Question Detail
Question
A parallel plate capacitor has an area of 2000 cm², and the plates are 1 cm apart. The original potential difference between them, V₀, is 3000 volts, and it decreases to 1000 volts when a sheet of dielectric is inserted between the plates. Compute the induced charge Qind on each face of the dielectric.
Correct Answer
Option D is the correct answer.
Detailed Explanation
Given:Area A = 2000 cm² = 2000 × 10⁻⁴ m² = 0.2 m²Distance d = 1 cm = 0.01 mOriginal potential difference V₀ = 3000 VPotential difference with dielectric V = 1000 VPermittivity of free space ε₀ = 8.854 × 10⁻¹² F/mFirst, calculate the dielectric constant (K). When a dielectric is inserted into a capacitor disconnected from a voltage source, the charge remains constant, but the voltage decreases. The relationship is V = V₀/K, so K = V₀/V = 3000 V / 1000 V = 3.Next, calculate the original charge (Q) on the capacitor plates:C₀ = ε₀A/d = (8.854 × 10⁻¹² F/m) * (0.2 m²) / (0.01 m) = 1.7708 × 10⁻¹⁰ FQ = C₀V₀ = (1.7708 × 10⁻¹⁰ F) * (3000 V) = 5.3124 × 10⁻⁷ CFinally, calculate the induced charge (Q_ind) on the dielectric faces:Q_ind = Q * (1 - 1/K) = (5.3124 × 10⁻⁷ C) * (1 - 1/3)Q_ind = (5.3124 × 10⁻⁷ C) * (2/3) ≈ 3.5416 × 10⁻⁷ CTo match the options, convert to 10⁻⁸ C:3.5416 × 10⁻⁷ C = 35.416 × 10⁻⁸ C.Therefore, the closest option is 35.4 × 10⁻⁸ C.
Hint
First, find the dielectric constant K. Then, calculate the original charge on the capacitor plates using C₀ = ε₀A/d and Q = C₀V₀. Finally, use the formula for induced charge Qind = Q(1 - 1/K).