JEE Main 2024 Physics question
A capacitor has air as dielectric medium and two conducting plates of area 12 cm² and they are 0.6 cm apart. When a slab of dielectric having area 12 cm² and 0.6 cm thickness is inserted between the plates, one of the conducting plates has to be moved by 0.2 cm to keep the capacitance same as in previous case. The dielectric constant of the slab is: (Given ε₀ = 8.834 × 10⁻¹² F/m)
Answer: (A)
Solution
The plate is moved away by 0.2 cm, so the gap is 0.8 cm: ε₀A/(0.2 + 0.6/K) = ε₀A/0.6 ⇒ 0.2 + 0.6/K = 0.6 ⇒ K = 1.5.