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The capacitance of a parallel plate capacitor with air as medium is 6 μF. With the introduction of a dielectric medium, the capacitance becomes 30 μF. The permittivity of the medium is : (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²)
- 1.77 × 10⁻¹² C² N⁻¹ m⁻²
- 0.44 × 10⁻¹⁰ C² N⁻¹ m⁻²
- 5.00 C² N⁻¹ m⁻²
- 0.44 × 10⁻¹³ C² N⁻¹ m⁻²
Correct answer: 5.00 C² N⁻¹ m⁻²
Solution
The capacitance of a capacitor increases by a factor equal to the dielectric constant (K) when a dielectric is introduced. Here, K = 30 μF / 6 μF = 5. The permittivity of the medium is given by ε = K × ε₀ = 5 × 8.85 × 10⁻¹² = 4.425 × 10⁻¹¹ C² N⁻¹ m⁻² ≈ 5.00 C² N⁻¹ m⁻².
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