Multi-Concept Equilibrium: Dielectrics and Buoyancy

In JEE Advanced, physical phenomena rarely occur in isolation. A standard electrostatics problem in a vacuum becomes significantly more complex when moved into a physical medium. The student must account for both the electrical properties of the medium (permittivity) and its mechanical properties (density and buoyancy).

A classic framework for this is the suspended pith ball experiment.

The Setup in Air (Vacuum)

Consider two identical small spheres, each of mass $m$, volume $V$, density $\rho$, and carrying a charge $+q$. They are suspended from a common rigid support by two massless, unstretchable strings of length $L$. Due to electrostatic repulsion, they separate and hang in equilibrium at an angle $\theta$ with the vertical.

Let the distance between the spheres be $r$. In this state, three forces act on each sphere:

1. Weight ($W$): Acting downwards, $W = mg = V\rho g$.
2. Electrostatic Force ($F_e$): Acting horizontally outward, $F_e = \frac{1}{4\pi\epsilon_0}\frac{q^2}{r^2}$.
3. Tension ($T$): Acting along the string.

At equilibrium, we resolve the Tension vector:

• Vertical balance: $T \cos\theta = mg$
• Horizontal balance: $T \sin\theta = F_e$

Dividing the horizontal equation by the vertical equation gives the fundamental equilibrium condition in air:

$\tan\theta = \frac{F_e}{mg}$

The System Submerged in a Dielectric Liquid

Now, the entire system is submerged in an insulating liquid of density $\sigma$ and dielectric constant $K$ (also known as relative permittivity, $\epsilon_r$).

Two massive changes occur simultaneously:

1. The Electrical Shift (Dielectric Weakening)

The presence of the dielectric medium polarizes and creates an opposing internal electric field, weakening the net electrostatic force between the charges. The new force $F_e'$ becomes:

$F_e' = \frac{F_e}{K}$

2. The Mechanical Shift (Buoyancy)

The fluid exerts an upward buoyant force ($F_B$) on the spheres according to Archimedes' Principle. The apparent weight of the sphere ($W_{app}$) decreases:

$F_B = V \sigma g$

$W_{app} = mg - F_B = V\rho g - V\sigma g$

We can factor out $V\rho g$ (which is $mg$) to express apparent weight purely in terms of densities:

$W_{app} = mg \left(1 - \frac{\sigma}{\rho}\right)$

The New Equilibrium

Let the new angle with the vertical be $\theta'$. The new force balance equations are $T' \sin\theta' = F_e'$ and $T' \cos\theta' = W_{app}$.

Dividing these gives the new equilibrium condition:

$\tan\theta' = \frac{F_e'}{W_{app}} = \frac{F_e / K}{mg (1 - \sigma/\rho)}$

The Classic JEE Constraint: "Angle Remains Unchanged"

The most common iteration of this problem in competitive exams states: "When the system is submerged, the angle of divergence remains the same. Find the dielectric constant $K$."

If $\theta' = \theta$, then $\tan\theta' = \tan\theta$. We equate our two conditions:

$\frac{F_e}{mg} = \frac{F_e}{K \cdot mg \left(1 - \frac{\sigma}{\rho}\right)}$

Canceling $F_e$ and $mg$ from both sides:

$1 = \frac{1}{K \left(1 - \frac{\sigma}{\rho}\right)}$

Solving for the dielectric constant $K$:

$K = \frac{1}{1 - \frac{\sigma}{\rho}} = \frac{\rho}{\rho - \sigma}$

Key Takeaway for Students: This elegant final result ($K = \frac{\rho}{\rho - \sigma}$) shows that if the geometry of the system doesn't change upon submersion, the dielectric constant of the liquid is strictly a function of the mass density of the sphere and the mass density of the liquid. The actual magnitude of the charge $q$ and the length of the string $L$ become completely irrelevant!