JEE Challenger
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PhyHardAdv 2020 • Paper 2

Charged Spheres Equilibrium in Dielectric Medium

Consider two identical non-conducting solid spheres of equal mass and charge suspended in air from a single point using two non-conducting massless strings of the same length. At equilibrium, the angle subtended between the strings is α\alpha. The spheres are subsequently immersed in a dielectric liquid of mass density 800 kg m−3800\text{ kg m}^{-3} and dielectric constant 2121. If the angle between the strings is unchanged after immersion, which of the following statements is/are correct?

Options

A

The electric force between the spheres remains unchanged

B

The electric force between the spheres decreases

Correct
C

The mass density of the spheres is 840 kg m−3840\text{ kg m}^{-3}

Correct
D

The tension in the strings holding the spheres remains unchanged

Topics & Concepts

Electric Charges and FieldsCoulomb's LawDielectrics

Step-by-Step Solution

To determine which of the statements are correct, we analyze the conditions for equilibrium in both air and the dielectric liquid.

1. Equilibrium in Air

Let each sphere have mass mm, volume VV, mass density ρ\rho, and charge qq. When suspended in air, the forces acting on each sphere are:

  1. Downward gravitational force: W=mg=ρVgW = mg = \rho V g
  2. Horizontal electrostatic repulsive force: Fe=14πε0q2r2F_e = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}
  3. Tension along the string: TT

If each string makes an angle θ=α2\theta = \frac{\alpha}{2} with the vertical, the equations of equilibrium are: Tcos⁡(α2)=mgT \cos\left(\frac{\alpha}{2}\right) = mg Tsin⁡(α2)=FeT \sin\left(\frac{\alpha}{2}\right) = F_e

Dividing the two equations yields: tan⁡(α2)=Femg— (1)\tan\left(\frac{\alpha}{2}\right) = \frac{F_e}{mg} \quad \text{--- (1)}

2. Equilibrium in the Dielectric Liquid

When the spheres are immersed in the dielectric liquid of density σ=800 kg m−3\sigma = 800\text{ kg m}^{-3} and dielectric constant K=21K = 21:

  • The electrostatic force between the spheres is reduced due to the permittivity of the medium: Fe′=FeK=Fe21F_e' = \frac{F_e}{K} = \frac{F_e}{21} Since K>1K > 1, the electric force between the spheres reduces. Therefore, Option (B) is correct, and Option (A) is incorrect.

  • The effective downward force acting on each sphere becomes: Weff=mg−B=ρVg−σVg=mg(1−σρ)W_{\text{eff}} = mg - B = \rho V g - \sigma V g = mg \left(1 - \frac{\sigma}{\rho}\right) where B=σVgB = \sigma V g is the buoyant force.

  • Let T′T' be the new tension in each string. Since the angle α\alpha remains unchanged, the distance rr between the spheres is also unchanged. The equations of equilibrium in the liquid are: T′cos⁡(α2)=mg(1−σρ)T' \cos\left(\frac{\alpha}{2}\right) = mg \left(1 - \frac{\sigma}{\rho}\right) T′sin⁡(α2)=Fe′=FeKT' \sin\left(\frac{\alpha}{2}\right) = F_e' = \frac{F_e}{K}

Dividing these equations gives: tan⁡(α2)=Fe/Kmg(1−σρ)— (2)\tan\left(\frac{\alpha}{2}\right) = \frac{F_e / K}{mg \left(1 - \frac{\sigma}{\rho}\right)} \quad \text{--- (2)}

3. Determining the Density of the Spheres

Equating equation (1) and equation (2): Femg=FeKmg(1−σρ)\frac{F_e}{mg} = \frac{F_e}{K mg \left(1 - \frac{\sigma}{\rho}\right)}

Simplifying: 1=1K(1−σρ)1 = \frac{1}{K \left(1 - \frac{\sigma}{\rho}\right)}

K(1−σρ)=1  ⟹  1−σρ=1KK \left(1 - \frac{\sigma}{\rho}\right) = 1 \implies 1 - \frac{\sigma}{\rho} = \frac{1}{K}

σρ=1−1K=K−1K\frac{\sigma}{\rho} = 1 - \frac{1}{K} = \frac{K - 1}{K}

ρ=σ(KK−1)\rho = \sigma \left(\frac{K}{K - 1}\right)

Substituting the given values σ=800 kg m−3\sigma = 800\text{ kg m}^{-3} and K=21K = 21: ρ=800×(2121−1)=800×2120=840 kg m−3\rho = 800 \times \left(\frac{21}{21 - 1}\right) = 800 \times \frac{21}{20} = 840\text{ kg m}^{-3}

Therefore, Option (C) is correct.

4. Change in Tension

Comparing the tensions in air (TT) and in the liquid (T′T'): T=mgcos⁡(α/2)T = \frac{mg}{\cos(\alpha/2)}

T′=mg(1−σρ)cos⁡(α/2)=mgKcos⁡(α/2)=TK=T21T' = \frac{mg\left(1 - \frac{\sigma}{\rho}\right)}{\cos(\alpha/2)} = \frac{mg}{K \cos(\alpha/2)} = \frac{T}{K} = \frac{T}{21}

Since T′=T21T' = \frac{T}{21}, the tension in the strings decreases and does not remain unchanged. Therefore, Option (D) is incorrect.

Conclusion

  • Correct Options: B and C
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