}{a^2} + 4\gamma a = 0

4\gamma a = K\frac{q^2}{a^2}

4\gamma a^3 = K q^2

a^3 = \frac{K}{4}\frac{q^2}{\gamma}

Hence,

a = k\left(\frac{q^2}{\gamma}\right)^{1/3}

where $k$ is a constant.

Therefore,

N=3

Comparison with stored answer

Stored correct answer: $3$

Our derived answer matches it.

More from Electrostatics

View written solutionFree

Correct answer: 3

Forces trying to expand the square

Each corner has a charge $+q$ . Consider one side of the square.

Electrostatic potential energy of the four charges

There are:

$4$ pairs along the sides, each separated by $a$
$2$ pairs along the diagonals, each separated by $a\sqrt{2}$

So,

U_e = 4\cdot \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a} + 2\cdot \frac{1}{4\pi\varepsilon_0}\frac{q^2}{a\sqrt{2}}

U_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{a}(4+\sqrt{2})

Let

K = \frac{1}{4\pi\varepsilon_0}(4+\sqrt{2})

Then

U_e = K\frac{q^2}{a}

Surface energy of soap film

A soap film has two surfaces, so the surface energy is

U_s = 2\gamma \times \text{area} = 2\gamma a^2

Equilibrium condition

Total energy:

U(a)=K\frac{q^2}{a}+2\gamma a^2

At equilibrium,

\frac{dU}{da}=0

So,

- K\frac{q^2}{a^2} + 4\gamma a = 0

4\gamma a = K\frac{q^2}{a^2}

4\gamma a^3 = K q^2

a^3 = \frac{K}{4}\frac{q^2}{\gamma}

Hence,

a = k\left(\frac{q^2}{\gamma}\right)^{1/3}

where $k$ is a constant.

Therefore,

N=3

Comparison with stored answer

Stored correct answer: $3$

Our derived answer matches it.

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