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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)

Asked in NEET 2020 · Dielectric slab

• (1) 0.44×10⁻¹³ C² N⁻¹ m⁻²
• (2) 1.77×10⁻¹² C² N⁻¹ m⁻²
• (3) 0.44×10⁻¹⁰ C² N⁻¹ m⁻²
• (4) 5.00 C² N⁻¹ m⁻²

Step-by-step solution

Why the other options are wrong

• (1) No single slip gives this; it is a thousand times smaller than the correct 0.44×10⁻¹⁰, and a permittivity below ε₀ would make the capacitance fall, not rise.
• (2) Divides by K instead of multiplying: (8.85×10⁻¹²)/5=1.77×10⁻¹².
• (4) That is the dielectric constant K=5, a pure number, not the permittivity.

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