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Class 12 › Physics › Electric Charge and Field
CBSE2024Class Class 12 · Physics2 Marks · Short✅ Verified
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.
✅ Answer & Solution
<b>Given:</b><br>• q<sub>A</sub> = q<sub>B</sub> = 6.5 × 10⁻⁷ C<br>• r = 50 cm = 0.5 m<br>• Dielectric constant K = 80<br>• k = 9 × 10⁹ N·m²/C²<br><br><b>Step 1: Calculate force in vacuum/air</b><br>F<sub>vac</sub> = k·q<sub>A</sub>·q<sub>B</sub>/r²<br><br>F<sub>vac</sub> = (9 × 10⁹) × (6.5 × 10⁻⁷)² / (0.5)²<br><br>F<sub>vac</sub> = (9 × 10⁹) × (4.225 × 10⁻¹³) / 0.25<br><br>F<sub>vac</sub> = 3.8025 × 10⁻³ / 0.25<br><br><b>F<sub>vac</sub> = 1.521 × 10⁻² N</b><br><br><b>Step 2: Force in water (dielectric medium)</b><br>F<sub>water</sub> = F<sub>vac</sub>/K<br><br>F<sub>water</sub> = 1.521 × 10⁻² / 80<br><br><b>F<sub>water</sub> ≈ 1.9 × 10⁻⁴ N</b><br><br><b>Final Answer:</b><br>The electrostatic force of repulsion between the spheres in water = <b>1.9 × 10⁻⁴ N</b>
