Irodov 3.87: Buoyancy Pays for the Weaker Repulsion – Amit & Deepali Sinha Foundation

Problem Irodov 3.87

Two small identical balls carry charges of the same sign and hang from a common point on insulating threads of equal length. When the space around them is filled with kerosene (permittivity $\varepsilon = 2.0$ , density $\rho_0 = 0.80$ g/cm $^3$ ), the angle between the threads does not change. Find the density of the material the balls are made of.

(a) In air: the two charged balls hang apart, the threads diverging at the angle 2θ. — (a) In air: the two charged balls hang apart, the threads diverging at the angle $2\theta$ .

(b) Immersed in kerosene ε: the same angle 2θ. — (b) Immersed in kerosene $\varepsilon$ : the same angle $2\theta$ .

Given

kerosene, $\varepsilon = 2.0$ (table value)
kerosene density $\rho_0 = 0.80$ g/cm $^3$ (table value)
the same angle between the threads in air and in kerosene

Find

the density

\rho

of the balls

Solution

Step 1Equilibrium in air

Let each ball have mass $m$ , volume $V$ and density $\rho$ , and let $\theta$ be the angle of each thread from the vertical. A ball is held by the tension along its thread, its weight, and the horizontal repulsion $F$ of the other ball. Balancing the horizontal and vertical components and dividing one by the other:

\tan\theta = \frac{F}{mg} \tag{1}

Step 2Equilibrium in kerosene

The angle is the same, so the balls sit at the same separation. A liquid dielectric filling all the space round two small charges weakens their interaction by the factor $\varepsilon$ , so the repulsion becomes $F' = F/\varepsilon$ . The permittivity and the density of kerosene are tabulated material constants, $\varepsilon = 2.0$ and $\rho_0 = 0.80$ g/cm $^3$ . The liquid also pushes each ball up with the buoyant force $F_{\mathrm{A}} = \rho_0Vg$ .

The right ball in kerosene: tension T along the thread, weight m g, buoyant force F_A up and the weakened repulsion F', in proportion for the drawn angle. — The right ball in kerosene: tension $\vect T$ along the thread, weight $m\vect g$ , buoyant force $\vect F_{\mathrm{A}}$ up and the weakened repulsion $\vect F'$ , in proportion for the drawn angle.

The vertical balance now has the weight less the buoyant force, which with $m = \rho V$ is a reduced weight; the same division then gives the angle:

\begin{align*} mg - \rho_0Vg &= mg\left(1 - \frac{\rho_0}{\rho}\right) \\ \tan\theta &= \frac{F/\varepsilon}{mg\,(1 - \rho_0/\rho)} \tag{2}\end{align*}

Step 3The density that keeps the angle

Setting (1) equal to (2), the charge, the mass and the angle all cancel:

\begin{aligned} \frac{F}{mg} &= \frac{F}{\varepsilon\, mg\,(1 - \rho_0/\rho)}\\ \varepsilon\left(1 - \frac{\rho_0}{\rho}\right) &= 1 \end{aligned}

Solving for $\rho$ and putting in kerosene’s values:

\rho = \frac{2.0 \times 0.80\text{ g/cm}^3}{2.0 - 1}

\rho = \frac{\varepsilon\rho_0}{\varepsilon - 1} = 1.6\text{ g/cm}^3

(3)

Check

The balls are denser than kerosene, as they must be to hang below the support at all: $1.6 > 0.80$ . In the limit $\varepsilon \to 1$ the liquid would not weaken the repulsion, and only a ball with no buoyancy correction, $\rho \to \infty$ , could keep the angle; for very large $\varepsilon$ the repulsion vanishes and $\rho \to \rho_0$ , a neutrally buoyant ball that needs no repulsion to stay where it is.

Remark

The liquid weakens both things that set the angle: the repulsion by $1/\varepsilon$ and the effective weight by $1 - \rho_0/\rho$ . The angle survives exactly when those factors agree, so an unchanged angle fixes the density from the liquid’s $\varepsilon$ and $\rho_0$ alone, independent of the charges and the thread length.