The capacitance of a parallel plate capacitor with air as medium is 6 μ F. With the introduction of a dielectric medium, the capacitance becomes 30 μ F. The permittivity of the medium is: (ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²) (a) 1.77 × 10⁻¹² C² N⁻¹ m⁻² (b) 0.44 × 10⁻¹⁰ C² N⁻¹ m⁻² (c) 5.00 C² N⁻¹ m⁻² (d) 0.44 × 10⁻¹³ C² N⁻¹ m⁻²

Answer: (b). K = C/C₀ = 30/6 = 5. ε = Kε₀ = 5 × 8.85 × 10⁻¹² = 4.425 × 10⁻¹¹ ≈ 0.44 × 10⁻¹⁰ C²N⁻¹m⁻². So the answer is (b).

The capacitance of a parallel plate capacitor with air as medium is 6… | NEET 2020 (Phase 1)

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Dielectrics

The capacitance of a parallel plate capacitor with air as medium is 6 $\mu$ F. With the introduction of a dielectric medium, the capacitance becomes 30 $\mu$ F. The permittivity of the medium is: $\left(\varepsilon_0 = 8.85 \times 10^{-12}\ C^2 N^{-1} m^{-2}\right)$

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Answer: (b)

$K = \frac{C}{C_0} = \frac{30}{6} = 5$ .
$\varepsilon = K\varepsilon_0 = 5 \times 8.85 \times 10^{-12} = 4.425 \times 10^{-11} \approx 0.44 \times 10^{-10}\ C^2N^{-1}m^{-2}$ .
So the answer is (b).

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