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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · 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

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

Answer: 3

Solution

The electrostatic force trying to expand the film is proportional to

\frac{q ^{2}}{a ^{2}}

(Coulomb repulsion).

The surface tension force trying to contract the film is proportional to

γ a

(force per unit length times perimeter).

At equilibrium:

\frac{q ^{2}}{a ^{2}} \propto γ a

q^{2} \propto γ a^{3}

a^{3} \propto \frac{q ^{2}}{γ}

a \propto (\frac{q ^{2}}{γ})^{1/3}

Therefore,

N = 3

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Integer answer

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