Question

Two electrons in air are at a distance of 1 m. Gravitational and electrostatic forces acting between them are \( F_1 \) and \( F_2 \) respectively. If the dielectric constant of the medium between them becomes infinite, then:

• A. \( F_1 \) remains unchanged and \( F_2 \) becomes halved.
• B. \( F_1 \) remains unchanged but \( F_2 \) becomes zero.
• C. \( F_1 = 0 \) and \( F_2 = \infty \).
• D. Both \( F_1 \) and \( F_2 \) will remain unchanged.

Hint:

Coulomb force in a medium is divided by the dielectric constant \( K \), while gravitation ignores \( K \) entirely. Let \( K\to\infty \).

Answer 1

The Correct Option is B

Step 1: Write the ratio of the forces to a medium factor.
Let the forces in air be \(F_1\) (gravitational) and \(F_2^{air}\) (electrostatic). Only the electrostatic force carries the medium constant \(K\) in its denominator: \(F_2 = F_2^{air}/K\).

Step 2: Reason physically about gravity.
Gravity is a mass-mass interaction. A dielectric between the charges polarises electrically but does nothing to the masses, so the gravitational pull \(F_1\) is completely independent of \(K\) and stays fixed at \(G m_e^2/r^2\).

Step 3: Push the dielectric constant to infinity.
A medium with \(K \to \infty\) behaves like a perfect conductor that fully screens the electric field between the charges. Substituting: \(F_2 = F_2^{air}/\infty = 0\). The electric force is entirely cancelled.

Step 4: Match to an option.
Gravitational force unchanged, electrostatic force reduced to zero. That is exactly statement (ii).

\[\boxed{F_1 = \text{const},\ F_2 \to 0}\]