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: Identify the two forces.
The gravitational force between the two electrons is \( F_1 = G\dfrac{m_e^2}{r^2} \), and the electrostatic (Coulomb) force is \( F_2 = \dfrac{1}{4\pi\varepsilon_0 K}\cdot\dfrac{e^2}{r^2} \), where \( K \) is the dielectric constant of the medium.

Step 2: Effect of the medium on gravitation.
Gravitational force depends only on the masses and their separation. It does not depend on the electrical properties of the medium. So changing \( K \) has no effect on \( F_1 \); it remains unchanged.

Step 3: Effect of the medium on the electrostatic force.
The Coulomb force in a medium is \( F_2 = \dfrac{F_2^{air}}{K} \). As the dielectric constant becomes infinite, \( K \to \infty \), so \[ F_2 = \frac{F_2^{air}}{\infty} \to 0. \]
Step 4: Conclusion.
\( F_1 \) stays the same and \( F_2 \) drops to zero, which matches option (ii). Option (i) is wrong because the force is divided by \( K \), not halved (halving needs \( K=2 \)). Options (iii) and (iv) are wrong because gravitation is never affected by \( K \) and \( F_2 \) cannot become infinite.

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