Question Detail
Question
The plates of a parallel plate air capacitor are charged to 100 V. A 2mm plate is inserted between the plates of the capacitor. To maintain the same potential difference, the distance between the capacitor plates is increased by 1.6mm. The dielectric constant of the plate is
Correct Answer
Option B is the correct answer.
Detailed Explanation
Let the initial separation between the plates be d0. The initial capacitance is C0 = ε0A / d0.When a dielectric plate of thickness t is inserted, the effective separation becomes deff = (d0 - t) + t/K, where K is the dielectric constant.The capacitance with the dielectric is C = ε0A / deff.The problem states that a plate of thickness t = 2 mm is inserted. To maintain the same potential difference (V), the plates are moved apart by Δd = 1.6 mm.This means the final plate separation is df = d0 + Δd.For the potential difference to remain the same (assuming it's connected to a battery, or the charge is adjusted accordingly to keep V constant), the equivalent effective distance with the dielectric must be equal to the original air gap d0.So, the new effective separation must be d0.d0 = (df - t) + t/Kd0 = (d0 + Δd - t) + t/KSubtract d0 from both sides:0 = Δd - t + t/Kt - Δd = t/KK = t / (t - Δd)Substitute the given values:t = 2 mmΔd = 1.6 mmK = 2 mm / (2 mm - 1.6 mm)K = 2 mm / 0.4 mmK = 5The dielectric constant of the plate is 5.
Hint
For a parallel plate capacitor, when a dielectric slab is inserted and the potential difference is maintained, the original effective air gap must be equivalent to the new effective gap considering the dielectric and any change in physical separation. The effective thickness of the dielectric is t/K.