Two insulated charged copper spheres A and B each having charge of ...

Question

Two insulated charged copper spheres A and B each having charge of 6.5 × 10⁻⁷ C are separated by a distance 50 cm. If they are placed in water of dielectric constant 80, then find the electrostatic force of repulsion between them.

Accepted Answer

Given:
• q_A = q_B = 6.5 × 10⁻⁷ C
• r = 50 cm = 0.5 m
• Dielectric constant K = 80
• k = 9 × 10⁹ N·m²/C²

Step 1: Calculate force in vacuum/air
F_vac = k·q_A·q_B/r²

F_vac = (9 × 10⁹) × (6.5 × 10⁻⁷)² / (0.5)²

F_vac = (9 × 10⁹) × (4.225 × 10⁻¹³) / 0.25

F_vac = 3.8025 × 10⁻³ / 0.25

F_vac = 1.521 × 10⁻² N

Step 2: Force in water (dielectric medium)
F_water = F_vac/K

F_water = 1.521 × 10⁻² / 80

F_water ≈ 1.9 × 10⁻⁴ N

Final Answer:
The electrostatic force of repulsion between the spheres in water = 1.9 × 10⁻⁴ N

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